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495 | OPTICS | In 1845, Faraday studied the influence of electricity and magnetism on polarized light and discovered that heavy glass, originally non-optically active, exhibited optical activity under the influence of a strong magnetic field, causing the plane of polarization of polarized light to rotate. This was the first time humanity recognized the relationship between electromagnetic phenomena and light phenomena. The magneto-optical rotation effect later became known as the Faraday Effect.
In the magneto-optical rotation effect, the rotation angle of the vibration plane \(\beta\) is proportional to the distance \(d\) that light travels through the medium and is also proportional to the magnetic induction intensity \(B\) within the medium, expressed as:
\[
\beta = vBd
\]
where \(v\) is the proportionality constant, known as the Verdet constant, which depends on the properties of the medium and is also related to the wavelength of the incident light. Assume that the medium can be modeled using the Lorentz model, where the medium is composed of atoms, with \(N\) atoms per unit volume, and each atom contains \(Z\) electrons in the outer shell. The outer electrons deviate from their equilibrium positions and are subject to linear restoring forces: \(\vec{f} = -k\vec{r}\), with the harmonic resonance frequency of the outer shell electrons being \(\omega_0\). The interaction between the incident light and the medium exclusively involves the outer shell electrons.
Circularly polarized light can be regarded as the result of the synthesis of two linearly polarized light waves with the same frequency and direction, mutually perpendicular vibration axes, and a stable phase relationship. The two linearly polarized light waves have the same amplitude and a phase difference of \(\pm\pi/2\), representing left-handed and right-handed circularly polarized light, respectively. Similarly, a linearly polarized light wave can be decomposed into left-handed and right-handed circularly polarized components.
Based on these models, calculate the Verdet constant \(\beta\) for incident light with an angular frequency of \(\omega\). Assume that the medium does not absorb the incident light and its refractive index near the angular frequency \(\omega\) is \(n\). | Solution: The propagation of light waves in a medium under an applied magnetic field involves the motion of electrons:
\[m\ddot{\vec{r}} = -k\vec{r} - g\vec{v} - e(\vec{E} + \vec{v} \times \vec{B})\]
\(\vec{E}\) is the electric field intensity of the incident light wave; \(\vec{B}\) is the magnetic induction intensity applied to the medium; \(g\) is the damping term, and since the medium does not absorb the incident light, it exhibits weak damping and low loss, thus neglecting the damping term:
\[m\ddot{\vec{r}} = -k\vec{r} - e(\vec{E} + \vec{v} \times \vec{B})\]
Let \(\vec{B}\) be along the positive \(z\)-axis, which is also the propagation direction of the light wave, thus:
\[
\begin{cases}
m\ddot{x} + e\dot{y}B + kx = -eE_x \\
m\ddot{y} - e\dot{x}B + ky = -eE_y
\end{cases}
\]
Assuming the incident wave is a steady-state wave, the relationship between the electric field and time: \(e^{-i\omega t}\)
The relationship between the perturbation term of the forced vibration of the outer electrons and time: \(e^{-i\omega t}\)
Thus: \(\frac{\partial}{\partial t} \equiv -i\omega\), \(\frac{\partial^2}{\partial t^2} \equiv -\omega^2\), therefore:
\[
\begin{cases}
(\omega_0^2 - \omega^2)r_x - i\omega\Omega r_y = -\frac{e}{m}E_x \\
(\omega_0^2 - \omega^2)r_y + i\omega\Omega r_x = -\frac{e}{m}E_y
\end{cases}
\]
Where:
\(\Omega = \frac{eB}{m}\), \(\omega_0 = \sqrt{\frac{k}{m}}\)
For left-hand circularly polarized light, \(E_+ = E_x + iE_y\), then: \(r_+ = r_x + ir_y\)
\[(\omega_0^2 - \omega^2 - \omega\Omega)r_+ = -\frac{e}{m}E_+\]
Solving this, we get:
\[r_+ = -\frac{e}{m(\omega_0^2 - \omega^2 - \omega\Omega)}E_+\]
The polarization vector: The atomic number density of the medium is \(N(1/m^3)\), with each atom providing \(Z\) electrons weakly bound in the outer layer;
\[\vec{P}_+ = NZ(-e)\vec{r}_+ = \frac{NZ\frac{e^2}{m}}{\omega_0^2 - \omega^2 - \omega\Omega}E_+\]
Since \(P = \varepsilon_0\chi E\), thus:
\[\chi_+ = \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 - \omega\Omega}\]
The refractive index for left-hand circularly polarized light is:
\[n_+^2 = \varepsilon_+ = (1 + \chi_+) = 1 + \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 - \omega\Omega}\]
For right-hand circularly polarized light \(E_- = E_x - iE_y\), then: \(r_- = r_x - ir_y\)
\[(\omega_0^2 - \omega^2 + \omega\Omega)r_- = -\frac{e}{m}E_-\]
Solving this, we get:
\[r_- = -\frac{e}{m(\omega_0^2 - \omega^2 + \omega\Omega)}E_-\]
\[\vec{P}_- = NZ(-e)\vec{r}_- = \frac{NZ\frac{e^2}{m}}{\omega_0^2 - \omega^2 + \omega\Omega}E_-\]
Thus:
\[\chi_- = \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 + \omega\Omega}\]
The refractive index for right-hand circularly polarized light is:
\[n_-^2 = \varepsilon_- = (1 + \chi_-) = 1 + \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 + \omega\Omega}\]
That is, the refractive indices for left-hand and right-hand circularly polarized light are different:
\[n_{\pm}^2 = \varepsilon_{\pm} = (1 + \chi_{\pm}) = 1 + \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 \mp \omega\Omega}\]
Any linearly polarized light can be decomposed into two circularly polarized lights with the same frequency, equal amplitude, and a stable phase relationship. When linearly polarized light passes through a medium with an external magnetic field aligned with the direction of propagation, the rotation angle of the polarization plane is:
\[\beta = \frac{1}{2}(\alpha_+ - \alpha_-) = \frac{\pi}{\lambda}(n_+ - n_-)d\]
where \(d\) is the propagation length of light in the medium.
Find \((n_+ - n_-)\)
\[
\begin{align*}
n_+^2 - n_-^2 &= (n_+ - n_-)(n_+ + n_-)\\
&= NZ\frac{e^2}{\varepsilon_0 m} \cdot \frac{2\omega\Omega}{(\omega_0^2 - \omega^2)^2 - (\omega\Omega)^2}
\end{align*}
\]
Generally, the cyclotron frequency \(\Omega\) is much smaller than the optical frequency \(\omega\), and the optical frequency \(\omega\) is not near the resonance frequency \(\omega_0\), hence:
\[n_+ + n_- \approx 2n\]
\[n^2 \approx 1 + NZ\frac{e^2}{\varepsilon_0 m} \cdot \frac{1}{\omega_0^2 - \omega^2}\]
Therefore:
\[n_+ - n_- = NZ\frac{e^2}{\varepsilon_0 mn} \cdot \frac{\omega\Omega}{(\omega_0^2 - \omega^2)^2}\]
Thus:
\[
\begin{align*}
\beta &= \frac{\pi}{\lambda}(n_+ - n_-)d = \frac{\omega}{2c}NZ\frac{e^2}{\varepsilon_0 mn} \cdot \frac{\omega\Omega}{(\omega_0^2 - \omega^2)^2}Bd = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2} \cdot Bd\\
&= vBd
\end{align*}
\]
The Verdet constant is obtained as:
\[v = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2}\] | \[v = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2}\] |
600 | OPTICS | The Yang's double slit interference experiment consists of three parts: light source, double slit, and receiving screen. In the experiment, we generated a line light source by placing a point light source behind the slit $S_0$, and the light was then projected onto the receiving screen through the double slits $S_1$ and $S_2$, forming interference fringes. Now we will perform the following actions on this device:
Place polarizer $P_0$ next to slit $S_0$, and polarizers $P_1$ and $P_2$ next to slits $S_1$ and $S_2$, respectively, so that the transmission direction of $P_0$ is parallel to $P_1$ and forms a $\theta=60\degree$ angle with $P_2$.
Try to determine the contrast of the stripes on the receiving screen at this time. | The magnitude of electric field intensity is set to A after passing through polarizer P0.
The optical amplitude through P1 is A, and the optical amplitude through P2 is $\ frac 12 $A.
Let the direction parallel to P0 be the x direction and the direction perpendicular to it be the y direction.
standard
$$
E_{1x}=A,E_{2x}=\frac{1}{4} A,E_{2y}=\frac {\sqrt{3}}{4} A
$$
$E2 {1x} $and $E2 {2x} $are coherently superimposed, with a total light intensity of
$$
I=I_{0}\left(\frac 54 +\frac 12 \cos\delta\right)
$$
The lining ratio is
$$
\gamma=\frac 25
$$ | $$\gamma = \frac{2}{5}$$ |
285 | MECHANICS | A small bug with a mass of $m$ crawls on a disk with a radius of $2R$. Relative to the disk, its crawling trajectory is a circle of radius $R$ that passes through the center of the disk. The disk rotates with a constant angular velocity $\omega$ about an axis passing through its center and perpendicular to the plane of the disk. The bug's angular crawling velocity relative to the disk is in the same direction as the disk's angular velocity and has the same magnitude. Solve for the maximum force $F_{max}$ between the bug and the disk required to maintain this motion (neglecting gravity). | The problem can use the acceleration transformation formula for rotating reference frames, where the actual acceleration of the small insect
$$
\overrightarrow{a}=\overrightarrow{a}+2\overrightarrow{\omega}\times\overrightarrow{v}+\dot{\overrightarrow{\omega}}\times\overrightarrow{r}-\omega^{2}\overrightarrow{r}
$$
Here, $\vec{a}^{\prime}$ is the acceleration relative to the rotating system, and since the rotating system is a uniform circular motion,
$$
\dot{a_{\mathrm{\scriptsize~=~}}}\omega^{2}R
$$
is the speed relative to the rotating system, with a magnitude of $\omega R$. Thus, the magnitude of the Coriolis acceleration is:
$$
2\omega v^{'}=2\omega^{2}R
$$
The last term is the centripetal acceleration, which is directly proportional to the distance from the insect to the center of the disk,
$$
\omega^{2}r=2\omega^{2}R\cos\left(\frac{\theta}{2}\right)
$$
(Analysis of each acceleration, 7 points)
These three acceleration components are combined to derive the total acceleration magnitude as:
$$
a=\omega^{2}R{\sqrt{\left(3+2\cos\left({\frac{\theta}{2}}\right)^{2}\right)^{2}+4\cos\left({\frac{\theta}{2}}\right)^{2}\sin\left({\frac{\theta}{2}}\right)^{2}}}=\omega^{2}R{\sqrt{9+16\cos\left({\frac{\theta}{2}}\right)^{2}}}
$$
which is a constant, so the maximum force is
$$
F_{m a x}=5m\omega^{2}R
$$ | $$F_{\max} = 5m\omega^2R$$ |
321 | MECHANICS | The civilization of *Three-Body* changes the values of fundamental physical constants, such as the Planck constant, inside the "water drop" to alter the range of strong forces.
What kind of effects would this produce? This problem provides a suitable discussion about this question.
Hideki Yukawa pointed out that the propagator of nuclear forces is the $\pi$ meson. One can equivalently simplify the effect by considering that protons and protons, protons and neutrons, and neutrons and neutrons have an identical strong-force potential energy. For a force field with a propagator of mass, Yukawa potential and its range can be expressed as:
$$
U = -\frac{A\mathrm{e}^{-\frac{2\pi m c r}{h}}}{r}\quad,\quad\lambda = \frac{h}{2\pi m c}
$$
For the deuteron (the strong-force binding between a proton and a neutron), consider two particles rapidly rotating around the center of mass. If the distance between the proton and the neutron is precisely $r$ and they do not collide, find the angle rotated around the center of mass during one complete cycle of small radial oscillation. Express the answer in terms of $\lambda$ and $r$. | $$
F=-\frac{A(1+\frac{2\pi m c r}{h})e^{-\frac{2\pi m c r}{h}}}{r^{2}}
$$
Newton's Law
$$
m\omega^{2}\frac{r}{2}=\frac{A(1+\frac{2\pi m c r}{h})e^{-\frac{2\pi m c r}{h}}}{r^{2}}
$$
Effective Potential Energy
$$
V_{eff}=-\frac{A e^{-\frac{2\pi m c r}{h}}}{r}+\frac{L^{2}}{2\cdot m/2\cdot r^{2}}
$$
Solving the Second Derivative
$$
\frac{d^{2}V_{eff}}{d r^{2}}=\frac{A e^{-\frac{r}{\lambda}}}{r^{3}}\left(1+\frac{r}{\lambda}-\frac{r^{2}}{\lambda^{2}}\right)
$$
Angular Frequency
$$
\omega_{0}=\sqrt{\frac{2\frac{d^{2}V_{eff}}{d r^{2}}}{m}}
$$
Geometric Relation
$$
\varphi=2\pi\cdot\frac{\omega}{\omega_{0}}
$$
Resulting
$$
\varphi=2\pi\sqrt{\frac{1+r/\lambda}{1+r/\lambda-r^2/\lambda^2}}
$$ | $$\varphi = 2\pi \sqrt{\frac{1 + \frac{r}{\lambda}}{1 + \frac{r}{\lambda} - \frac{r^2}{\lambda^2}}}$$ |
674 | MECHANICS | A coin is placed at rest on the edge of a smooth tabletop, with only a small portion of its right side extending beyond the edge. The coin can be considered as a uniform disk with mass $m$, radius $r$, and gravitational acceleration $g$. A vertical impulse $I$ is applied to the right side of the coin. During its subsequent motion after the impact, the coin might fly off the table at some point. Determine the minimum initial velocity of the center of mass, $v_{0, \min}$, required for the coin to leave the table, expressed in terms of $g$ and $r$.",
| The moment of inertia of the coin about the axis is \[J=\frac14mr^2\] According to the impulse and impulse-momentum theorem, about the center of mass we have \[I+I_N=m v_0, Ir-I_Nr=J\omega_0\] There is also the velocity relation \[v_0=\omega_0 r\] Therefore, we have \[v_0=\frac{8I}{5m}, \omega_0=\frac{8I}{5mr}\] After being struck, assume the angle between the coin and the horizontal direction is \(\phi\). Because the tabletop is smooth, the horizontal coordinate of the center of mass remains unchanged, so \[\vec r_c=(0,r\sin\phi)\] Thus, we have \[E_k=\frac 12m\dot y_c^2+\frac12 J\dot\phi ^2=\frac12mr^2\left(\cos^2\phi+\frac 14\right)\dot\phi^2\] \[E_p=mgy_c=mgr\sin\phi\] Thus, mechanical energy is conserved \[\frac 18mr^2(4\cos^2\phi+1)\dot\phi^2+mgr\sin\phi=\frac{8I^2}{5m}\] From this, we can obtain \[\dot\phi^2=\frac{64I^2-40m^2gr\sin\phi}{5m^2r^2(4\cos^2\phi+1)}\] Since \(\frac{\mathrm d\dot\phi^2}{\mathrm d\phi}=2\ddot\phi\), we have \[\ddot\phi = \frac{4\cos\phi}{5r^2(4\cos^2\phi+1)^2}\left(64\frac{I^2}{m^2}\sin\phi-5gr(5+4\sin^2\phi)\right)\] Hence, we can obtain \[\begin{align*}\ddot y_c&=r(\ddot\phi\cos\phi-\dot\phi^2\sin\phi)\\ &=\frac{4\cos^2\phi}{5r(4\cos^2\phi+1)^2}\left(64\frac{I^2}{m^2}\sin\phi-5gr(5+4\sin^2\phi)\right)-\frac{(64 I^2-40 m^2gr\sin\phi)\sin\phi}{5m^2r(4\cos^2\phi+1)}\end{align*}\] When there is no support force from the tabletop, the coin leaves the table, at that moment \[m\ddot y_c=-mg\] At this point \(\phi=\theta\), \[20\sin^2\theta-64\frac{I^2}{m^2gr}\sin\theta+25=0\] It can be seen that the solution for \(\sin\theta\) can take all values in \((0,1]\), which means for it to jump, we have \[I=m\sqrt{gr}\left(\sqrt{\frac{20\sin^2\theta+25}{64\sin\theta}}\right)_{\min}=\frac{3\sqrt5}{8}m\sqrt{gr}\] Thus we have \[v_{0,\min}=\frac{3\sqrt 5}{5}\sqrt{gr}\]
| $$\frac{3\sqrt{5}}{5}\sqrt{gr}$$ |
83 | ELECTRICITY |
---
A concentric spherical shell with inner and outer radii of $R_{1}=R$ and $R_{2}=2^{1/3}R$, and magnetic permeability $\mu=3\mu_{0}$, is placed in an external uniform magnetic field with magnetic flux density $B_{0}$ and the magnetic field generated by a fixed magnetic dipole located at the center of the sphere. The magnetic dipole has a magnetic moment $m$ that is aligned with the external magnetic field. The system reaches magnetostatic equilibrium. To solve this problem, we can use the method of magnetic scalar potential. The concept is to express the magnetic field intensity as ${\pmb{H}}=-\nabla\varphi$, where $\varphi$ is referred to as the magnetic scalar potential. This approach applies to all three regions: the space inside the shell, the space outside the shell, and the region within the shell itself. The relationship between magnetic flux density and magnetic field intensity is $B=\mu H$.
If the magnetic moment is rotated by an angle $\alpha$, calculate the torque it experiences. The answer should be expressed in terms of $m$, $B_0$, and $\alpha$. Please verify the result before outputting.
--- | (1)
Let:
$$
\varphi={\left\{\begin{array}{l l}{A_{1}{\frac{r}{R}}\cos\theta+B_{1}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(r<R)}
{A_{2}{\frac{r}{R}}\cos\theta+B_{2}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(R<r<k R)}\ {A_{3}{\frac{r}{R}}\cos\theta+B_{3}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(r>k R)}\end{array}\right.}
$$
(Tangential) continuity $(k=2^{1/3})$
$$
\begin{array}{c}{{A_{1}+B_{1}=A_{2}+B_{2}}}\ {{\qquad}}\ {{k A_{2}+k^{-2}B_{2}=k A_{3}+k^{-2}B_{3}}}\end{array}
$$
Magnetic induction normal continuity:
$$
\mu_{0}A_{1}-2\mu_{0}B_{1}=\mu A_{2}-2\mu B_{2}
$$
$$
k\mu A_{2}-2k^{-2}\mu B_{2}=k\mu_{0}A_{3}-2k^{-2}\mu_{0}B_{3}
$$
According to the boundary condition at infinity:
$$
A_{3}=-\frac{B_{0}R}{\mu_{0}}
$$
According to the boundary condition at the magnetic dipole:
$$
B_{1}=\frac{m}{4\pi R^{2}}
$$
The solution is:
$$
\begin{array}{c}{{A_{2}=\displaystyle\frac{21}{31}A_{3}+\displaystyle\frac{6}{31}B_{1}}}\ {{{}}}\ {{B_{2}=\displaystyle\frac{6}{31}A_{3}+\displaystyle\frac{15}{31}B_{1}}}\ {{{}}}\ {{A_{1}=\displaystyle\frac{27}{31}A_{3}-\displaystyle\frac{10}{31}B_{1}}}\ {{{}}}\ {{B_{3}=\displaystyle-\frac{14}{31}A_{3}+\displaystyle\frac{27}{31}B_{1}}}\end{array}
$$
Substituting:
$$
\varphi=\left\{\begin{array}{l l}{\left(-\frac{27}{31}\frac{B_{0}R}{\mu_{0}}-\frac{10}{31}\frac{m}{4\pi R^{2}}\right)\frac{r}{R}\cos\theta+\frac{m}{4\pi R^{2}}\frac{R^{2}}{r^{2}}\cos\theta}&{(r<R)}\ {\left(-\frac{21}{31}\frac{B_{0}R}{\mu_{0}}+\frac{6}{31}\frac{m}{4\pi R^{2}}\right)\frac{r}{R}\cos\theta+\left(-\frac{6}{31}\frac{B_{0}R}{\mu_{0}}+\frac{15}{31}\frac{m}{4\pi R^{2}}\right)\frac{R^{2}}{r^{2}}\cos\theta}&{(R<r<k R)}\ {-\frac{B_{0}R}{\mu_{0}}\frac{r}{R}\cos\theta+\left(\frac{14}{31}\frac{B_{0}R}{\mu_{0}}+\frac{27}{31}\frac{m}{4\pi R^{2}}\right)\frac{R^{2}}{r^{2}}\cos\theta}&{(r>k R)}\end{array}\right.
$$
(2)
Magnetic field induced by external magnetic field:
$$
B^{\prime}=\frac{27}{31}B_{0}
$$
Thus, torque:
$$
{\cal M}=\underline{{{m\times B}}}
$$
Result:
$$
M=-{\frac{27}{31}}m B_{0}\sin\alpha
$$ | $$ M = -\frac{27}{31} m B_0 \sin \alpha $$ |
54 | MECHANICS | Three small balls are connected in series with three light strings to form a line, and the end of one of the strings is hung from the ceiling. The strings are non-extensible, with a length of $l$, and the mass of each small ball is $m$.
Initially, the system is stationary and vertical. A hammer strikes one of the small balls in a horizontal direction, causing the ball to acquire an instantaneous velocity of $v_0$. Determine the instantaneous tension in the middle string when the topmost ball is struck. (The gravitational acceleration is $g$.) | Strike the top ball, the acceleration of the top ball is:
$$a=\frac{v_0^2}{l}$$
Switching to the reference frame of the top ball, let the tensions from top to bottom be $T_1, T_2, T_3$. Applying Newton's second law to the middle ball:
$$T_2-T_3-mg-m\frac{v_0^2}{l}=m\frac{v_0^2}{l}$$
The acceleration of the bottom ball is: $$a=2\frac{v_0^2}{l}$$
Applying Newton's second law to the bottom ball: $$T_3-mg=2m\frac{v_0^2}{l}$$
Thus, we have: $$T_2=2mg+4m\frac{v_0^2}{l}$$ | $$T_2 = 2mg + 4m \frac{v_0^2}{l}$$ |
10 | ELECTRICITY | The region in space where $x > 0$ and $y > 0$ is a vacuum, while the remaining region is a conductor. The surfaces of the conductor are the $xOz$ plane and the $yOz$ plane. A point charge $q$ is fixed at the point $(a, b, c)$ in the vacuum, and the system has reached electrostatic equilibrium. Find the magnitude of electric field intensity on the surface of the conductor at the $xOz$ plane, ${E}(x, +0, z)$. | The infinite conductor and the point at infinity are equipotential, $U{=}0$, with the boundary condition that the tangential electric field $E_{\tau}{=}0$. By the uniqueness theorem of electrostatics, the induced charges on the conductor can be equivalently replaced by image charges to determine the contribution to the electric potential in the external space of the conductor. The top view of the image charges is shown below.
Therefore, the answer is
\(\vec{E}\left(x,+0,z\right)=\left(\frac{-2q}{4\pi\varepsilon_{0}}\frac{b}{r_{1}^{3}}{+}\frac{2q}{4\pi\varepsilon_{0}}\frac{b}{r_{2}^{3}}\right)\hat{y}=\frac{q b}{2\pi\varepsilon_{0}}\biggl[\bigl[(a+x)^{2}+(z-c)^{2}+b^{2}\bigr]^{\frac{-3}{2}}-\bigl[(x-a)^{2}+(z-c)^{2}+b^{2}\bigr]^{\frac{-3}{2}}\biggr]\hat{y},\) \(\vec{E}\left(+0,y,z\right)=(-\frac{2q}{4\pi\varepsilon_{0}}\frac{a}{r_{1}^{3}}+\frac{2q}{4\pi\varepsilon_{0}}\frac{a}{r_{2}^{3}})\hat{x}=\frac{q a}{2\pi\varepsilon_{0}}\Bigg[\Big[(y+b)^{2}+\big(z-c\big)^{2}+a^{2}\Big]^{\frac{-3}{2}}-\Big[\big(y-b\big)^{2}+\big(z-c\big)^{2}+a^{2}\Big]^{\frac{-3}{2}}\Bigg]\hat{x}\) | $$\frac{q b}{2\pi \varepsilon_0} \left[ ((a+x)^2 + (z-c)^2 + b^2)^{-3/2} - ((x-a)^2 + (z-c)^2 + b^2)^{-3/2} \right]$$ |
33 | THERMODYNAMICS | Consider an infinite-length black body with inner and outer cylinders, which are in contact with heat sources at temperatures $T_{1}$ and $T_{2}$, respectively; assume that the temperature of the heat sources remains constant. Let the inner cylinder have a radius $r$, the outer cylinder have a radius $R$, and the distance between the axes of the inner and outer cylinders be $b$, with $r < b < R$ and $r + b < R$. Find the power $p(\theta)$ absorbed per unit area from the heat source at angle $\theta$ on the surface of the outer cylinder (i.e., the power density at $\theta$), where $\theta$ is the angle between the line connecting a point on the surface of the outer cylinder and the center of the outer cylinder, and the line connecting the centers of the inner and outer cylinders. The Stefan-Boltzmann constant is denoted as $\sigma$. | In the case of non-coaxial, we examine the long strip-shaped area element at the outer wall reaching $\theta + \mathrm{d} \theta$:
The heat flux projected onto the inner axis is:
$$
\frac{d S}{2}\sigma T_{2}^{4}\int_{\mathrm{arcsin}\left({\frac{b \sin\theta}{D}}\right)-\mathrm{arcsin}({\frac{r}{D}})}^{\mathrm{arcsin}\left({\frac{b \sin\theta}{D}}\right)+\mathrm{arcsin}({\frac{r}{D}})}\cos\theta \, d\theta
$$
The above equation equals $\sigma T_{2}^{4}\frac{r(R-b \cos \theta)}{R^{2}+b^{2}-2R b \cos \theta}\mathrm{d}S$, where dS is the area of the strip-shaped area element. According to the zeroth law of thermodynamics, the heat flux sent from the inner cylinder to the strip-shaped area element at $\theta + \mathrm{d} \theta$ on the outer cylinder is $\begin{array}{r}{\sigma T_{1}^{4}\frac{r(R-b \cos\theta)}{R^{2}+b^{2}-2R b \cos\theta}\mathrm{d} S}\end{array}$: Similarly, according to the zeroth law of thermodynamics,
It is known that the heat flux sent from other places on the outer cylinder to this area element is
$$
\begin{array}{r}{(\sigma T_{2}^{4}-\sigma T_{2}^{4}\frac{r(R-b}{R^{2}+b^{2}-2R b \cos \theta})\mathrm{d} S}\end{array}
$$
It can be seen that the heat absorbed by the unit area element at $\theta$ from the heat source with temperature $T_{2}$ is equal to
$$
\begin{array}{r}{(\sigma T_{2}^{4}-\sigma T_{1}^{4})\frac{r(R-b)}{R^{2}+b^{2}-2R b \cos \theta}}\end{array}
$$ | $$
p(\theta) = (\sigma T_2^4 - \sigma T_1^4) \frac{r(R - b \cos \theta)}{R^2 + b^2 - 2Rb \cos \theta}
$$ |
41 | MECHANICS | The rocket ascends vertically from the Earth's surface with a constant acceleration $a = k_1g_0$, where $g_0$ is the gravitational acceleration at the Earth's surface. Inside the rocket, there is a smooth groove containing a testing apparatus. When the rocket reaches a height $h$ above the ground, the pressure exerted by the instrument on the bottom of the groove is $k_2$ times what it was at the time of liftoff. Given that the radius of the Earth is $R$, find $h$. | At the moment of takeoff, $N_1=(1+k_1)mg_0$, where $m$ is the mass of the tester.
When ascending to a height $h$, the gravitational acceleration changes to $g=g_0(\dfrac{R}{R+h})^2$
The pressure at this time is $N_2=mg_0(\dfrac{R}{R+h})^2+k_1mg_0$
Since $N_2=k_2N_1$, solving the equations together gives
$h=R(\dfrac{1}{\sqrt{k_2(1+k_1)-k_1}}-1)$ | $$R\left(\frac{1}{\sqrt{k_2(1+k_1)-k_1}}-1\right)$$ |
453 | ELECTRICITY | Most media exhibit absorption of light, where the intensity of light decreases as it penetrates deeper into the medium.
Let monochromatic parallel light (with angular frequency $\omega$) pass through a uniform medium with a refractive index of $n$. Experimental results show that, over a reasonably wide range of light intensities, after passing through a small thickness, the decrease in light intensity is proportional to both the light intensity itself and this small thickness. The proportionality coefficient is denoted as $\alpha$.
Next, we will estimate the coefficient $\alpha$ based on the classical oscillator model of atoms. In this model, the atomic nucleus can be considered stationary, while the electron, when in motion, is bound by the nucleus. This binding force can be approximated as a linear restoring force: $-m\omega_{0}^{2}x$. Here, $\mathsf{m}$ is the mass of the electron, $\omega_{0}$ is the natural angular frequency (describing the strength of the restoring force), and $x$ is the distance between the electron and the nucleus. Simultaneously, when the electron is in motion, there often exists a damping force: $-m\gamma v$. However, since the damping is usually quite weak, it can be approximately assumed that the electron undergoes simple harmonic motion within any given period.
Additionally, it is important to know that the power of radiation emitted by a charged particle in accelerated motion is given by:
$$
P = {\frac{e^{2}{\overline{{{\dot{v}}^{2}}}}}{6\pi\varepsilon_{0}c^{3}}}
$$
Based on this model, derive the expression for $\alpha$.
**Hint**: The primary interaction between electromagnetic waves and the medium originates from the contribution of the electric field component. $e$ is the magnitude of the electron charge, $\overline{{{\dot{v}}^{2}}}$ represents the average value of the square of the acceleration, $\gamma$ is a positive constant, $\varepsilon_{0}$ is the vacuum permittivity, $v$ is the velocity of the electron, $c$ is the speed of light in vacuum, and $N$ is the number density of electrons. | The dynamics equation for electrons:
$$
m\ddot{x}=-m\gamma\dot{x}-m\omega_{0}^2x-eE_0e^{i\omega t}
$$
Substituting $x=\tilde{x}e^{i\omega t}$ yields $\tilde{x}=\frac{-eE_0/m}{-\omega^2+\omega_0^2+i\gamma \omega}$
$$
\tilde{v}=i\omega \tilde{x}=\frac{-i\omega eE_0/m}{-\omega^2+\omega_0^2+i\gamma \omega}
$$
$$
\dot{\tilde{v}}=\frac{-\omega^2 eE_0/m}{-\omega^2+\omega_0^2+i\gamma \omega}
$$
$$
\overline{\dot{v^2}}=\frac{\frac{1}{2}(\omega^2 eE_0/m)^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
$$
P=-\frac{e^2\overline{\dot{v^2}}}{6\pi\epsilon_0 c^3}=-\frac{1}{12\pi\epsilon_0 c^3}\frac{(\omega^2 e^2 E_0/m)^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
The energy flux density of the electromagnetic wave:
$$
I=\frac{1}{2}nc\epsilon_0E_0^2
$$
Assuming the number density of electrons is \(N\), then the energy loss per unit area in a length \(dx\) is \(NPdx=dI\).
Also, from $-dI=\alpha Idx$, we have
$$
\alpha=-\frac{dI}{Idx}=-\frac{NP}{I}
$$
Thus,
$$
\alpha=\frac{N}{6\pi n\epsilon_0^2c^4}\frac{(\frac{\omega^2 e^2}{m})^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$ | $$
\alpha=\frac{N}{6\pi n\epsilon_0^2c^4}\frac{(\frac{\omega^2 e^2}{m})^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
|
281 | THERMODYNAMICS | Due to the imbalance of molecular forces on the surface layer of a liquid, tension will arise along any boundary of the surface layer. In this context, the so-called "capillary phenomenon" occurs, affecting the geometric shape of a free liquid surface. Assume a static large container exists in a uniform atmosphere. The container holds a liquid in static equilibrium, and its side walls are vertical planes. The contact angle between the liquid and the side wall is $\theta$ $(0 < \theta < \frac{\pi}{2})$, the density of the liquid is $\rho$, its surface tension coefficient is $\sigma$, atmospheric pressure is $p_{0}$, and the gravitational acceleration is $g$.
Find the additional height $h$ of the liquid surface rise at the junction between the container wall and the liquid surface due to the capillary phenomenon, expressing the result using $\theta, \rho, \sigma, g$.
Hint: The formula for the radius of curvature in a two-dimensional Cartesian coordinate system is: $R = {\frac{(1+z^{\prime}(x)^{2})^{\frac{3}{2}}}{\mid z^{\prime\prime}(x)\mid}}$. | Choose $x\sim x+\mathrm{d}x.$ , analyze the surface microelement of the liquid located at a height between $z\sim z+\mathrm{d}z$, considering the liquid pressure $p_{i}$ at that point, it is evident that:
$$
p_{0}=p_{i}+\rho g z
$$
Next, consider the forces perpendicular to the liquid surface. From the Laplace's additional pressure formula:
$$
p_{i}+\frac{\sigma}{R}=p_{0}
$$
$R$ is the radius of curvature of the liquid surface equation at that point. By combining equations (5) and (6), we obtain:
$$
z=z(x)=\cfrac{\sigma}{\rho g R}=\cfrac{\sigma z^{\"v}(x)}{\rho g(1+z^{\prime}(x)^{2})^{\frac{3}{2}}}
$$
Note that here, due to the concave shape of the liquid surface, $z"(x)$ is positive.
Integrate equation (7):
$$
\int_{h}^{0}z\mathrm{d}z=\frac{\sigma}{\rho g}\int_{h}^{0}\frac{z^{\prime}(x)}{\left(1+z^{\prime}(x)^{2}\right)^{\frac{3}{2}}}\mathrm{d}z=\frac{\sigma}{\rho g}\int_{z^{\prime}(0)}^{z^{\prime}(\infty)}\frac{\mathrm{d}z^{\prime}(x)}{\left(1+z^{\prime}(x)^{2}\right)^{\frac{3}{2}}}=\frac{\sigma}{\rho g}\int_{\frac{1}{\sqrt{1+z\left(\left(x\right)^{2}\right)}}}^{\frac{1}{\sqrt{1+z\left(0\right)^{2}}}}\mathrm{d}\left(\frac{1}{\sqrt{1+z^{\prime}(x)^{2}}}\right)
$$
Here, the boundary conditions are specifically pointed out: $z(0)=h,z(\infty)=0,z^{\prime}(\infty)=0$ (the liquid surface approaches horizontal infinitely far away). Thus, $h$ can be expressed as:
$$
\frac{1}{2}h^{2}=\frac{\sigma}{\rho g}[1-\frac{1}{\sqrt{1+z^{\prime}(0)^{2}}}]
$$
The contact angle $\theta$ satisfies:
$$
z^{\prime}(0)=\tan(\frac{\pi}{2}-\theta)=-\cot\theta
$$
Substitute to calculate, and the final result is:
$$
h={\sqrt{\frac{2\sigma}{\rho g}(1-\sin\theta)}}
$$ | $$\sqrt{\frac{2\sigma}{\rho g}(1-\sin \theta)}$$ |
95 | MECHANICS | At the North Pole, the gravitational acceleration is considered a constant vector of magnitude $g$, and the Earth's rotation angular velocity $\Omega$ is along the $z$-axis pointing upward (the $z$-axis direction is vertically upward). A mass $M$ carriage has a mass $m=0.1M$ particle suspended from its ceiling by a light string of length $l$. It is known that $\frac{g}{l} = \omega_{0}^{2} = 10\Omega^{2}$. Initially, the particle is at rest, hanging vertically. The carriage slides frictionlessly along a track in the $x$-direction. Suddenly, the carriage is acted upon by a constant forward force $F$ in the $x$-direction. Determine the expression for the $x$-coordinate of the particle relative to the carriage's reference frame as a function of time $t$. Use the small-angle approximation. At $t=0$, $x=0$. Express the answer in terms of $M$ and $\Omega$. Consider the effects of the Coriolis force caused by the Earth's rotation in this scenario. | The two components of the force exerted on the small ball can be approximated as:
$$
(F_{x},F_{y})=\left(-\frac{m g}{l}x,-\frac{m g}{l}y\right)
$$
Assume the coordinate of the carriage is $x$. Then the Coriolis force acting on the small ball is:
$$
(C_{x},C_{y})=\Big(2m\Omega\dot{y},-2m\Omega(\dot{X}+\dot{x})\Big)
$$
Newton's laws:
$$
F_{\pi}+C_{x}=m\left(\ddot{X}+\ddot{x}\right)
$$
$$
F_{y}+C_{y}=m{\ddot{y}}
$$
$$
F-F_{x}=M\ddot{X}
$$
Rearranging:
$$
\ddot{X}-\frac{m g}{M l}x=\frac{F}{M}
$$
$$
\ddot{X}+\ddot{x}+\frac{g}{l}x-2\Omega\dot{y}=0
$$
$$
\ddot{y}+\frac{g}{l}y+2\Omega(\dot{X}+\dot{x})=0
$$
We find the following particular solutions $\scriptstyle X={\frac{1}{2}}a t^{2}$ ${\pmb y}=b t$ ${\pmb x}={\pmb c}$
$$
\begin{array}{c}{{a-\displaystyle\frac{m}{M}\omega_{0}^{2}c=\displaystyle\frac{F}{M}}}\ {{}}\ {{a+\omega_{0}^{2}c-2\Omega b=0}}\ {{}}\ {{\omega_{0}^{2}b+2\Omega a=0}}\end{array}
$$
From this, we obtain:
$$
X_{0}=\frac{25F}{57M}t^{2}\quad,\quad y_{0}=-\frac{10F}{57M\Omega}t\quad,\quad x_{0}=-\frac{7F}{57M\Omega^{2}}
$$
The final solution should be $X=X_{0}+X_{1}$, ${\mathfrak{y}}={\mathfrak{y}}_{0}+{\mathfrak{y}}_{1}$ ${\boldsymbol{x}}={\boldsymbol{x}}_{0}+{\boldsymbol{x}}_{1}$, where $X_{1}, y_{1}, x_{1}$ satisfy the corresponding homogeneous equations. Let:
$$
X_{1}, x_{1}, y_{1}=A, B, C\mathrm{e}^{i\omega t}
$$
Substituting these into the equations results in the matrix equation:
$$
\left[\begin{array}{c c c}{{-\omega^{2}}}&{{-m\omega_{0}^{2}/M}}&{{0}}\ {{-\omega^{2}}}&{{-\omega^{2}+\omega_{0}^{2}}}&{{-2\mathrm{i}\Omega\omega}}\ {{2\mathrm{i}\Omega\omega}}&{{2\mathrm{i}\Omega\omega}}&{{-\omega^{2}+\omega_{0}^{2}}}\end{array}\right]\left[\begin{array}{c}{{A}}\ {{B}}\ {{C}}\end{array}\right]=\left[\begin{array}{l}{{0}}\ {{0}}\ {{0}}\ {{0}}\end{array}\right]
$$
Taking the determinant of the coefficient matrix equal to zero allows it to be factored:
$$
\omega^{2}\left(\omega^{2}-6\Omega^{2}\right)\left(\omega^{2}-19\Omega^{2}\right)=0
$$
The case $\omega^{2}=0$ corresponds to the trivial solution:
$$
X_{1}, x_{1}, y_{1}=C_{1}, 0, 0
$$
The solution for $\omega=\sqrt{6}\Omega$ is:
$$
X_{1}, x_{1}, y_{1}=C_{2}, -6C_{2}, \frac{5\sqrt{6}}{2}\mathrm{i}C_{2}\mathrm{e}^{\mathrm{i}\omega t}
$$
The solution for $\omega=19\Omega$ is:
$$
X_{1}, x_{1}, y_{1}=C_{3}, -19C_{3}, -4\sqrt{19}\mathrm{i}C_{3}\mathrm{e}^{\mathrm{i}\omega t}
$$
By superimposing these three solutions and considering the initial conditions:
$$
C_{1}+C_{2}+C_{3}=0
$$
$$
-6C_{2}-19C_{3}+c=0
$$
$$
-\frac{5\sqrt{6}}{2}\cdot\sqrt{6}\Omega\cdot C_{2}+4\sqrt{19}\cdot\sqrt{19}\Omega\cdot C_{3}+b=0
$$
We obtain:
$$
C_{1}=\frac{59F}{3249M\Omega^{2}}\quad,\quad C_{2}=-\frac{2F}{117M\Omega^{2}}\quad,\quad C_{3}=-\frac{5F}{4693M\Omega^{2}}
$$
Finally, substituting these into the equations and taking the real part:
$$
\boxed{x=-\frac{7F}{57M\Omega^{2}}+\frac{4F}{39M\Omega^{2}}\cos{\sqrt{6}}\Omega t+\frac{5F}{247M\Omega^{2}}\cos{\sqrt{19}}\Omega t}
$$ | $$
\boxed{x=-\frac{7F}{57M\Omega^2}+\frac{4F}{39M\Omega^2}\cos(\sqrt{6}\Omega t)+\frac{5F}{247M\Omega^2}\cos(\sqrt{19}\Omega t)}
$$ |
101 | MECHANICS | The problem discusses a stick figure model. The stick figure's head is a uniform sphere with a radius of $r$ and a mass of $m$, while the rest of the body consists of uniform rods with negligible thickness and a mass per unit length of $\lambda$. All parts are connected by hinges. Specifically, the torso and both arms have a length of $l$, while the legs have a length of $1.2l$. $\theta$ represents the angle between the arms and the torso, and $\varphi$ represents the angle between the legs and the extended torso line (not an obtuse angle). The stick figure is composed of a head, a torso, two arms, and two legs. Initially, the stick figure is in a state where $\theta=\textstyle{\frac{\pi}{2}}$ and $\varphi={\frac{\pi}{4}}$. The head has a mass of $m=0.6\lambda l$, with the result expressed in terms of $\lambda$.
The stick figure is a simulation of a real biological body, so if necessary, the connections between the torso and the arms/legs, in addition to providing interaction forces, can also exert a couple (i.e., a torque), even if there is no interaction force between them. This net torque is independent of the choice of reference point. In the following discussion, this will be referred to simply as a "couple."
During a certain free-fall motion, the stick figure, overwhelmed with panic, forgets to bring its legs together and swings its arms and torso with an amplitude of $\theta_{0}$ (small angle) and an angular frequency of $\omega$. As a result, its legs passively oscillate near $\varphi={\frac{\pi}{4}}$. Find the amplitude of resonance for the legs (seek the steady-state solution starting from the initial position). Assume that the stick figure can actively generate a couple for the arms but not for the legs. |
(1) The forces are already marked on the diagram. List Newton's laws and rotational equations, assuming the displacement of the driven object is $\pmb{x}$, positive upwards.
Vertical direction of the leg
$$
T_{3}=1.2\lambda l\times\left(0.6l\times{\frac{d^{2}}{d t^{2}}}\left({\mathrm{cos}}\varphi\right)-{\ddot{x}}\right)
$$
Vertical direction of the arm
$$
T_{1}=\lambda l\left(\ddot{x}-0.5l\times\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\theta\right)\right)
$$
Vertical direction of the torso (head)
$$
2\left(T_{3}-T_{1}\right)=(m+\lambda l)\ddot{x}
$$
Horizontal direction of the leg
$$
-T_{4}=1.2\lambda l\times0.6l\times\frac{d^{2}}{d t^{2}}\left(\mathrm{sin}\varphi\right)
$$
Horizontal direction of the arm
$$
-T_{2}=\lambda l\times0.5l\times{\frac{d^{2}}{d t^{2}}}\left(\mathrm{sin}\theta\right)
$$
Angular momentum theorem for the leg
$$
T_{3}\times0.6l\mathrm{sin}\varphi+T_{4}\times0.6l\mathrm{cos}\varphi-M=\frac{1}{12}\times1.2\lambda l\times(1.2l)^{2}\ddot{\varphi}
$$
Angular momentum theorem for the arm
$$
T_{2}\times0.5l\mathrm{cos}\theta-T_{1}\times0.5l\mathrm{sin}\theta=\frac{1}{12}\times\lambda l\times l^{2}\ddot{\theta}
$$
Substitute in, we have
$$
\frac{1}{12}l^{2}{\ddot{\theta}}+0.25l^{2}{\mathrm{cos}}\theta\frac{d^{2}}{d t^{2}}{\mathrm{sin}}\theta+\frac{T_{1}}{\lambda}\times0.5{\mathrm{sin}}\theta=0
$$
By combining the previous equations, we get
$$
T_{1}=\left({\frac{6}{25}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\varphi\right)-{\frac{1}{3}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\theta\right)\right)\lambda l^{2}
$$
By substituting, we obtain the final expression
$$
\left(\frac{1}{3}-\frac{1}{12}\mathrm{sin}^{2}\theta\right)\ddot{\theta}-\frac{1}{12}\mathrm{sin}\theta\mathrm{cos}\theta\dot{\theta}^{2}=\frac{3}{25}\omega^{2}\mathrm{sin}\theta\mathrm{cos}\varphi
$$
Therefore,
$$
\begin{array}{c}{{A(\theta)=\displaystyle\frac{1}{3}-\frac{1}{12}\mathrm{sin}^{2}\theta}}\ {{{}}}\ {{B(\theta)=-\displaystyle\frac{1}{12}\mathrm{sin}\theta\mathrm{cos}\theta}}\ {{{}}}\ {{C(\theta)=\displaystyle\frac{3}{25}\omega^{2}\mathrm{sin}\theta}}\end{array}
$$
Substituting back into the previous angular momentum theorem, we have
$$
\begin{array}{r l r}{M=\frac{9}{25}\mathrm{sin}\varphi\lambda l^{3}\left(\frac{12}{25}\mathrm{cos}\varphi\omega^{2}+\frac{1}{3}\mathrm{cos}\theta\dot{\theta}^{2}+\frac{1}{3}\mathrm{sin}\theta\ddot{\theta}\right)}&{{}=\frac{12}{25}\frac{\mathrm{sin}\varphi}{\mathrm{sin}\theta}\ddot{\theta}}\end{array}
$$
There still remain the first five equations from the first question and the modified angular momentum theorem
$$
T_{3}\times0.6l\mathrm{sin}\varphi+T_{4}\times0.6l\mathrm{cos}\varphi=\frac{1}{12}\times1.2\lambda l\times(1.2l)^{2}\ddot{\varphi}
$$
Substitute and combine, we get
$$
T_{3}\times0.6l\mathrm{sin}\varphi-1.2\times0.6\lambda l^{2}{\frac{d^{2}}{d t^{2}}}(\mathrm{sin}\varphi)\times0.6l\mathrm{cos}\varphi={\frac{1}{12}}\times1.2\lambda l\times(1.2l)^{2}\ddot{\varphi}
$$
$$
T_{3}=\lambda l\left(\ddot{x}-0.5l\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\theta\right)\right)+\frac{1}{2}(m+\lambda l)\ddot{x}
$$
Combining, we get
$$
T_{3}=-\frac{1}{5}\lambda l^{2}\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\theta\right)+\frac{54}{125}\lambda l^{2}\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\varphi\right)
$$
Substituting the above equation, we obtain the final expression
$$
\left({\frac{54}{125}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\varphi\right)-{\frac{1}{5}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\theta\right)\right)\mathrm{sin}\varphi-{\frac{18}{25}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{sin}\varphi\right)\mathrm{cos}\varphi={\frac{6}{25}}{\ddot{\varphi}}
$$
Let = +, then retain terms up to the first order of and ΞΈ, we get
$$
\ddot{\delta}=-\frac{25\sqrt{2}}{204}\omega^{2}\theta_{0}\mathrm{cos}\omega t
$$
That is
$$
A={\frac{25{\sqrt{2}}}{204}}\theta_{0}
$$ | $$
A = \frac{25 \sqrt{2}}{204} \theta_0
$$ |
630 | THERMODYNAMICS | In this problem, we study a simple \"gas-fueled rocket.\" The main body of the rocket is a plastic bottle, which can take off after adding a certain amount of fuel gas and igniting it. It is known that the external atmospheric pressure is $P_{0}$, and the initial pressure of the gas in the bottle is $P_{0}$, with a temperature of $T_{0}$. The proportion of the molar amount of fuel gas is $\alpha$, and the rest is air. Alcohol is chosen as the fuel, with the reaction equation in air as: $C_{2}H_{5}OH(g) + 3O_{2}(g) \rightarrow 3H_{2}O(g) + 2CO_{2}(g)$. Assume that the air contains only $N_{2}$ and $O_{2}$, all gases are considered as ideal gases, and none of the vibrational degrees of freedom are excited. It is known that under conditions of pressure $P_{0}$ and temperature $T_{0}$, the reaction heat is $-\lambda$ (enthalpy change for $1\mathrm{mol}$ of ethanol reacting under isothermal and isobaric conditions). Ignore heat loss, assume the reaction is complete, and that there is air remaining, with the gas volume remaining constant throughout. Find the pressure $P_{1}$ of the gas inside the bottle after the gas has completely reacted upon ignition. | Let the initial moles of air be $n_{0}$, and the moles of alcohol be $$ n_{x}=\frac{\alpha n_{0}}{1-\alpha} $$ The ideal gas equation of state is $$ {P_{0}V_{0}=\left(n_{x}+n_{0}\right)R T_{0}} \\ {P_{0}V_{1}=\left(2n_{x}+n_{0}\right)R T_{0}} $$ According to the definition of enthalpy change, we have $$ \Delta{{H}_{0}}=\Delta{{U}_{0}}+\Delta\left(P V\right)=\Delta{{U}_{0}}+{{P}_{0}}\left({{V}_{1}}-{{V}_{0}}\right) $$ It follows that $$ \Delta U_{0}=-n_{x}\lambda-n_{x}RT_{0} $$ Since heat loss is neglected and no gas escapes from the bottle, the reaction is adiabatic and the volume does not change, giving us $$ \Delta U=\Delta U_{\mathrm{0}}+\Big[\frac{5}{2}R\cdot\big(n_{0}-n_{x}\big)+3R\cdot 3n_{x}\Big]\big(T_{1}-T_{0}\big)=0 $$ We obtain $$ T_{1}=T_{0}+\frac{2n_{x}\left(\lambda+R T_{0}\right)}{R\left(5n_{0}+13n_{x}\right)}=T_{0}+\frac{2\alpha\left(\lambda+R T_{0}\right)}{R\left(5+8\alpha\right)} $$ The ideal gas equation of state is $$ {P_{0}V_{0}=\left(n_{x}+n_{0}\right)R T_{0}} \\ {P_{1}V_{0}=\left(2n_{x}+n_{0}\right)R T_{1}} $$ We find $$ P_{1}=P_{0}\frac{(1+\alpha)(5+10\alpha+2\alpha(\frac{\lambda}{R T_{0}}))}{5+8\alpha} $$ | $$P_0 \frac{(1+\alpha)(5+10\alpha+2\alpha \frac{\lambda}{R T_0})}{5+8\alpha}$$ |
134 | MECHANICS | A fox is escaping along a straight line $AB$ at a constant speed $v_{1}$. A hound is pursuing it at a constant speed $v_{2}$, always aimed at the fox. At a certain moment, the fox is at $F$ and the hound is at $D$. $FD \bot AB$ and $FD = L$. Find the magnitude of the hound's acceleration at this moment. | To solve $\textcircled{1}$, the hunting dog performs uniform circular motion. Thus, the tangential acceleration is zero, while the normal acceleration changes. Assume that during a very short period of time $\mathrm{d}t$ after the given moment, the curvature radius of the trajectory of the hunting dog is $\rho,$ then the normal acceleration is
$$
a_{n}={\frac{v_{2}^{2}}{\rho}}.
$$
During the time interval $\mathrm{d}t$, the fox and the hunting dog respectively reach positions $F^{\prime}$ and $D^{\prime}$ (see Figure 1.7(a) in the problem). The angle turned by the hunting dog's direction of motion is
$$
\alpha={\frac{\widehat{D D^{\prime}}}{\rho}}={\frac{v_{2}\mathrm{d}t}{\rho}}.
$$
However, the distance traveled by the fox is
$$
\begin{array}{r}{\overline{{F F^{\prime}}}=v_{1}\mathrm{d}t\approx L\tan\alpha\approx\alpha L,}\end{array}
$$
which implies
$$
\alpha={\frac{v_{1}\mathrm{d}t}{L}}.
$$
Thus,
$$
\frac{\upsilon_{2}\mathrm{d}t}{\rho}=\frac{\upsilon_{1}\mathrm{d}t}{L},
$$
and from this, the curvature radius can be obtained as
$$
\rho=\frac{L v_{2}}{v_{1}}.
$$
Therefore, the magnitude of the acceleration is
$$
a_{n}={\frac{v_{2}^{2}}{\rho}}={\frac{v_{1}v_{2}}{L}}.
$$
It should be noted that the acceleration obtained this way corresponds only to the specific moment mentioned in the problem, not to the acceleration at any arbitrary position during the chasing process. | $$\frac{v_1 v_2}{L}$$ |
650 | MECHANICS | Two parallel light strings, each with length $L$, are horizontally separated by a distance $d$. Their upper ends are connected to the ceiling, and the lower ends are symmetrically attached to a uniformly distributed smooth semicircular arc. The radius of the semicircle is $R$, and its mass is $m$. The gravitational acceleration is $g$. A small ball with the same mass $m$ is placed at the bottom of the semicircle. The entire system undergoes small oscillations in the plane in some manner, with a total energy of $E_{0}$. The potential energy zero point is chosen such that the energy of the system is zero when it is vertically stable. Find the maximum $d$, denoted as $d_{max}$, that allows the system to have zero tension in one of the strings at some moment.
| Let the angle between the rope and the vertical be $\theta$, and the angle between the line connecting the small ball and the center of the circle and the vertical be $\varphi$. First, consider the kinetic equations. The kinetic energy $$ T={\frac{1}{2}}m L^{2}{\dot{\theta}}^{2}+{\frac{1}{2}}m(L{\dot{\theta}}+R{\dot{\varphi}})^{2} $$ The potential energy $$ V=m g L\theta^{2}+{\frac{1}{2}}m g R\varphi^{2} $$ Thus, the equations of motion are $$ \begin{array}{c}{{2m L^{2}\ddot{\theta}+m L R\ddot{\varphi}+2m g L\theta=0}}{{\phantom{-}}}{{m R^{2}\ddot{\varphi}+m L R\ddot{\theta}+m g R\varphi=0}}\end{array} $$ Consider the torque at the midpoint of the endpoints of the two ropes on the ceiling $$ \tau=2m g L\theta+m g R\varphi+m L^{2}\ddot{\theta}+m L(L-\left(1-\frac{2}{\pi}\right)R)\ddot{\theta}+m L R\ddot{\varphi} $$ Substituting the equations of motion, simplifying gives $$ \tau=2m g\left(1-\frac{2}{\pi}\right)R\theta+\frac{2}{\pi}m g R\varphi $$ To require that one rope has no tension, then $\tau=m g d$. Keeping $E_{0}$ constant and maximizing $d$ is equivalent to minimizing $E_{0}$ under the condition. Obviously, at this time one should have $$ \dot{\theta}=\dot{\varphi}=0 $$ because the velocity terms do not contribute to the torque and only increase the energy. Therefore, at the critical point $$ E_{0}=m g L\theta^{2}+{\frac{1}{2}}m g R\varphi^{2} $$ According to the Cauchy inequality $$ \theta^{2}+\frac{1}{2}m g R\varphi^{2}\Bigg)\left(\left(\frac{2m g\left(1-\frac{2}{\pi}\right)R}{\sqrt{m g L}}\right)^{2}+\left(\frac{\frac{2}{\pi}m g R}{\sqrt{\frac{1}{2}m g R}}\right)^{2}\right)\geq\left(2m g\left(1-\frac{2}{\pi}\right)R\theta+\frac{2}{\pi}m g R\varphi\right)^{2} $$ Substituting the torque, the energy has: $$ d_{\operatorname*{max}}=\frac{2}{\pi}\sqrt{\frac{E_{0}R}{m g}}\sqrt{(\pi-2)^{2}\frac{R}{L}+2} $$
| $$\frac{2}{\pi}\sqrt{\frac{E_0 R}{m g}}\sqrt{(\pi-2)^2\frac{R}{L}+2}$$ |
332 | MECHANICS | There is a uniform thin spherical shell, with its center fixed on a horizontal axis and able to rotate freely around this axis. The spherical shell has a mass of $M$ and a radius of $R$. There is also a uniform thin rod, with one end smoothly hinged to the axis at a distance $d$ from the center of the sphere, and the other end resting on the spherical shell. The thin rod has a mass of $m$ and a length of $L$ ($d-R<L<\sqrt{d^{2}-R^{2}}$). Initially, the point of contact and the line connecting the sphere's center lie in the same vertical plane as the rod. Now, a small perturbation is applied, causing the system to begin rotating as a whole without relative sliding around the horizontal axis. When the entire system rotates by an angle $\varphi$, the spherical shell and the rod just begin to experience relative sliding. Determine the static friction coefficient $\mu$ between the spherical shell and the rod, expressed as a function of $\varphi$. | Introducing simplified parameters $$ \begin{array}{l}{\displaystyle{\beta=\cos^{-1}\frac{L^{2}+d^{2}-R^{2}}{2L d}}}{\displaystyle{\gamma=\cos^{-1}\frac{R^{2}+d^{2}-L^{2}}{2R d}}}\end{array} $$ The coordinates of the contact point \( B \) are written as $$ \overrightarrow{O B}=(\cos\beta,\sin\beta\sin\varphi,\sin\beta\cos\varphi)L $$ Thus, the rod direction vector is written as $$ \hat{\boldsymbol{l}}=(\cos\beta,\sin\beta\sin\varphi,\sin\beta\cos\varphi) $$ Support force direction vector Assume the direction vector of the friction force is $$ \pmb{\hat{n}}=(-\cos\gamma,\sin\gamma\sin\varphi,\sin\gamma\cos\varphi) $$ $$ {\hat{f}}=\cos\theta{\frac{{\hat{l}}\times{\widehat{\pmb{n}}}}{\left|{\hat{l}}\times{\widehat{\pmb{n}}}\right|}}+\sin\theta{\frac{{\widehat{\pmb{n}}}\times({\hat{l}}\times{\widehat{\pmb{n}}})}{\left|{\widehat{\pmb{n}}}\times({\hat{l}}\times{\widehat{\pmb{n}}})\right|}} $$ That is $$ \hat{\boldsymbol f}=\cos\theta\left(0,-\cos\varphi,\sin\varphi\right)+\sin\theta\left(\sin\gamma,\cos\gamma\sin\varphi,\cos\gamma\cos\varphi\right)\qquad\mathrm{(}||\boldsymbol\varphi||\overset{\leq}{\sin\theta}|\leq\sin\gamma\sin\varphi\mathrm{(}||\boldsymbol\varphi||,\cos\varphi| $$ For the spherical shell, by the angular momentum theorem $$ {\frac{2}{3}}M R^{2}{\frac{d^{2}\varphi}{d t^{2}}}=R\cos\theta\sin\gamma f $$ For the rod, by the angular momentum theorem $$ \left\{\begin{array}{c}{{\frac{1}{3}m L^{2}\sin\beta^{2}\frac{d^{2}\varphi}{d t^{2}}=\frac{1}{2}m g L\sin\beta\sin\varphi-L\cos\theta\sin\beta f\qquad\quad\bigtriangledown}}{{\frac{1}{3}m L^{2}\sin\beta\cos\beta(\frac{d\varphi}{d t})^{2}=\frac{1}{2}m g L\cos\beta\cos\varphi-L\sin(\beta+\gamma)N-L\sin\theta\cos(\beta+\gamma)f\qquad\quad\bigtriangledown}}\end{array}\right. $$ From Equation $\textcircled{1}$ It follows that $$ {\begin{array}{r l r}{f\cos\theta={\frac{M m g R^{2}\sin\varphi}{2M R^{2}+m L^{2}\sin\beta^{2}}}}&{\quad{\textcircled{14}}}{{\frac{d^{2}\varphi}{d t^{2}}}={\frac{3m g L\sin\beta\sin\varphi}{4M R^{2}+2m L^{2}\sin\beta^{2}}}}&{\quad{\textcircled{15}}}\end{array}} $$ From Equation $\textcircled{1}$ $$ (\frac{d\varphi}{d t})^{2}=\frac{3m g L\sin\beta(1-\cos\varphi)}{2M R^{2}+m L^{2}\sin\beta^{2}} $$ $$ N=\frac{m g\left[M R^{2}(\cos\beta\cos\varphi-\tan\theta\sin\varphi\cos(\beta+\gamma))+m L^{2}\sin\beta^{2}\cos\beta(\frac{3}{2}\cos\varphi-1)\right]}{\sin(\beta+\gamma)(2M R^{2}+m L^{2}\sin\beta^{2})} $$ Thus $$ \mu=\frac{f}{N}=\frac{M R^{2}\sin\varphi\sin(\beta+\gamma)}{\cos\theta\left[M R^{2}(\cos\beta\cos\varphi-\tan\theta\sin\varphi\cos(\beta+\gamma))+m L^{2}\sin\beta^{2}\cos\beta(\frac{3}{2}\cos\varphi-1)\right]} $$ To happen relative slipping, the above expression should take the minimum value $$ \mu={\frac{M R^{2}\sin\varphi\sin(\beta+\gamma)}{\sqrt{\cos\beta^{2}\left[M R^{2}\cos\varphi+m L^{2}\sin\beta^{2}({\frac{3}{2}}\cos\varphi-1)\right]^{2}+[M R^{2}\sin\varphi\cos(\beta+\gamma)]^{2}}}} $$ | $$
\mu = \frac{MR^2 \sin\varphi \sin\left(\arccos\frac{L^2 + d^2 - R^2}{2L d} + \arccos\frac{R^2 + d^2 - L^2}{2R d}\right)}{\sqrt{\left(\frac{L^2 + d^2 - R^2}{2 L d}\right)^2 \left[ M R^2 \cos\varphi + m L^2 \sin^2\left(\arccos\frac{L^2 + d^2 - R^2}{2L d}\right) \left(\frac{3}{2} \cos\varphi - 1\right)\right]^2 + \left[ M R^2 \sin\varphi \cos\left(\arccos\frac{L^2 + d^2 - R^2}{2L d} + \arccos\frac{R^2 + d^2 - L^2}{2R d}\right)\right]^2 }}
$$ |
250 | ELECTRICITY |
There is now an electrolyte with thickness $L$ in the $z$ direction, infinite in the $x$ direction, and infinite in the $y$ direction. The region where $y > 0$ is electrolyte 1, and the region where $y < 0$ is electrolyte 2. The conductivities of the two dielectrics are $\sigma_{1}, \sigma_{2}$, and the dielectric constants are $\varepsilon_{1}, \varepsilon_{2}$, respectively. On the $xOz$ interface of the two dielectrics, two cylindrical holes with a radius $R$ are drilled in the $z$ direction, spaced $2D (D > R, R, D \ll L)$ apart, with centers located on the interface as long straight cylindrical holes. Two cylindrical bodies $\pm$ are inserted into the holes, with the type of the cylinders given by the problem text below.
The cylindrical bodies $\pm$ are metal electrodes filling the entire cylinder. Initially, the system is uncharged, and at $t=0$, a power source with an electromotive force $U$ and internal resistance $r_{0}$ is used to connect the electrodes. Find the relationship between the current through the power source and time, denoted as $i(t)$. | Given the potential difference $u$, it can be seen:
$$
\varphi_{+}=u/2,\varphi_{-}=-u/2,\lambda=\frac{2\pi(\varepsilon_{1}+\varepsilon_{2})\varphi_{+}}{2\xi_{+}}=\frac{\pi(\varepsilon_{1}+\varepsilon_{2})u}{\operatorname{arccosh}(\mathrm{D/R})}
$$
Select a surface encapsulating the cylindrical surface and examine Gauss's theorem. For the positive electrode, it is easy to see:
$$
\iint\vec{E}\cdot d\vec{S}=L\oint\vec{E}\cdot\hat{n}d l=\frac{\lambda L}{(\varepsilon_{1}+\varepsilon_{2})/2}=\frac{2\pi u L}{2\mathrm{arccosh}(D/R)}
$$
Since the above potential distribution is deemed directly applicable for the calculation of current, the total current flowing out of the positive electrode is:
$$
I=\iint\sigma\vec{E}\cdot d\vec{S}=\frac{\sigma_{1}+\sigma_{2}}{2}\times\frac{2\pi u L}{2\mathrm{arccosh}(D/R)}
$$
Given the current $i$ passing through the power source, this current changes the net charge:
$$
{\frac{d(\lambda L)}{d t}}=i-I=i-{\frac{2\pi u L}{2\mathrm{arccosh}(D/R)}}={\frac{\pi(\varepsilon_{1}+\varepsilon_{2})L}{2\mathrm{arccosh}(D/R)}}{\frac{d u}{d t}}
$$
According to the loop voltage drop equation:
$$
\begin{array}{l l}{{U=r_{0}i+u\rightarrow u=U-r_{0}i}}\ {{\rightarrow i-{\displaystyle\frac{\pi(\sigma_{1}+\sigma_{2})L}{2\mathrm{arccosh}(D/R)}}}(U-r_{0}i)=-{\displaystyle\frac{\pi(\varepsilon_{1}+\varepsilon_{2})L}{2\mathrm{arccosh}(D/R)}}r_{0}{\displaystyle\frac{d i}{d t}}}\ {{\rightarrow{\displaystyle\frac{d i}{d t}}={\displaystyle\frac{(\sigma_{1}+\sigma_{2})U}{r_{0}(\varepsilon_{1}+\varepsilon_{2})}}-\left(\frac{\sigma_{1}+\sigma_{2}}{\varepsilon_{1}+\varepsilon_{2}}+{\displaystyle\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\varepsilon_{1}+\varepsilon_{2})}}\right)i}}\end{array}
$$
At time $t=0$, all current should preferentially enter the capacitor. At this time, the initial current is $U/r_{0}$, and this differential equation yields:
$$
i(t)=\frac{U}{r_{0}\left(1+\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\sigma_{1}+\sigma_{2})}\right)}\left\{2+\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\sigma_{1}+\sigma_{2})}-\exp{\left[-\left(\frac{\sigma_{1}+\sigma_{2}}{\varepsilon_{1}+\varepsilon_{2}}+\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\varepsilon_{1}+\sigma_{2})}\right)\right]}\right\},
$$ | $$
i(t)=\frac{U}{r_0\left(1+\frac{2\arccosh(D/R)}{\pi r_0 L(\sigma_1+\sigma_2)}\right)}\left(2+\frac{2\arccosh(D/R)}{\pi r_0 L(\sigma_1+\sigma_2)}-\exp\left[-\left(\frac{\sigma_1+\sigma_2}{\varepsilon_1+\varepsilon_2}+\frac{2\arccosh(D/R)}{\pi r_0 L(\varepsilon_1+\varepsilon_2)}\right)t\right]\right)
$$ |
409 | ELECTRICITY | In electromagnetism, we often study the problem of electromagnetic field distribution in a region without charge or current distribution. In such cases, the electromagnetic field will become a tubular field. The so-called tubular field is named because of the nature of the velocity field of an incompressible fluid at every instant. Its characteristic can be described using the language of vector analysis as the field's divergence being equal to zero. Another perspective is that if we take the field lines of a vector field $\pmb{F}$ (the tangent line at each point being the direction of field strength) and select a flux tube along a set of field lines passing through a certain cross-section, then at any cross-section of the flux tube, the field flux:
$$
\int F\cdot\mathrm{d}S=\Phi
$$
will be a conserved quantity. This problem will investigate a special case of a tubular field that is rotationally symmetric around the $z$ axis: using oblate spheroidal coordinates to construct a unique tubular field as an electric field or magnetic field. The so-called oblate spheroidal coordinate system is similar to spherical coordinates, and it is generated by the following coordinate transformations:
$$
x= a \mathrm{ch}\mu \cos \nu \cos \varphi
$$
$$
y= a \mathrm{ch}\mu \cos \nu \sin \varphi
$$
$$
z=a\mathrm{sh}\mu\sin\nu
$$
We want the shape of the field lines of an electric field or magnetic field to precisely follow its generatrix direction, i.e., the tangent direction of the curve where field strength changes with $\mu$ while $\nu,\varphi$ remain fixed (upwards when $\relax z>0$).
Find the charge distribution in the $xy$ plane that can produce this electric field distribution. Assume the field strength near the origin is $E_0$. The dielectric constant is $\varepsilon_{0}$. | First, we determine the equation of the constant-$\mu$ line. Now, using the trigonometric identity $\sin^{2}\nu+\cos^{2}\nu=1$, we eliminate the parameter $\nu$:
$$
{\frac{x^{2}}{a^{2}{\mathrm{ch}}^{2}\mu}}+{\frac{z^{2}}{a^{2}{\mathrm{sh}}^{2}\mu}}=1
$$
Its physical significance is that, for an electric field, it forms a surface of equal potential. This is based on the fact that equipotential surfaces should always be perpendicular to the electric field lines. The constant-$\mu$ lines, being ellipses, satisfy this condition: their tangent directions are aligned with the external bisectors of angles formed by lines connecting the two foci, and they are perpendicular to the tangent directions of the hyperbolic electric field lines, which correspond to the internal bisectors.
Next, we determine an inverse transformation. For a point in the first quadrant with coordinates $(x, z)$, the distances to the two foci are:
$$
r_{+}=\sqrt{(x+a)^{2}+z^{2}}\quad,\quad r_{-}=\sqrt{(x-a)^{2}+z^{2}}
$$
Thus, the semi-major axis lengths of the ellipse and hyperbola are respectively:
$$
r_{+}+r_{-}=2a\mathrm{ch}\mu
$$
$$
r_{+}-r_{-}=2a\cos\nu
$$
These two equations help convert Cartesian coordinates into generalized coordinates $\mu$ and $\nu$. Next, we utilize the following property: the infinitesimal displacement caused by an increase in $\mu$ between equipotential surfaces is the derivative of the coordinates with respect to $\mu$:
$$
\mathrm{d}x=a\mathrm{sh}\mu\cos\nu\mathrm{d}\mu\quad,\quad\mathrm{d}z=a\mathrm{ch}\mu\sin\nu\mathrm{d}\mu
$$
We proceed to calculate the length of this displacement:
$$
\mathrm{d}s={\sqrt{\mathrm{d}x^{2}+\mathrm{d}z^{2}}}=a{\sqrt{\mathrm{ch}^{2}\mu-\cos^{2}\nu}}\mathrm{d}\mu
$$
Next, we consider the electric potential difference caused by displacements between one ellipse and another. This difference is equal to the electric field strength multiplied by the displacement length. Because the ellipses represent equipotential surfaces, this product is constant everywhere. In particular, taking $\nu=90^{\circ}$ places this point exactly on the $z$-axis, and the electric field strength at this location has already been computed in (2). We obtain the following equation:
$$
{\frac{E_{0}a^{2}}{a^{2}+a^{2}\mathrm{{sh}}^{2}\mu\sin^{2}{90^{\circ}}}}\cdot a\mathrm{{ch}}\mu\mathrm{{d}}\mu=E\cdot a{\sqrt{\mathrm{{ch}}^{2}\mu-\cos^{2}\nu}}\mathrm{{d}}\mu
$$
This gives the distribution of electric field strength. To determine the direction of electric field lines, we only need to locate the bisector of the angle formed by lines connecting the two foci:
$$
\theta={\frac{\theta_{+}+\theta_{-}}{2}}
$$
$$
\tan\theta_{\pm}={\frac{z}{x\pm a}}
$$
We organize the result:
$$
f(x,z)={\frac{2a^{2}}{\left[{\sqrt{(x+a)^{2}+z^{2}}}+{\sqrt{(x-a)^{2}+z^{2}}}\right]^{2}}}\left[\left[{\frac{(x+a)^{2}+z^{2}}{(x-a)^{2}+z^{2}}}\right]^{\frac{1}{4}}+\left[{\frac{(x-a)^{2}+z^{2}}{(x+a)^{2}+z^{2}}}\right]^{\frac{1}{4}}\right]~.
$$
With $z=0$, $x=r$, we obtain:
$$
f(r)=\frac{2a^{2}}{(r+a+|r-a|)\sqrt{|r^{2}-a^{2}|}}
$$
For the electric field, a charged circular disk with radius $a$ is required to generate it. In this case, the field strength at $r<a$ is:
$$
E=E_{0}f(r)={\frac{E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$
By symmetry and Gauss's law:
$$
E=\frac{\sigma}{2\varepsilon_{0}}
$$
Thus, the surface charge density on the disk is:
$$
\boxed{\sigma=\frac{2\varepsilon_{0}E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$ | $$
\boxed{\sigma=\frac{2\varepsilon_{0}E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$ |
133 | MECHANICS | A particle undergoes planar motion, where the $x$ component of its velocity $v_{x}$ remains constant. The radius of curvature at this moment is $R$. Determine the acceleration at this moment. | To solve for the radius of curvature, it's more appropriate to use the natural coordinate system. In this coordinate system, the tangential and normal components of acceleration are represented as:
$$
a_{\tau} = {\frac{\mathrm{d}v}{\mathrm{d}t}}, \quad a_{n} = {\frac{v^{2}}{R}}.
$$
Also,
$$
v_{x} = v \cos\theta,
$$
where $\theta$ can be referenced in Figure 1.5. If we take the time derivative of equation (2) and consider that $v_{x}$ is a constant, we get:
$$
{\frac{\mathrm{d}v}{\mathrm{d}t}} = v {\dot{\theta}} \tan\theta.
$$
Since $\dot{\theta} = \frac{\mathrm{d}\theta}{\mathrm{d}s} \frac{\mathrm{d}s}{\mathrm{d}t} = \frac{v}{R}$, where $s$ is the arc length coordinate, substituting this relation into equation (3) gives:
$$
a_{\tau} = {\frac{\mathrm{d}v}{\mathrm{d}t}} = {\frac{v^{2}}{R}} \tan\theta.
$$
From equations (1), (2), and (4), the magnitude of the particle's acceleration can be expressed as:
$$
a = \sqrt{a_{n}^{2} + a_{\tau}^{2}} = \frac{v^{2}}{R} \frac{1}{\cos\theta} = \frac{v^{3}}{R v_{x}}.
$$ | $$\frac{v^3}{R v_x}$$ |
336 | MECHANICS | A homogeneous solid small elliptical cylinder is placed inside a thin circular cylinder. The semi-major axis of the elliptical cylinder's cross section is $a$, the semi-minor axis is $b$, and its mass is $m$. The inner radius of the circular cylinder is $R$, and its mass is $M$. The central axes of both the circular cylinder and the elliptical cylinder are horizontal. The magnitude of gravitational acceleration is $g$. The circular cylinder is fixed and cannot rotate. Initially, the elliptical cylinder is in equilibrium at the bottom of the circular cylinder with its minor axis aligned vertically. Assuming the elliptical cylinder undergoes rolling without slipping inside the circular cylinder, derive the frequency $f$ of the small oscillations of the elliptical cylinder's center of mass near this equilibrium position. | The radius of curvature at the contact point is
$$
\rho_{1}=\frac{a^{2}}{b}
$$
Thus, in geometric terms, it can be equivalent to a cylinder whose center of mass is offset by $(\rho_{1}-b)$. Let the angle between the contact point and the cylinder center line be $\theta$, and the self-rotation angle of the elliptical cylinder be $\varphi$.
No-slip rolling gives
$$
R\theta=\rho_{1}(\theta+\varphi)
$$
That is
$$
\varphi=({\frac{b R}{a^{2}}}-1)\theta
$$
Thus the potential energy of the system is
$$
E_{p}=m g[(R-\rho_{1})(1-\cos\theta)+(\rho_{1}-b)(1-\cos\varphi)]
$$
Under the small angle approximation
$$
E_{p}={\frac{1}{2}}m g\left[(R-{\frac{a^{2}}{b}})+({\frac{a^{2}}{b}}-b)({\frac{b R}{a^{2}}}-1)^{2}\right]\theta^{2}={\frac{1}{2}}m g(b+R-{\frac{b^{2}R}{a^{2}}})({\frac{b R}{a^{2}}}-1)\theta^{2}
$$
The stability condition for a single degree of freedom conservative system:
$$
\frac{d^{2}E_{p}}{d\theta^{2}}>0
$$
That is
$$
R>{\frac{a^{2}}{b}}
$$
The moment of inertia of the elliptical cylinder about its central axis is
$$
I={\frac{1}{4}}m(a^{2}+b^{2})
$$
Thus, the kinetic energy of the system is
$$
E_{k}=\frac{1}{2}(I+m b^{2})\dot{\varphi}^{2}=\frac{1}{2}\frac{m(a^{2}+5b^{2})}{4}(\frac{b R}{a^{2}}-1)^{2}\dot{\theta}^{2}
$$
Therefore, the frequency of oscillation is
$$
f={\frac{1}{2\pi}}{\sqrt{\frac{4g(b+R-{\frac{b^{2}R}{a^{2}}})}{(a^{2}+5b^{2})({\frac{b R}{a^{2}}}-1)}}}
$$ | $$
f=\frac{1}{2\pi}\sqrt{\frac{4g(b+R-\frac{b^2R}{a^2})}{(a^2+5b^2)\left(\frac{bR}{a^2}-1\right)}}
$$ |
88 | MODERN | In a certain atom $A$, there are only two energy levels: the lower energy level $A_{0}$ is called the ground state, and the higher energy level $A^{\star}$ is called the excited state. The energy difference between the excited state and the ground state is $E_{0}$. When the atom is in the ground state, its rest mass is $m_{0}$; when it is in the excited state, due to its intrinsic instability, it will transition to the ground state while emitting photons externally. It is known that the probability of transition from the excited state to the ground state in unit time in the atom's rest reference frame is $\lambda$, and the probability of emitting photons in all directions is equal. At the initial moment, the total number of atoms is $N_{0}(N_{0}\gg1)$, and the experimental reference frame $S$ and the atom's proper reference frame $S^{\prime}$ are time-synchronized. All $N_{0}$ atoms are in the excited state and have a common velocity $\pmb{v}$ in the $+\hat{\pmb{x}}$ direction relative to the $S$ frame.
Considering $m_{0}c^{2}\gg E_{0}$, at this time the recoil effect on the atom due to photon emission can be ignored. Try to solve the angular distribution of the light emission power $w(\theta)$ from the atom at time $\pmb{t}$ in the laboratory reference frame $S$. (Calculate the emission power per unit solid angle in the direction of angle $\pmb{\theta}$ relative to the $+\hat{\pmb{x}}$ axis, rather than the received power at an infinite distance. The speed of light in vacuum is $c$ (known). The answer should be expressed using $\lambda, N_0, E_0, v, \theta, \pi, t, c$. Please check and output the final answer. | Consider the reference frame $S^{\prime}$:
$$
h\nu_{0}=E_{0}
$$
Let the remaining number of atoms at time $t^{\prime}$ be $N^{\prime}$:
$$
-\mathrm{d}N^{\prime}=\lambda N^{\prime}\mathrm{d}t^{\prime}
$$
Solving for $N^{\prime}$, we get:
$$
N^{\prime}=N_{0}e^{-\lambda t^{\prime}}
$$
The total power emitted by photons:
$$
P^{\prime}=-\dot{N}^{\prime}h\nu_{0}=\lambda N_{0}E_{0}e^{-\lambda t^{\prime}}
$$
Consider the relativistic transformation of energy, momentum, and time between $s^{\prime}$ and $s$:
$$
\mathrm{d}t=\gamma\mathrm{d}t^{\prime}
$$
$$
\mathrm{d}E_{l i g h t}=\gamma(\mathrm{d}E_{l i g h t}^{\prime}+v\mathrm{d}p_{l i g h t}^{\prime})=\gamma\mathrm{d}E_{l i g h t}^{\prime}
$$
Using the relations:
$$
\begin{array}{r}{\mathrm{d}E_{l i g h t}=P\mathrm{d}t}\ {\mathrm{d}E_{l i g h t}^{\prime}=P^{\prime}\mathrm{d}t^{\prime}}\end{array}
$$
We obtain:
$$
P=P^{\prime}=\lambda N_{0}E_{0}e^{-\lambda\frac{\epsilon}{\gamma}}
$$
Where:
$$
\gamma={\frac{1}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}
$$
Considering that number density is Lorentz invariant, let the number of photons emitted in $S$, within $\theta \sim \theta + \mathrm{d}\theta$, be $\dot{n}$.
From the transformation of the photon energy and momentum, the relationship between $\pmb\theta^{\prime}$ and $\pmb\theta$ can be expressed as:
$$
\dot{n}\mathrm{d}t\times2\pi\sin\theta\mathrm{d}\theta=\frac{\dot{N}^{\prime}}{4\pi}\mathrm{d}t^{\prime}\times2\pi\sin\theta^{\prime}\mathrm{d}\theta^{\prime}
$$
$$
\cos\theta^{\prime}=\frac{\cos\theta-\beta}{1-\beta\cos\theta}
$$
And the transformation of time:
$$
\mathrm{d}t^{\prime}=\sqrt{1-\beta^{2}}\mathrm{d}t
$$
Substituting into the equations, we get:
$$
\dot{n}=\frac{\dot{N}^{\prime}}{4\pi}\cdot\frac{\mathrm{d}\cos\theta^{\prime}}{\mathrm{d}\cos\theta}\cdot\frac{\mathrm{d}t^{\prime}}{\mathrm{d}t}=\frac{\dot{N}^{\prime}}{4\pi}\cdot\frac{(1-\beta^{2})^{\frac{3}{2}}}{(1-\beta\cos\theta)^{2}}
$$
Considering the frequency transformation of photons:
$$
\nu=\frac{\sqrt{1-\beta^{2}}}{1-\beta\cos\theta}\nu_{0}
$$
We further obtain the power distribution of the emitted light:
$$
w={\frac{\mathrm{d}P}{\mathrm{d}\Omega}}={\dot{n}h\nu}={\frac{P}{4\pi}}{\frac{(1-\beta^{2})^{2}}{(1-\beta\cos\theta)^{3}}}={\frac{\lambda N_{0}E_{0}}{4\pi}}{\frac{(1-\beta^{2})^{2}}{(1-\beta\cos\theta)^{3}}}\cdot e^{-\lambda{\frac{4}{7}}}
$$
Rewriting $\beta$ in terms of velocity:
In the relativistic framework, $\beta = \frac{v}{c}$ (where $v$ is the velocity and $c$ is the speed of light in a vacuum). Replacing $\beta$ and $\lambda$ with velocity-related quantities, the formula becomes:
$$
w=\frac{\mathrm{d}P}{\mathrm{d}\Omega}=\dot{n}h\nu=\frac{P}{4\pi}\frac{(1 - (\frac{v}{c})^{2})^{2}}{(1 - \frac{v}{c}\cos\theta)^{3}}=\frac{\lambda N_{0}E_{0}}{4\pi}\frac{(1 - (\frac{v}{c})^{2})^{2}}{(1 - \frac{v}{c}\cos\theta)^{3}}\cdot e^{-\lambda\frac{4}{7}}
$$ | $$
w=\frac{\lambda N_0 E_0}{4\pi}\frac{\left(1-\left(\frac{v}{c}\right)^2\right)^2}{\left(1-\frac{v}{c}\cos\theta\right)^3}e^{-\frac{4\lambda}{7}}
$$ |
105 | ELECTRICITY | This problem aims to guide everyone to discover a very interesting way to understand electromagnetic fields. It is known that the Maxwell equations in vacuum are
$$
\nabla\cdot{\pmb{E}}=\frac{\rho}{\varepsilon_{0}}
$$
$$
\nabla\times{\pmb{{E}}}=-\frac{\partial{\pmb{{B}}}}{\partial t}
$$
$$
\nabla\cdot\pmb{B}=0
$$
$$
\nabla\times\boldsymbol{B}=\mu_{0}\boldsymbol{j}+\varepsilon_{0}\mu_{0}\frac{\partial\boldsymbol{E}}{\partial t}
$$
To make the equations of electromagnetic fields more symmetrical, we introduce magnetic charge. Magnetic charge $\pmb{g}$ will generate a static magnetic field, and the motion of magnetic charge will generate a static electric field. It takes the following form:
$$
\boldsymbol{B}={\frac{g \boldsymbol e_{r}}{4\pi r^{2}}}
$$
$$
\pmb{{E}}=\frac{\lambda g\pmb{{v}}\times\pmb{\boldsymbol{e}}_{r}}{4\pi c r^{2}}
$$
Find $\lambda$ by conservation of magnetic charge. | $$
\pmb{\varepsilon}\rightarrow\pmb{\varepsilon}
$$
It can be obtained:
$$
{\pmb B}\rightarrow\mu_{0}{\pmb B}c
$$
$$
\nabla\cdot\pmb{\cal E}=\rho
$$
$$
\nabla\times\pmb{\mathcal{E}}=-\frac{1}{c}\frac{\partial\pmb{\mathcal{B}}}{\partial t}
$$
$$
\nabla\cdot\pmb{B}=0
$$
$$
\nabla\times\pmb{B}=\frac{1}{c}\pmb{j}+\frac{1}{c}\frac{\partial\pmb{E}}{\partial t}
$$
Following the position where the charge density term appears, add the divergence term of the magnetic field and the curl term of the electric field, we obtain:
$$
\begin{array}{c}{\boldsymbol{\nabla\cdot E}=\rho}\ {\boldsymbol{\nabla\times E}=\lambda\frac{k}{c}-\frac{1}{c}\frac{\partial\boldsymbol{B}}{\partial t}}\ {\boldsymbol{\nabla\cdot B}=\omega}\ {\boldsymbol{\nabla\times B}=\frac{1}{c}\boldsymbol{j}+\frac{1}{c}\frac{\partial E}{\partial t}}\end{array}
$$
The conservation of magnetic charge implies:
$$
\frac{\partial\omega}{\partial t}+\nabla\cdot\boldsymbol{k}=0
$$
Since the curl of the electric field does not have divergence, the identity is obtained:
$$
\lambda\frac{\nabla\cdot\pmb{k}}{c}-\frac{1}{c}\frac{\partial\nabla\cdot\pmb{B}}{\partial t}=0
$$
Substituting the divergence expression of the magnetic field and comparing it with the conservation of magnetic charge, we find:
$$
\lambda=-1
$$ | $$\lambda = -c$$ |
617 | MECHANICS | Consider each domino as a uniform, smooth slender rod with height $h$ and mass $m$ (neglecting thickness and width), connected to the ground with a smooth hinge. All rods are initially vertical to the ground and arranged evenly in a straight line, with the distance between adjacent rods being $l$, where $l \ll h$. The coefficient of restitution between the rods is $e = 0$. For convenience, assume that forces between the rods only occur at the instant of collision, and are otherwise negligible. Assume there are infinitely many rods, all initially standing upright. At a certain moment, a disturbance is applied to the first rod, causing it to fall with a certain initial angular velocity. The angular velocity of each rod at the instant it is hit by the previous rod may vary, but as the dominoes collectively fall forward, this post-collision angular velocity will tend toward a certain value. Determine this value $\omega_{\infty}$.
| Suppose the initial angular velocity of the $n$th rod when it falls is $\omega _{n}$. From the beginning of the fall until it collides with the next rod, the angle through which the rod rotates is denoted as $\theta$. Since $l\ll h$, we have $$\theta\approx\tan\theta\approx\sin\theta=\frac lh$$ Rigid body dynamics: $$\frac12mgh\theta=\frac13mh^2\ddot{\theta}$$ Thus: $\ddot{\theta}-\rho^2\theta=0$ where $\rho=\sqrt{\frac{3g}{2h}}$ The general solution is: $\theta(t)=C_1\cosh\rho t+C_2\sinh\rho t$ With initial conditions $$\theta(0)=0,\quad\dot{\theta}(0)=\omega_n$$ Therefore, $$\theta(t)=\frac{\omega_n}{\rho}\sinh\rho t$$ The angle rotated by the $n$th rod before colliding with the $n+1$th rod: $$\theta_0=\frac{l}{h}$$ Time required: $$t_n=\dfrac{1}{\rho}\sinh^{-1}\dfrac{\rho l}{h\omega_n}$$ Using properties of hyperbolic functions $$\cosh^2x-\sinh^2x=1$$ We obtain $$\dot{\theta}(t_n)=\sqrt{\omega_n^2+\frac{\rho^2l^2}{h^2}}$$ At the moment of collision between the $n$th rod and the $n+1$th rod, the force and impulse must be along the horizontal direction. Let $\omega_n^{\prime}$ be the angular velocity of the $n$th rod after the collision. Angular momentum: $$\dot{\theta}(t_n)=\omega_n^{\prime}+\omega_{n+1}$$ Inelastic collision: $\sqrt{h^2-l^2}\omega_{n+1}-h\omega^{\prime}\cdot\frac{\sqrt{h^2-l^2}}h=e\left[h\dot{\theta}(t_n)\cdot\frac{\sqrt{h^2-l^2}}h-0\right]=0$ Solving yields $$\omega_{n+1}^{2}=\frac14\omega_{n}^{2}+\frac14\frac{\rho^{2}l^{2}}{h^{2}}\\\omega_{n+1}^{2}-\frac13\frac{\rho^{2}l^{2}}{h^{2}}=\frac14\left(\omega_{n}^{2}-\frac13\frac{\rho^{2}l^{2}}{h^{2}}\right)$$ Solving this sequence gives the angular velocity of the $n$th rod just after collision $$\omega_n=\sqrt{\frac{\omega_0^2}{4^n}+\frac{\rho^2l^2}{3h^2}\left(1-\frac1{4^n}\right)}$$ In this formula, $\omega_0$ is the initial angular velocity when the first rod starts to fall. Let $n\to\infty$ to obtain $$\omega_\infty=\frac{\rho l}{\sqrt{3}h}=\frac lh\sqrt{\frac g{2h}}$$ | $$\omega_\infty = \frac{l}{h} \sqrt{\frac{g}{2h}}$$ |
702 | ELECTRICITY | To establish a rectangular coordinate system in an infinitely large three-dimensional space, there are two uniformly positively charged infinite lines at $(a,0)$ and $(-a,0)$, parallel to the $z$-axis, with a charge line density of $\lambda$; and two uniformly negatively charged lines at $(0,a)$ and $(0,-a)$, also parallel to the $z$-axis, with a charge line density of $-\lambda$. For each line, the point at a distance $a$ from the line can be taken as where the electric potential is zero. If a positive ion with a mass of $m$ and a charge of $q$ is placed near the origin, and it is restricted to move only within the $xy$ plane, and somehow all four charged lines are made to rotate counterclockwise around the origin with an angular velocity $\Omega$, with the angular velocity vector directed along the $z$-axis, given that the vacuum permittivity is $\varepsilon_{0}$, find the minimum value of $\Omega$, denoted as $\Omega_{\min}$, required for the ion to maintain stable equilibrium near the origin. | Near the origin, the electric potential can be expressed as $$ \begin{array}{r l} U&=-\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{(a-x)^{2}+y^{2}}}{a}-\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{(a+x)^{2}+y^{2}}}{a}\\ &+\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{x^{2}+(a-y)^{2}}}{a}+\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{x^{2}+(a+y)^{2}}}{a}\end{array} $$ Using the formula given in the problem, it can be considered that $x\cdot y$ are both small quantities, thus we have $$ \begin{array}{r l} \frac{\lambda}{2\pi\varepsilon_{0}}\ln{\frac{\sqrt{(a-x)^{2}+y^{2}}}{a}} & =\frac{\lambda}{4\pi\varepsilon_{0}}\ln\frac{a^{2}-2a x+x^{2}+y^{2}}{a^{2}} \\ & =\frac{\lambda}{4\pi\varepsilon_{0}}\ln\left(1+{\frac{x^{2}+y^{2}-2a x}{a^{2}}}\right) \\ & =\frac{\lambda}{4\pi\varepsilon_{0}}\left[{\frac{x^{2}+y^{2}-2a x}{a^{2}}}-{\frac{1}{2}}\left({\frac{x^{2}+y^{2}-2a x}{a^{2}}}\right)^{2}\right] \\ & =\frac{\lambda}{4\pi\varepsilon_{0}}\left({\frac{x^{2}+y^{2}-2a x}{a^{2}}}-{\frac{2x^{2}}{a^{2}}}\right) \\ & =\frac{\lambda\left(-x^{2}+y^{2}-2a x\right)}{4\pi\varepsilon_{0}a^{2}} \end{array} $$ By proceeding similarly, we obtain the approximate results of the other terms, thus we have $$ {\begin{array}{r l}\\&{U(x,y)=-{\frac{\lambda\left(-x^{2}+y^{2}-2a x\right)}{4\pi\varepsilon_{0}a^{2}}}-{\frac{\lambda\left(-x^{2}+y^{2}+2a x\right)}{4\pi\varepsilon_{0}a^{2}}}}\\ &{\qquad+{\frac{\lambda\left(x^{2}-y^{2}-2a y\right)}{4\pi\varepsilon_{0}a^{2}}}+{\frac{\lambda\left(x^{2}-y^{2}+2a y\right)}{4\pi\varepsilon_{0}a^{2}}}}\\ &{\qquad={\frac{\lambda\left(x^{2}-y^{2}\right)}{\pi\varepsilon_{0}a^{2}}}}\end{array}} $$ After obtaining the electric potential, we can find the components of the electric field intensity along the $x$ and $y$ directions, obtaining $$ {\begin{array}{r}{E_{x}=-{\cfrac{\partial U(x,y)}{\partial x}}=-{\cfrac{2\lambda x}{\pi\varepsilon_{0}a^{2}}}}\ {E_{y}=-{\cfrac{\partial U(x,y)}{\partial y}}={\cfrac{2\lambda y}{\pi\varepsilon_{0}a^{2}}}}\end{array}} $$ Therefore, the equation of motion for the positive ion is $$ \begin{array}{l}{{m{\ddot{x}}+\displaystyle\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}}x=0}}\ {{m{\ddot{y}}-\displaystyle\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}}y=0}}\end{array} $$ It is obvious that in the $x$ direction it is stable, whereas in the $y$ direction it is unstable. Choose the $\Omega$ system that rotates with the four lines, in this system, the four lines are stationary, so the form of the electric force near the origin remains unchanged. Parameterize the electric force, letting $$ \omega_{0}^{2}={\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}m}} $$ Considering the Coriolis force and inertial centrifugal force, the particle's dynamic equation in this system is $$ \begin{array}{c}{m\ddot{x}=m\left(-\omega_{0}^{2}+\Omega^{2}\right)x+2m\Omega\dot{y}}\ {m\ddot{y}=m\left(\omega_{0}^{2}+\Omega^{2}\right)x-2m\Omega\dot{x}}\end{array} $$ Which can be simplified to $$ \begin{array}{r}{\ddot{x}-2\Omega\dot{y}+\left({\omega_{0}^{2}-\Omega^{2}}\right)x=0}\ {\ddot{y}+2\Omega\dot{x}-\left({\omega_{0}^{2}+\Omega^{2}}\right)x=0}\end{array} $$ To discuss the stability of this two-degree-of-freedom system, consider the normal solution, thus we have $$ \begin{array}{r}{x=A e^{i\omega t}}\ {y=B e^{i\omega t}}\end{array} $$ Substitute into the above equations to obtain $$ \begin{array}{r}{\left(\omega_{0}^{2}-\Omega^{2}-\omega^{2}\right)A-2i\omega\Omega B=0}\ {2i\omega\Omega A-\left(\omega_{0}^{2}+\Omega^{2}+\omega^{2}\right)B=0}\end{array} $$ The characteristic equation is $$ -\left(\omega_{0}^{2}-\Omega^{2}-\omega^{2}\right)\left(\omega_{0}^{2}+\Omega^{2}+\omega^{2}\right)-4\omega^{2}\Omega^{2}=0 $$ Solving for the angular frequencies of the normal modes gives $$ \begin{array}{r}{\omega_{1}=\sqrt{\Omega^{2}-\omega_{0}^{2}}}\ {\omega_{2}=\sqrt{\Omega^{2}+\omega_{0}^{2}}}\end{array} $$ The existing normal modes of the system must all be stable for the system to be stable. Therefore, the condition is $$ \Omega>\omega_{0}=\sqrt{\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}m}} $$ | $$\sqrt{\frac{2q\lambda}{\pi\varepsilon_0 a^2 m}}$$ |
367 | MECHANICS | There is a smooth elliptical track, and the equation of the track satisfies $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$). On the track, there is a small charged object that can move freely along the track, with mass $m$ and charge $q$. Now, a point charge with charge $Q$ (same sign as $q$) is placed at the origin. Find the period $T$ of the small oscillations of the charged object around its stable equilibrium position. | It is evident that at this point, the small object is in a stable equilibrium position at the two vertices of the major axis of the elliptical orbit. When the small object experiences a slight deviation along the orbit from the stable equilibrium position, as shown in the figure below, the distance from the small object to the origin O is \( r \), and the angle relative to the x-axis is \( \theta \) (\( \theta << 1 \)).
From the elliptical orbit equation, differentiating both sides, we have:
\[ \frac{2x dx}{a^2} + \frac{2y dy}{b^2} = 0 \]
This leads to the elliptic tangent slope at the point:
\[ \tan \alpha = \frac{a^2}{b^2} \tan \theta \]
The angle between the direction of the electrostatic force on the small object and the normal is:
\[ \beta = \alpha - \theta \]
Since the distance of the small object deviating from the stable equilibrium position is very small, we have \(\theta, \alpha, \beta << 1\), which implies:
\[ \beta = \alpha - \theta \approx \left( \frac{a^2}{b^2} - 1 \right) \theta \]
Magnitude of the electrostatic force on the small object:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2} \]
Where \( r \) satisfies:
\[ \frac{r^2 \cos^2 \theta}{a^2} + \frac{r^2 \sin^2 \theta}{b^2} = 1 \]
Thus, we obtain:
\[ r = \sqrt{\frac{a^2 b^2}{b^2 + (a^2 - b^2) \sin^2 \theta}} = a \left[ 1 + \frac{a^2 - b^2}{b^2} \sin^2 \theta \right]^{-\frac{1}{2}} \approx a \left[ 1 + \frac{a^2 - b^2}{b^2} \theta^2 \right]^{\frac{1}{2}} \]
\[ \approx a \left( 1 - \frac{a^2 - b^2}{2b^2} \theta^2 \right) \]
And the tangential component of the electrostatic force:
\[ F_t = -F \sin \beta \approx -\frac{Qq}{4\pi \epsilon_0 a^2} \left( 1 - \frac{a^2 - b^2}{2b^2} \theta^2 \right)^{-2} \cdot \left( \frac{a^2}{b^2} - 1 \right) \theta \]
\[ \approx -\frac{Qq (a^2 - b^2)}{4\pi \epsilon_0 a^2 b^2} \theta \]
Velocity of the small object:
\[ v = \dot{r} + r \dot{\theta} = -\frac{a (a^2 - b^2)}{2b^2} 2 \theta \dot{\theta} + a \left( 1 - \frac{a^2 - b^2}{2b^2} \theta^2 \right) \dot{\theta} \approx a \dot{\theta} \]
\[ F_t = m a_t = m \dot{v} = m a \ddot{\theta} \]
\[ ma \ddot{\theta} = -\frac{Qq (a^2 - b^2)}{4\pi \epsilon_0 a^2 b^2} \theta \]
This is the standard equation of simple harmonic motion, with the angular frequency of motion being:
\[ \omega = \sqrt{\frac{Qq (a^2 - b^2)}{4\pi \epsilon_0 m a^3 b^2}} \]
Period of the small vibrations:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4\pi \epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} \] | $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4\pi \epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} $$ |
497 | MECHANICS |
---
There is a uniform spring with a spring constant of \(k\) and an original length of \(L\), with a mass of \(m\). One end is connected to a wall, and the other end is connected to a mass \(M\) block, placed on a horizontal smooth surface for simple harmonic motion. It is known that \(m \ll M\).
The spring, after being used for a long time, experiences corrosion, causing the spring constant to decrease and the mass to increase. It is approximately assumed that the spring constant and mass uniformly increase or decrease throughout the spring.
A new spring initially has an amplitude of \(A_0\). After a period of time, the spring mass becomes \(\gamma_1\) times its original mass, and the spring constant becomes \((1 - \gamma_2\frac{m}{M})\) times its original value. Under the following model, find the change in amplitude \(\Delta A\).
Assume that the additional mass added to the spring during corrosion moves at the same speed as the spring, meaning the added mass has the same velocity as the point where it is added.
--- | First, based on the analysis of the change in the system's mechanical energy, through the integration of energy relationships and separation of variables, the relationship between the energy ratio and the stiffness coefficient as well as mass is obtained:
\(\ln\frac{E}{E_0} = \frac{1}{2}\ln\frac{k'}{k} + \frac{1}{2}\ln\frac{M'}{M_0'}\), taking the exponential on both sides gives \(\frac{E}{E_0} = \sqrt{\frac{k'}{k}\frac{M'}{M_0'}}\).
From \(E = \frac{1}{2}k'A^2\) and \(E_0 = \frac{1}{2}kA_0^2\), it follows that \(\frac{\frac{1}{2}k'A^2}{\frac{1}{2}kA_0^2} = \sqrt{\frac{k'}{k}\frac{M'}{M_0'}}\).
Substituting \(k' = (1 - \gamma_2\frac{m}{M})k\), \(M' = M + \frac{1}{3}\gamma_1 m\), \(M_0' = M + \frac{1}{3}m\), and approximating using \(m \ll M\):
\(\frac{A^4}{A_0^4} = \frac{M + \frac{1}{3}\gamma_1 m}{M + \frac{1}{3}m} \cdot \frac{1}{1 - \gamma_2\frac{m}{M}} \approx \left(1 + \frac{\gamma_1 - 1}{3}\frac{m}{M}\right)\left(1 + \gamma_2\frac{m}{M}\right) \approx 1 + \left(\frac{\gamma_1 - 1}{3} + \gamma_2\right)\frac{m}{M}\).
Expanding \(A^4 \approx A_0^4\left[1 + \left(\frac{\gamma_1 - 1}{3} + \gamma_2\right)\frac{m}{M}\right]\) to the fourth root gives \(A \approx A_0\left[1 + \frac{1}{4}\left(\frac{\gamma_1 - 1}{3} + \gamma_2\right)\frac{m}{M}\right]\).
Thus, the change in amplitude \(\Delta A = A - A_0\) is:
\[
\Delta A = \frac{m A_0}{4M}\left[\frac{1}{3}(\gamma_1 - 1) + \gamma_2\right]
\] | \[
\Delta A = \frac{m A_0}{4M}\left[\frac{1}{3}(\gamma_1 - 1) + \gamma_2\right]
\] |
457 | MECHANICS | A regular solid uniform $N$-sided polygonal prism, with a mass of $m$ and the distance from the center of its end face to a vertex as $l$, is resting on a horizontal table. The axis of the prism is horizontal and points forward. A constant horizontal force $F$ acts on the center of the prism, perpendicular to its axis and large enough to cause the prism to start rotating. As a result, the prism will roll to the right while undergoing completely inelastic collisions with the table. It is known that the coefficient of static friction with the ground is sufficiently large. After a sufficiently long time, the angular velocity after each collision becomes constant. Find this angular velocity. | For a regular $N$-sided polygon with the distance from its center to a vertex being $l$, the moment of inertia about the centroid is:
$$
I_{o}=\frac{m l^{2}}{2}(\frac{1}{3}\sin^{2}\frac{\pi}{N}+\cos^{2}\frac{\pi}{N})
$$
The moment of inertia about a vertex is:
$$
I=\frac{m l^{2}}{2}(\frac{1}{3}\sin^{2}\frac{\pi}{N}+\cos^{2}\frac{\pi}{N})+m l^{2}
$$
After sufficient time has passed:
$$
\frac{1}{2}I(\Omega^{2}-\omega^{2})=F\cdot 2l\sin\theta
$$
$$
I_{o}\Omega+m\Omega a\cos{2\theta} a=\mathbf{I}\omega
$$
Where:
$$
\theta = \frac{\pi}{N}
$$
After collision,
$$
{\omega}=\sqrt{\frac{8F\sin\displaystyle\frac{\pi}{N}(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2}}{m l(2+\displaystyle\frac{1}{3}\sin^{2}\displaystyle\frac{\pi}{N}+\cos^{2}\displaystyle\frac{\pi}{N})((9+7\tan^{2}\displaystyle\frac{\pi}{N})^{2}-(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2})}}
$$ | $$
{\omega}=\sqrt{\frac{8F\sin\displaystyle\frac{\pi}{N}(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2}}{m l(2+\displaystyle\frac{1}{3}\sin^{2}\displaystyle\frac{\pi}{N}+\cos^{2}\displaystyle\frac{\pi}{N})((9+7\tan^{2}\displaystyle\frac{\pi}{N})^{2}-(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2})}}
$$ |
432 | ELECTRICITY | A homogeneous sphere with a mass of $m$ and a radius of $R$ carries a uniform charge of $Q$ and rotates around the $z$-axis (passing through the center of the sphere) with a constant angular velocity $\omega$. The formula for calculating the magnetic moment is $\sum_{i} I_{i} S_{i}$, where $I_{i}$ represents the current in a current loop and $S_{i}$ is the area vector of the current loop. The sphere is placed on an infinitely large superconducting plate, with the $z$-axis oriented vertically and perpendicular to the plane of the plate. The gravitational acceleration is $g$. When the system reaches a stable state (the sphere rotates but its center remains stationary), determine the distance $h$ from the center of the sphere to the surface of the plate. Assume the parameters satisfy the conditions for stable levitation, and neglect dissipative effects such as air resistance.
The vacuum permittivity is given as $\epsilon_0$, and the vacuum permeability is given as $\mu_0$. | Using the method of electric imaging and magnetic imaging, the electric force is given by
$$
F_{e}={\frac{-Q^{2}}{4\pi\varepsilon_{0}(2h)^{2}}}={\frac{-Q^{2}}{16\pi\varepsilon_{0}h^{2}}}
$$
Magnetic force is given by
$$
F_{m}=-{\frac{d}{d x}}\left({\frac{2\mu_{0}\mu^{2}}{4\pi x^{3}}}\right)\mid_{x=2h}={\frac{3\mu_{0}Q^{2}R^{4}\omega^{2}}{800\pi h^{4}}}
$$
Force balance condition is
$$
F_{e}+F_{m}=m g
$$
That is
$$
\frac{3\mu_{0}Q^{2}R^{4}\omega^{2}}{800\pi h^{4}}=m g+\frac{Q^{2}}{16\pi\varepsilon_{0}h^{2}}
$$
Solving for \( h \), we get
$$
h={\sqrt{\frac{-Q^{2}+{\sqrt{Q^{4}+{\frac{96\pi\varepsilon_{0}^{2}m g\mu_{0}Q^{2}R^{4}\omega^{2}}{25}}}}}{32\pi\varepsilon_{0}m g}}}
$$ | $$
h={\sqrt{\frac{-Q^{2}+{\sqrt{Q^{4}+{\frac{96\pi\varepsilon_{0}^{2}m g\mu_{0}Q^{2}R^{4}\omega^{2}}{25}}}}}{32\pi\varepsilon_{0}m g}}}
$$
|
284 | OPTICS | Fiber-optic communication has greatly advanced our information technology development. Below, we will briefly calculate how the light carrying information propagates through a rectangular optical fiber. Consider a two-dimensional waveguide (uniform in the direction perpendicular to the paper), extending along the $x$ direction, with variations in refractive index in the $y$ direction. The middle layer is called the core, with a thickness of $b$ and refractive index $n$; the top and bottom layers are called the cladding, with a refractive index of $n+\delta n$. Assume the light entering the fiber is a monochromatic wave with a wavelength of $\lambda$.
When reflecting at the waveguide interface, the light transitioning from $n_{1}$ to $n_{2}$ has a reflection coefficient:
$$
r={\frac{n_{1}\cos\theta_{1}-n_{2}\cos\theta_{2}}{n_{1}\cos\theta_{1}+n_{2}\cos\theta_{2}}}
$$
where $\theta_1$ is the angle of incidence and $\theta_2$ is the angle of refraction. According to theoretical calculations, determine the phase shift $\varphi$ of the light after its first reflection, given that the angle of incidence is $\theta$, assuming total internal reflection occurs. Assume the plane wave phase is $\omega t-kx$. | Considering that the optical signal should propagate without loss, the light transmission in the optical fiber should involve total internal reflection at the interface. Assume the incident electric field is represented by ${\widetilde{E}}_{i}$, and the reflected electric field by $\widetilde{E}_{r}$; we have $\widetilde{E_{r}}=r\cdot\widetilde{E_{i}}$. By substituting Snell's law
$$
n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}
$$
into the expression for the reflection coefficient, we obtain:
$$
r={\frac{n_{1}\cos\theta_{1}-{\sqrt{{n_{2}}^{2}-{n_{1}}^{2}\sin^{2}\theta_{1}}}}{n_{1}\cos\theta_{1}+{\sqrt{{n_{2}}^{2}-{n_{1}}^{2}\sin^{2}\theta_{1}}}}}
$$
Total internal reflection condition (no loss) requires:
$$
n_{2}\leqslant n_{1}\sin\theta_{1}
$$
Thus, the reflection coefficient can be expressed in the form of a complex number:
$$
r={\frac{n_{1}\cos\theta_{1}-j{\sqrt{{n_{1}}^{2}\sin^{2}\theta_{1}-{n_{2}}^{2}}}}{n_{1}\cos\theta_{1}+j{\sqrt{{n_{1}}^{2}\sin^{2}\theta_{1}-{n_{2}}^{2}}}}}={\frac{{n_{1}}^{2}\cos2\theta+{n_{2}}^{2}}{{n_{1}}^{2}-{n_{2}}^{2}}}-j{\frac{2n_{1}\cos\theta_{1}{\sqrt{{n_{1}}^{2}\sin^{2}\theta_{1}-{n_{2}}^{2}}}}{{n_{1}}^{2}-{n_{2}}^{2}}}
$$
Using the double angle formula of tangent, the phase shift of the reflection coefficient $\widetilde{r}=r_{0}\cdot e^{j\varphi}$ can be obtained:
$$
\varphi=2\arctan\left[{\frac{\sqrt{\sin^{2}\theta_{1}-\left(n_{2}/n_{1}\right)^{2}}}{\cos\theta_{1}}}\right]
$$
Substituting the refractive index distribution of the waveguide, the phase shift resulting from total internal reflection at the upper and lower interfaces can be obtained as:
$$
\varphi=2\arctan\left[\frac{\sqrt{\sin^{2}\theta-\left(1+\delta n/n\right)^{2}}}{\cos\theta}\right]
$$ | $$\varphi=2\arctan\left(\frac{\sqrt{\sin^2\theta-\left(1+\frac{\delta n}{n}\right)^2}}{\cos\theta}\right)$$ |
39 | MECHANICS | A rectangular wooden block of height $H$ and density $\rho_{1}$ is gently placed on the water surface. The density of the water is $\rho_{2}$, where $\rho_{1} < \rho_{2}$. The gravitational acceleration is $g$. Consider only the translational motion of the wooden block in the vertical direction, neglecting all resistance in the direction of motion and assuming the height of the water surface remains unchanged. Gently press down on the wooden block and then release it. Afterward, the wooden block is neither fully submerged nor does it detach from the water surface. Determine the period of motion of the wooden block $T$; | Taking the top at equilibrium as the origin, establish the coordinate system as shown.
Let the position of the top of the block be $y$. The net force on the block is $-\rho_{1}H S g + \rho_{2}(H-h_{0}-y)S g = -\rho_{2}S g y$. The net force on the block is a linear restoring force, and its motion is simple harmonic motion. The equation of motion for the block is
$$
\rho_{1}H S{\ddot{y}}+\rho_{2}S g y=0
$$
Therefore, the period of the block is
$$
T=2\pi{\sqrt{\frac{\rho_{1}H}{\rho_{2}g}}}
$$ | $$
T = 2\pi \sqrt{\frac{\rho_1 H}{\rho_2 g}}
$$ |
124 | MECHANICS | A wedge-shaped block with an inclination angle of $\theta$, having a mass of $M$, is placed on a smooth tabletop. Another small block of mass $m$ is attached to the top of the wedge using a spring with a spring constant $k$. The natural length of the spring is $L_{0}$, and the surfaces between the two blocks are frictionless. Now, the small block is released from rest at a position $L_{0}$ away from the top of the wedge, allowing it to slide freely downward. Find the oscillation period of the small block $m$. | Let's set the vertical and horizontal accelerations of the small wooden block $m$ as $a_{y}$ and $a\_x$, respectively. Assume that the horizontal acceleration of the wedge-shaped wooden block $M$ is $A_{x}$. We obtain:
$$
\begin{array}
{r l}{m a_{x}+M A_{x}=0.}&{(1)}\\
\ \frac{a_{x}-A_{x}}{a_{y}}=\cot \theta&{(2)}\\
\ {-N\sin\theta+F\cos\theta=M A_{x}}&{(3)}\\
\ {m g-N\cos\theta-F\sin\theta=m a_{y}}&{(4)}\\
\ {F=k(l-l_{0})=\frac{k x}{\cos\theta}\Bigl(1+\frac{m}{M}\Bigr)}&{(5)}
\end{array}
$$
From equations (1) and (2), we get:
$$
a_{y}=(a_{x}-A_{x})\tan\theta=a_{x}\left(1+\frac{m}{M}\right)\tan\theta
$$
Substitute into equation (4),
$$
\begin{array}
{r l}m g-N\cos\theta-F\sin\theta=m a_{x}\left(1+\frac{m}{M}\right)\tan\theta &{(6)}
\end{array}
$$
From equations (3) and (6), we can eliminate $N$, resulting in:
$$
m g\sin\theta-F=m a_{x}\left(1+\frac{m}{M}\right)\frac{\sin^{2}\theta}{\cos\theta}-M A_{x}\cos\theta=m a_{x}\left[\left(1+\frac{m}{M}\right)\frac{\sin^{2}\theta}{\cos\theta}+\cos\theta\right]
$$
Substitute $F$ from equation (5),
$$
m g\sin\theta-\frac{k x}{\cos\theta}\Bigl(1+\frac{m}{M}\Bigr)=m a_{x}\left[\Bigl(1+\frac{m}{M}\Bigr)\frac{\sin^{2}\theta}{\cos\theta}+\cos\theta\right]
$$
We find that there is a term $a_{x}\propto-x$ in the expression. Therefore, the angular frequency is:
$$
\begin{array}{c c c}{\displaystyle{\omega^{2}=\frac{\frac{k}{\cos\theta}\left(1+\frac{m}{M}\right)}{m\left[\left(1+\frac{m}{M}\right)\frac{\sin^{2}\theta}{\cos\theta}+\cos\theta\right]}=\frac{k\left(1+\frac{m}{M}\right)}{m[\left(1+\frac{m}{M}\right)\sin^{2}\theta+\cos^{2}\theta]}}}\ {\displaystyle{\Rightarrow\omega^{2}=\frac{k(M+m)}{m[M+m\sin^{2}\theta]}}}\
\end{array}
$$
Finally, we get the period:
$$
T=2\pi\sqrt{\frac{m(M+m\sin^{2}\theta)}{k(M+m)}}
$$ | $$
T=2\pi\sqrt{\frac{m(M+m\sin^2\theta)}{k(M+m)}}
$$ |
55 | ELECTRICITY | In a vacuum, there is an infinitely long, uniformly charged straight line fixed in place, with a charge line density of $\lambda$. Additionally, there is a dust particle with mass $m$, which can be considered as an isotropic, uniform dielectric sphere with a volume $V$, and a relative permittivity $\varepsilon_{r}$. It is given that the volume $V$ of the dielectric sphere is very small, the permittivity of vacuum is $\varepsilon_{0}$, and the following factors can be neglected: gravity, electromagnetic radiation caused by the motion of charges, and relativistic effects. Study the motion of the dust particle under the influence of the charged straight line:
Find the force acting on the particle and provide its magnitude. | Since the volume of the medium sphere $V$ is very small, we can approximate the external electric field $E$ at the medium sphere as a uniform field. Therefore, the medium sphere is uniformly polarized, and let the polarization intensity inside the sphere be $P$. The polarization surface charge density follows the cosine distribution as below (where $\alpha=0$ corresponds to the direction of the external field $E$):
$$
\sigma_{p}(\alpha)=P\cos{\alpha}
$$
The depolarization field generated by the polarized charges inside the medium sphere is a uniform field (the negative sign indicates the opposite direction to the external field $E$):
$$
E^{\prime}=-{\frac{P}{3\varepsilon_{0}}}
$$
The total electric field inside the sphere is $E+E^{\prime}$, and according to the polarization law, we have
$$
P=(\varepsilon_{r}-1)\varepsilon_{0}(E+E^{\prime})=(\varepsilon_{r}-1)\varepsilon_{0}\left(E-\frac{P}{3\varepsilon_{0}}\right)
$$
According to Gauss's theorem, the field strength generated by a charged line at the medium sphere is
$$
E(r)=\frac{\lambda}{2\pi\varepsilon_{0}r}
$$
By combining the above equations, we can solve for the required polarization intensity
$$
P(r)=\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}\frac{3\lambda}{2\pi r}
$$
The electric dipole moment of the medium sphere is
$$
p(r)=P(r)\cdot V=\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}\frac{3\lambda V}{2\pi r}
$$
Thus, the force on the particle is
$$
F(r)=p(r){\frac{\mathrm{d}E}{\mathrm{d}r}}=-{\frac{3\left(\varepsilon_{r}-1\right)V\lambda^{2}}{4\pi^{2}\left(\varepsilon_{r}+2\right)}}{\frac{1}{r^{3}}}
$$ | $$
-\frac{3(\varepsilon_r-1)V\lambda^2}{4\pi^2(\varepsilon_r+2)r^3}
$$ |
419 | OPTICS | Consider a thin layer with refractive index $ n_1 $ and thickness $ d $, sandwiched between a medium with refractive index $ n_2 $ on both sides (for simplicity, let $ n = \frac{n_1}{n_2} < 1 $). Now, suppose a beam of light with wavelength $ \lambda $ (wavelength inside $ n_2 $) is incident at an angle $ i_2 $, with a refraction angle $ i_1 $. Through multiple reflections and refractions, and given that the surface of the thin layer exhibits total internal reflection, determine the total reflectance $ R_s $ of the light intensity in this scenario, considering only the s-polarized (perpendicular to the plane of incidence) light intensity. Express the result using $ \lambda $, $ d $, $ i_2 $, and $ n$. | Calculate the amplitude after infinite reflections:
\[
\begin{align*}
E_{r}&=E_{0}\left(r + tr't\mathrm{e}^{\mathrm{j}2\delta}+t(r')^{3}t\mathrm{e}^{\mathrm{j}4\delta}+\cdots +t(r')^{2n - 1}t\mathrm{e}^{\mathrm{j}(2n - 2)\delta}+\cdots\right)\\
&=E_{0}\left(r+tr't\mathrm{e}^{\mathrm{j}2\delta}\left(1 + r'^{2}\mathrm{e}^{\mathrm{j}2\delta}+\cdots\right)\right)\\
&=E_{0}\left(r + tr't\mathrm{e}^{\mathrm{j}2\delta}\frac{1}{1 - r'^{2}\mathrm{e}^{\mathrm{j}2\delta}}\right)\\
&=E_{0}\left(r-\frac{(1 - r'^{2})r\mathrm{e}^{\mathrm{j}2\delta}}{1 - r'^{2}\mathrm{e}^{\mathrm{j}2\delta}}\right)\\
&=E_{0}\left(\frac{r(1 - \mathrm{e}^{\mathrm{j}2\delta})}{1 - r'^{2}\mathrm{e}^{\mathrm{j}2\delta}}\right)
\end{align*}
\]
Substitute the complex refractive index to obtain:
\[
R_{s}=\left(1+\left(\frac{2\mathrm{e}^{-\frac{2\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}{1 - \mathrm{e}^{-\frac{4\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}\frac{2\sqrt{(1 - \sin^{2}i_{2})(\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\right)^{2}\right)^{-1}
\] | \[
R_{s}=\left(1+\left(\frac{2\mathrm{e}^{-\frac{2\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}{1 - \mathrm{e}^{-\frac{4\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}\frac{2\sqrt{(1 - \sin^{2}i_{2})(\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\right)^{2}\right)^{-1}
\] |
458 | ADVANCED | The incompressible viscous fluid satisfies the Navier-Stokes equations:
$$
\frac{\partial \vec{v}}{\partial t} + (\vec{v} \cdot \nabla) \vec{v} = -\frac{1}{\rho} \nabla p + \frac{\mu}{\rho} \Delta \vec{v}
$$
where $\eta$ is the viscosity of the viscous fluid, $\rho$ is the density of the viscous fluid, and:
Using the Navier-Stokes equations, solve the following problem: An incompressible viscous fluid flows through a regular triangular pipe with a side length of $a$ and a length of $l$, with a pressure difference of $\Delta p$ between the two ends. Determine the volumetric flow rate $Q$. | Hypothesis:
$ v = \frac{\Delta p}{l} \frac{2}{\sqrt{30\eta}} h_1 h_2 h_3 $
It can be proved that:
$ \Delta(h_1 h_2 h_3) = -(h_1 + h_2 + h_3) = -\frac{\sqrt{3}}{2} a $
Therefore, the hypothesis satisfies the Navier-Stokes equations and boundary conditions and constitutes a solution. It is evident that the solution is unique, thus $ v = \frac{\Delta p}{l} \frac{2}{\sqrt{30\eta}} h_1 h_2 h_3 $ is the only correct solution.
Integrating yields the flow rate:
$ Q = \frac{\sqrt{3} a^4 \Delta p}{320 l} $ | $ Q = \frac{\sqrt{3} a^4 \Delta p}{320\eta l} $ |
206 | ELECTRICITY | In a zero-gravity space, two coaxial cone surfaces $A$ and $B$ are placed. Assume their common vertex is located at the origin of the coordinate system. The cylindrical coordinate equations are given as:
$$
A: r = z \tan\alpha_{1} \quad,\quad B: r = z \tan\alpha_{2}
$$
where $\alpha_{2} > \alpha_{1} (\alpha$ is the angle between the line connecting a point in space to the origin and the positive direction of the common axis $\hat{z}$).
Analysis of the electric potential in the space: It is known that the electric potentials on surfaces $A$ and $B$ satisfy $V_{A} = 0$, $V_{B} = V_{0}$. Solve for the potential distribution $V(\alpha)$ between the cone surfaces. | **Electric Field Distribution Analysis:**
Based on symmetry, select the cone vertex as the origin, and establish spherical coordinates $(R,\alpha,\theta)$ to solve the spatial electric field potential distribution. Here, $R=\sqrt{r^{2}+z^{2}}$ and $R\sin\alpha=r$, which correspond to cylindrical coordinates. The spherical coordinate form of Laplace's equation $\nabla^{2}V=0$ is written as:
$$
\frac{1}{R^{2}}\frac{\partial}{\partial R}(R^{2}\frac{\partial V}{\partial R})+\frac{1}{R^{2}\sin\alpha}\frac{\partial}{\partial\alpha}(\sin\alpha\frac{\partial V}{\partial\alpha})+\frac{1}{R^{2}\sin^{2}\alpha}\frac{\partial^{2}V}{\partial\theta^{2}}=0
$$
Considering symmetry, ${\mathrm{{.}}}V=V(\alpha)$, the equation simplifies to:
$$
\frac{1}{R^{2}\sin\alpha}\frac{\partial}{\partial\alpha}(\sin\alpha\frac{\partial V}{\partial\alpha})=0
$$
That is:
$$
\sin\alpha\frac{\partial V}{\partial\alpha}=A
$$
Where $A$ is a constant. Integrating both sides:
$$
V=\int{\frac{\mathrm{d}\alpha}{\sin\alpha}}=A\ln\tan{\frac{\alpha}{2}}+C
$$
Considering boundary conditions:
$$
V(\alpha_{1})=0,V(\alpha_{2})=V_{0}
$$
Substituting these into the solution yields:
$$
V=\frac{V_{0}}{\ln\frac{\tan\frac{\alpha_{2}}{2}}{\tan\frac{\alpha_{1}}{2}}}\mathrm{ln}\frac{\tan\frac{\alpha}{2}}{\tan\frac{\alpha_{1}}{2}}
$$ | $$
\frac{V_0}{\ln\left(\frac{\tan\left(\frac{\alpha_2}{2}\right)}{\tan\left(\frac{\alpha_1}{2}\right)}\right)}\ln\left(\frac{\tan\left(\frac{\alpha}{2}\right)}{\tan\left(\frac{\alpha_1}{2}\right)}\right)
$$ |
204 | ELECTRICITY | In modern plasma physics experiments, two methods are commonly used to confine negatively charged particles. In the following discussion, relativistic effects and contributions such as delayed potentials are not considered.
In space, uniformly charged rings with a radius of $R$ and a charge $Q_{0}$ are placed on planes $z=l$ and $z=-l$, respectively. Additionally, a particle with a charge of $-q$ and a mass of $m$ is located at the origin of the coordinate system. The permittivity of vacuum is given as $\varepsilon_{0}$.
Find the angular frequency $\omega_{z}$ corresponding to the perturbation stability of the particle in the $\hat{z}$ direction. | Analyzing the electric field at a small displacement $z$ from the equilibrium position along the $z$ axis
we obtain
$$
\vec{E_{z}}=\frac{Q}{4\pi\varepsilon_{0}}\left(\frac{l+z}{[R^{2}+(l+z)^{2}]^{\frac{3}{2}}}-\frac{l-z}{[R^{2}+(l-z)^{2}]^{\frac{3}{2}}}\right)
$$
$$
\vec{E_{z}}=\frac{Q}{2\pi\varepsilon_{0}}\frac{\left(R^{2}-2l^{2}\right)}{\left(R^{2}+l^{2}\right)^{\frac{5}{2}}}\vec{z}
$$
The equation of motion is
$$
m{\ddot{z}}=-{\frac{Q q}{2\pi\varepsilon_{0}}}{\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}z
$$
For stability, the condition of the simple harmonic oscillator equation must be satisfied
$$
R^{2}>2l^{2}
$$
Solving gives
$$
\omega_{z}=\sqrt{\frac{Q q}{2\pi\varepsilon_{0}m}\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}
$$ | $$
\sqrt{\frac{Q q}{2 \pi \varepsilon_0 m} \frac{R^2 - 2l^2}{(R^2+l^2)^{5/2}}}
$$ |
478 | ELECTRICITY | In an infinitely large, isotropic, linear dielectric medium, there exists a uniform external electric field $\vec{E}_{0}$. The vacuum permittivity is given as $\varepsilon_{0}$.
The dielectric medium is a liquid dielectric with a relative permittivity of $\varepsilon_{r}$. A solid, ideal conducting sphere with a radius of $R$ is placed inside the dielectric medium. The net charge carried by the conducting sphere is $Q$. It is assumed that the conducting sphere and the dielectric medium are in close contact, such that when considering forces, the free charges on the surface of the conducting sphere and the polarization charges at the interface of the dielectric medium must be regarded as a whole.
Determine the magnitude of the electrostatic force $F$ acting on the conducting sphere. | Spatial Electric Field Distribution
$$
\vec{E}=0,~0<r<R
$$
$$
\vec{E}=\vec{E}_{0}+R^{3}\frac{3\hat{n}(\hat{n}\cdot \vec{E}_{0})-\vec{E}_{0}}{r^{3}}+\frac{Q \hat{n}}{4\pi\varepsilon_{0}\varepsilon_{r}r^{2}},~r>R
$$
Total Surface Charge Density on the Conductor Sphere (including free charge and polarization charge):
$$
\sigma(\theta)=\frac{Q}{4\pi\varepsilon_{r} R^2}+3\varepsilon_{0}E_{0}\cos\theta
$$
Utilizing the technique where the electric field experienced by the surface charge is the average of the electric fields on both sides, the total electrostatic force obtained by integrating over the sphere's surface is:
$$
F=\frac{1}{\varepsilon_{r}}QE_{0}
$$ | $$
F=\frac{1}{\varepsilon_{r}}QE_{0}
$$ |
71 | ELECTRICITY | Given a particle with charge $q$ and mass $m$ moving in an electric field $\pmb{E}=E_{x}\pmb{x}+E_{z}\pmb{z}$ and a magnetic field $\pmb{B}=B\pmb{z}$. The initial conditions are: position $(x_{0},y_{0},z_{0})$ and velocity $(v_{\perp}\cos\delta,v_{\perp}\sin\delta,v_{z})$.
We know that a particle in a uniform magnetic field undergoes circular Larmor gyration, and the center of this rotation is called the guiding center. The drift velocity of this guiding center due to the electric field can be expressed using $\pmb{E},\pmb{B}$ and their magnitudes.
Problem: We discuss the case where the magnetic field is uniform, but the electric field is non-uniform. For simplicity, we assume $\pmb{E}$ is in the $\pmb{x}$ direction and varies sinusoidally in the $\pmb{y}$ direction.
$$
\pmb{E} \equiv E_{0}\cos(k y)\hat{\pmb{x}}
$$
In reality, such a charge distribution can occur in a plasma during wave propagation. Find the corresponding guiding center drift velocity. It is known that the electric field is very weak, and the particle's initial position is $(x_{0},y_{0},z_{0})$. Approximate to the lowest order that can distinguish from the uniform electric field case. | This problem has been modified; the original problem had four questions.
If there is an electric field present, we find that the motion of the particle will be a combination of two movements: the normal circular Larmor gyration and a drift towards the center of guidance. We can choose the $\pmb{x}$ axis along the direction of $\pmb{E}$, so $\pmb{E}_{y}=0$, and the velocity components related to transverse components can be treated separately. The equation of motion is
$$
m{\frac{d v}{d t}}=q({\pmb{E}}+v\times{\pmb{B}})
$$
Its $\pmb{z}$ component is
$$
\frac{d v_{z}}{d t}=\frac{q}{m}E_{z}
$$
Integrating yields
$$
v_{\tau}=\frac{q E_{z}}{m}t+v_{z0}
$$
This is a simple motion along the direction of $\pmb{B}$. The transverse components of the previous equation are
$$
\frac{d\upsilon_{x}}{d t}=\frac{q}{m}E_{x}\pm\omega_{c}\upsilon_{y}
$$
$$
\frac{d\upsilon_{y}}{d t}=0\pm\omega_{c}\upsilon_{x}
$$
where $\omega_{c}\equiv{\frac{|g|B}{m}}$ is the Larmor gyration angular frequency. Differentiating these two equations gives
$$
\ddot{v}_{x}=-\omega_{c}^{2}v_{x}
$$
$$
\ddot{v}_{y}=-\omega_{c}^{2}(v_{y}+\frac{E_{x}}{B})
$$
We can rewrite the above equations as
$$
\frac{d^{2}}{d t^{2}}(v_{y}+\frac{E_{x}}{B})=-\omega_{c}^{2}(v_{y}+\frac{E_{x}}{B})
$$
Therefore, if we use $\begin{array}{r}{v_{y}+\frac{E_{x}}{B}}\end{array}$ instead of ${\pmb v_{\pmb y}}$, the equation simplifies to the situation when the electric field is zero. Thus, the two equations can be replaced by
$$
v_{x}=v_{\perp}e^{i(\omega_{\mathrm{e}}t+\delta)}
$$
$$
\upsilon_{y}=-i\upsilon_{\perp}e^{i(\omega_{c}t+\delta)}-\frac{E_{x}}{B}
$$
where $\delta$ is the angle between the component of the particle's initial velocity in the $_{xy}$ plane ${\pmb v}_{\bot}$ and the $\pmb{x}$ axis (counter-clockwise direction). Further integration and taking the real part yields the equations of motion
$$
x=x_{0}+\frac{v_{\perp}}{\omega_{c}}\sin(\omega_{c}t+\delta)
$$
$$
y=y_{0}-\frac{E_{x}}{B}t+\frac{v_{\perp}}{\omega_{c}}[1-\cos(\omega_{c}t+\delta)]
$$
$$
z=\frac{q E_{z}}{2m}t^{2}+v_{z0}t+z_{0}
$$
From the equations of motion, it is observed that the Larmor motion of the particle is identical to the case without an electric field, but there is a drift superimposed in the $-y$ direction towards the center of guidance ${\pmb v}_{{\pmb g}{\pmb c}}$ (for $E_{z}>0)$.
To obtain the general formula for ${\pmb v}_{{\pmb g}{\pmb c}}$, we can solve the equation (1) using vector form. Since we already know that the term $\textstyle m{\frac{d v}{d t}}$ only gives rise to circular motion with frequency $\omega_{c}$, this term can be ignored in equation (1). Thus, equation (1) becomes
$$
\pmb{{ E}}+\pmb{{\upsilon}}\times\pmb{{ B}}=0
$$
By taking the cross product of $\pmb{B}$ with the above expression, we get
$$
\pmb{E}\times\pmb{B}=\pmb{B}\times(\pmb{v}\times\pmb{B})=\pmb{v}\pmb{B}^{2}-\pmb{B}(\pmb{v}\cdot\pmb{B})
$$
The transverse component of this equation is
$$
v_{\perp q c}=E\times B/B^{2}\equiv v_{E}
$$
We define this transverse component as $\pmb{v}_{E}$, which is the electric field drift towards the center of guidance.
The equation of motion for the particle is
$$
m{\frac{d{\boldsymbol{v}}}{d t}}=q[{\boldsymbol{E}}({\boldsymbol{y}})+{\boldsymbol{v}}\times{\boldsymbol{B}}]
$$
Decomposing into scalar form, we have
$$
\dot{v}_{x}=\frac{q B}{m}v_{y}+\frac{q}{m}E_{x}(y)
$$
$$
\dot{v}_{y}=-\frac{q B}{m}v_{x}
$$
Simplifying yields
$$
\ddot{\upsilon}_{x}=-\omega_{c}^{2}\upsilon_{x}\pm\omega_{c}\frac{\dot{E}_{x}(y)}{B}
$$
$$
\ddot{v}_{y}=-\omega_{c}^{2}v_{y}-\omega_{c}^{2}\frac{E_{x}(y)}{B}
$$
Here, $\pmb{E_{x}}(\pmb{y})$ is the electric field at the particle's location. To compute this value, we need to know the particle's trajectory, which is precisely what we attempt to solve initially. If the electric field is weak, as a first approximation, we can estimate $E_{x}(y)$ using the undisturbed trajectory.
The trajectory in the absence of an electric field is given by
$$
y=y_{0}\pm r_{L}\cos\omega_{c}t
$$
where $\begin{array}{r}{r_{L}=\frac{m v_{\perp}}{|q|B}}\end{array}$ is the Larmor gyration radius, and we obtain
$$
\ddot{v}_{y}=-\omega_{c}^{2}v_{y}-\omega_{c}^{2}\frac{E_{0}}{B}\cos k(y_{0}\pm r_{L}\cos\omega_{c}t)
$$
We seek a solution that is the sum of the gyration at $\omega_{c}$ and a stationary drift $_{v_{E}}$. Since we are interested in expressing $v_{E}$, we can average over one cycle to eliminate the gyration motion. Thus, the equation yields
$$
\bar{v}_{x}=0
$$
In the above equation, the average of the oscillating term $\ddot{v}_{y}$ is clearly zero, giving us
$$
\overline{{\ddot{v}}}_{y}=0=-\omega_{c}^{2}\overline{{v}}_{y}-\omega_{c}^{2}\frac{E_{0}}{B}\overline{{\cos k(y_{0}\pm r_{L}\cos\omega_{c}t)}}
$$
Upon expanding the small approximation and averaging, we find
$$
\overline{{v}}_{y}=-\frac{E_{x}(y_{0})}{B}(1-\frac{1}{4}k^{2}r_{L}^{2})
$$
Thus, due to non-uniformity, the usual $\pmb{{E}}\times\pmb{{B}}$ drift is modified to
$$
\boxed{v_{E}={\frac{E\times B}{B^{2}}}(1-{\frac{1}{4}}{(k \frac{m v_{\perp}}{\vert{q}\vert B})}^2)}
$$
The correction term represents the finite Larmor radius effect under the sinusoidal electric field distribution. | $$
v_E = \frac{E \times B}{B^2} \left(1 - \frac{1}{4} \left(k \frac{m v_\perp}{|q| B}\right)^2 \right)
$$ |
259 | MODERN | In the inertial frame \(S\), at time \(t = 0\), four particles simultaneously start from the origin and move in the directions of \(+x, -x, +y, -y\), respectively, with velocity \(v\). Consider another inertial frame \(S'\), which moves relative to \(S\) along the positive x-axis with velocity \(u\). At the initial moment, the two reference frames satisfy \(t = t' = 0\), where \(t'\) represents the time in the \(S'\) reference frame, and at the initial moment, the origins of the two reference frames coincide. Derive the relationship between the area of the quadrilateral formed by connecting the four particles in reference frame \(S'\) and time \(t'\). Consider the effects of special relativity, with the speed of light given as \(c\). | In the inertial frame \( S \), four particles start from the origin at \( t = 0 \), moving with velocity \( v \) in the \( +x, -x, +y, -y \) directions, respectively. Another inertial frame \( S' \) moves relative to \( S \) in the positive \( x \)-direction with velocity \( u \). At the initial moment, the origins of the two reference frames coincide, and their clocks are synchronized. We need to find the relationship between the area of the quadrilateral formed by connecting the four particles' positions and time \( t' \) in frame \( S' \), taking into account relativistic effects as per special relativity.
---
1. **Lorentz Transformation**:
- The coordinate transformation equations:
\[
x' = \gamma \left( x - ut \right), \quad t' = \gamma \left( t - \frac{ux}{c^2} \right)
\]
where \( \gamma = \frac{1}{\sqrt{1 - u^2/c^2}} \).
---
2. **Coordinate Transformation of Particles**:
- **Particle 1 (in the \( +x \) direction)**:
\[
x_1' = \gamma \frac{(v - u)t}{1 - uv/c^2}, \quad y_1' = 0
\]
Time \( t_1' = \gamma t \left(1 - \frac{uv}{c^2}\right) \)
- **Particle 2 (in the \( -x \) direction)**:
\[
x_2' = \gamma \frac{-(v + u)t}{1 + uv/c^2}, \quad y_2' = 0
\]
Time \( t_2' = \gamma t \left(1 + \frac{uv}{c^2}\right) \)
- **Particle 3 (in the \( +y \) direction)**:
\[
x_3' = -\gamma ut, \quad y_3' = \frac{vt}{\gamma}
\]
Time \( t_3' = \gamma t \)
- **Particle 4 (in the \( -y \) direction)**:
\[
x_4' = -\gamma ut, \quad y_4' = -\frac{vt}{\gamma}
\]
Time \( t_4' = \gamma t \)
---
3. **Coordinates at the Same Time \( t' \)**:
- Coordinates of Particle 1 at time \( t' \) in frame \( S' \):
\[
x_1' = \frac{(v - u)t'}{1 - uv/c^2}, \quad y_1' = 0
\]
- Coordinates of Particle 2 at time \( t' \) in frame \( S' \):
\[
x_2' = \frac{-(v + u)t'}{1 + uv/c^2}, \quad y_2' = 0
\]
- Coordinates of Particles 3 and 4 at time \( t' \) in frame \( S' \):
\[
x_3' = -ut', \quad y_3' = \frac{vt'}{\gamma}
\]
\[
x_4' = -ut', \quad y_4' = -\frac{vt'}{\gamma}
\]
---
4. **Calculation of the Area**:
- Coordinates of the four vertices of the quadrilateral:
- \( A \left( \frac{(v - u)t'}{1 - uv/c^2}, 0 \right) \)
- \( B \left( -ut', \frac{vt'}{\gamma} \right) \)
- \( C \left( -ut', -\frac{vt'}{\gamma} \right) \)
- \( D \left( \frac{-(v + u)t'}{1 + uv/c^2}, 0 \right) \)
- Length of the base (distance from \( A \) to \( D \)):
\[
\text{Base} = \left| \frac{(v - u)t'}{1 - uv/c^2} - \frac{-(v + u)t'}{1 + uv/c^2} \right| = 2vt' \frac{(1 - u^2/c^2)}{1 - (uv/c)^2}
\]
- Height (distance from \( B \) to the \( x \)-axis):
\[
\text{Height} = \frac{vt'}{\gamma}
\]
- Area formula:
\[
\text{Area} = \text{Base} \times \text{Height} = 2vt' \frac{(1 - u^2/c^2)}{1 - (uv/c)^2} \times \frac{vt'}{\gamma}
\]
Substituting \( \gamma = \frac{1}{\sqrt{1 - u^2/c^2}} \) and simplifying:
\[
\text{Area} = \frac{2v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}
\] | $$\frac{2 v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}$$ |
766 | MECHANICS | A small ring $A$ with mass $m$ is placed on a smooth horizontal fixed rod and connected to a small ball $B$ with mass $m$ by a thin string of length $l$. Initially, the string is pulled to a horizontal position, and then the system is released from rest. Find: When the angle between the string and the horizontal rod is $\theta$, what is the tension in the string?
| γSolutionγLet the angle between the rope and the rod be $\theta$, the velocity of the small ring A be $\boldsymbol{v}_{A}$, and the velocity of the small ball B relative to the small ring be $v'$. Then the velocity of the small ball B relative to the bottom surface $\boldsymbol{v}_{B}$ is given by the relative motion formula, that is $$ \boldsymbol{v}_{B} = \boldsymbol{v'} + \boldsymbol{v}_{A} \tag{1} $$ $$ \upsilon_{Br} = \upsilon_{A} - \upsilon^{\prime} \sin\theta \tag{2} $$ $$ \upsilon_{B} = \sqrt{(\upsilon_{A} - \upsilon^{\prime} \sin\theta)^2 + (\upsilon^{\prime} \cos\theta)^2} \tag{3} $$ By horizontal momentum conservation and mechanical energy conservation for the system, we have $$ m \upsilon_{A} + m \upsilon_{Br} = 0 \tag{4} $$ $$ \frac{1}{2} m \upsilon_{A}^2 + \frac{1}{2} m \upsilon_{B}^2 - m g l \sin\theta = 0 \tag{5} $$ Where the zero point of gravitational potential energy is taken at the release point of ball B. Substitute equations (2) and (3) into equations (4) and (5) respectively to obtain $$ m\upsilon_{A}+ m (\upsilon_{A} - \upsilon^{\prime} \sin\theta) = 0 \tag{6} $$ $$ \frac{1}{2} m \upsilon_{A}^2 + \frac{1}{2} m \left[(\upsilon_{A} - \upsilon^{\prime} \sin\theta)^2 + (\upsilon^{\prime} \cos\theta)^2\right] - m g l \sin\theta = 0 \tag{7} $$ Jointly solving equations (6) and (7), we obtain $$ \upsilon^{\prime 2} = \frac{4g l \sin\theta}{1 + \cos^2\theta} \tag{8} $$ $$ \upsilon_{A} = \frac{1}{2} \sin\theta \sqrt{\frac{4g l \sin\theta}{1 + \cos^2\theta}} \tag{9} $$ Let the angle between $ u_{B}$ and $ u^{\prime}$ be $a$. According to the geometric relationship shown in the figure, using the sine theorem, we have $$ \frac{\upsilon_{A}}{\sin a} = \frac{\upsilon^{\prime}}{\sin\left(\frac{\pi}{2} + \theta - a\right)} \tag{10} $$ Substituting equations (8) and (9) into equation (10), we can obtain $$ \frac{\upsilon^{\prime}}{\upsilon_{A}} = \frac{\cos(\theta - a)}{\sin a} = \frac{\cos\theta \cos a + \sin\theta \sin a}{\sin a} = \frac{2}{\sin\theta} \tag{11} $$ Therefore, we can solve to get $$ \cot a = \frac{1 + \cos^2\theta}{\sin\theta \cos\theta} $$ That is $$ \left\{ \begin{array}{l} \sin a = \frac{\sin\theta \cos\theta}{\sqrt{1 + 3\cos^2\theta}} \\ \cos a = \frac{1 + \cos^2\theta}{\sqrt{1 + 3\cos^2\theta}} \end{array} \right. \tag{12} $$ Let the tension in the rope at this time be $T$ and the acceleration of the small ring A be $a$, then we have $ T \cos\theta = m a \tag{13} $ Taking the small ring A as the reference frame, in this non-inertial frame, the forces acting on the small ball B are shown in the figure. In the figure $f_i = m a$, from the figure, we obtain the normal equation for the circular motion of the small ball B as $$ T + m a \cos\theta - m g \sin\theta = m \frac{\upsilon^{\prime 2}}{l} \tag{14} $$ Substituting equation (7) into equation (13), we obtain $$ T = \frac{5 + \cos^2\theta}{1 + \cos^2\theta} m g \sin\theta - m a \cos\theta \tag{15} $$ Jointly solving equations (12) and (14), we obtain $$ T = \frac{5 + \cos^2\theta}{(1 + \cos^2\theta)^2} m g \sin\theta \tag{16} $$
| $$
T = \frac{5 + \cos^2 \theta}{(1 + \cos^2 \theta)^2} m g \sin \theta
$$ |
716 | MECHANICS | The principle of a rotational speed measurement and control device is as follows. At point O, there is a positive charge with an electric quantity of Q. A lightweight, smooth-walled insulating thin tube can rotate around a vertical axis through point O in the horizontal plane. At a distance L from point O inside the tube, there is a photoelectric trigger control switch A. A lightweight insulating spring with a free length of L/4 is fixed at the O end, and the other end of the spring is connected to a small ball with mass m and positive charge q. Initially, the system is in static equilibrium. The thin tube rotates about a fixed axis under the action of an external torque, allowing the small ball to move within the thin tube. When the rotational speed $\omega$ of the thin tube gradually increases, the small ball reaches point A in the thin tube and just achieves radial equilibrium relative to the thin tube, triggering the control switch. The external torque instantaneously becomes zero, thus limiting excessive rotational speed; at the same time, the charge at point O becomes an equal amount of negative charge -Q. By measuring the position B of the radial equilibrium point of the small ball relative to the thin tube thereafter, the rotational speed can be determined. If the distance OB is measured to be $L/2$, determine the rotational speed $\omega$ of the thin tube when the ball is at point B. Express the result using the following physical quantities: Electric charge $Q$, ball's electric charge $q$, mass $m$, length $L$, and Coulomb's constant $k$.
| Let the angular velocity of the thin tube be $\omega_A$. When the small ball is in equilibrium at point A relative to the thin tube, we have: $$ k_0 \cdot \frac{3}{4}L - \frac{k q Q}{L^2} = m L \omega_A^2 \tag{1} $$ When the small ball is in equilibrium at point B ($OB = L/2$), with angular velocity $\omega_B$, we have: $$ k_0 \cdot \frac{1}{4}L + \frac{k q Q}{L^2 / 4} = \frac{L}{2} m \omega_B^2 \tag{2} $$ Conservation of angular momentum gives: $$ m L^2 \omega_A = m \cdot \frac{L^2}{4} \omega_B \tag{3} $$ From this, we obtain: $$ \omega_B = 4 \omega_A $$ Substituting the above into equation (2): $$ k_0 \cdot \frac{1}{4}L + \frac{k q Q}{L^2 / 4} = 8 L m \omega_A^2 \tag{4} $$ Simultaneously solving equations (1) and (4), we find the spring constant: $$ k_0 = \frac{48 k q Q}{23 L^3} \tag{5} $$ Substituting equation (5) into equation (2), we can find the angular velocity of the thin tube at this time: $$ \omega_B = 4 \sqrt{ \frac{13 k q Q}{23 m L^3} } \tag{6} $$
| $$\omega_B = 4 \sqrt{\frac{13 k q Q}{23 m L^3}}$$ |
103 | THERMODYNAMICS | Solving physics problems involves many techniques and methods: analogy, equivalence, diagrams, and so on. A smart person like you can definitely use these techniques and methods to solve the following problem:
In space, there is an infinite series of nodes, numbered in order as $\cdots -3, -2, -1, 0, 1, 2, 3, \cdots.$ From each node, there are three thermal resistances connected. The connection method is: for node ${\pmb n}$, there are thermal resistances $R$ to nodes ${\pmb n} \pm 1$ respectively. Additionally, if $m$ is even, ${n}$ is odd, and $n = m + 3$, there is a thermal resistance $R$ between node $m$ and node $n$. Find the equivalent thermal resistance $R_{04}$ between node 0 and node 4 when all other nodes remain adiabatic. | Noticing that it can be arranged in a zigzag pattern, the diagram is shown above:
First, consider $R_{02}$, which is relatively simple. Due to symmetry, the two nodes divide the network into two parts, each side being equivalent to $\scriptstyle{R^{\prime}}$. The self-similarity of $\scriptstyle{R^{\prime}}$ provides the equivalence shown on the right of the diagram above. Therefore, the 35 resistors are also $\scriptstyle{R^{\prime}}$, leading to:
$$
R^{\prime}=R+R//(R+R^{\prime})
$$
Discarding the negative root gives:
$$
\smash{R^{\prime}=\sqrt{3}R}
$$
Thus:
$$
R_{02}=\frac{\sqrt{3}}{2}R
$$
By treating $\scriptstyle{R^{\prime}}$ as equivalent to the series connection of $\scriptstyle{R^{\prime}-R}$ and $\scriptstyle{R}$, the equivalent network between 0123 is shown in the lower-left diagram above. Consequently:
$$
R_{01}=(R^{\prime}-R)//(R^{\prime}-R+R+R)
$$
$$
R_{03}=R//(R^{\prime}-R+R^{\prime}-R+R)
$$
The calculations yield:
$$
R_{01}=\frac{\sqrt{3}}{3}R
$$
$$
R_{03}=\left(1-\frac{\sqrt{3}}{6}\right)R
$$
Finally, the equivalent circuit diagram for $R_{04}$ is shown in the lower-right diagram above, which is a balanced bridge. Thus:
$$
R_{04}=2R//2R^{\prime}
$$
Resulting in:
$$
R_{04}=\left(3-\sqrt{3}\right)R
$$ | $$
R_{04} = (3 - \sqrt{3})R
$$ |
310 | MODERN | Consider an ideal mirror moving at relativistic velocity, with mass $m$ and area $S_{\circ}$. (The direction of photon incidence is the same as the direction of the mirror's motion.)
Now consider the case where the mirror is moving with an initial velocity $\beta_{0}c$. In this situation, the mirror is unconstrained by external forces, and photons are incident on it with constant power for a certain period of time, with energy $E$. Assuming the mirror's velocity after irradiation is $\beta_{1}\mathfrak{c}$, find the expression for $\beta_{1}$. | List the conservation of energy and momentum:
$$
E+{\frac{m c^{2}}{\sqrt{1-{\beta_{0}}^{2}}}}=E^{\prime}+{\frac{m c^{2}}{\sqrt{1-{\beta_{1}}^{2}}}}
$$
$$
\frac{E}{c}+\frac{m c\beta_{0}}{\sqrt{1-\beta_{0}{}^{2}}}=\frac{m c\beta_{1}}{\sqrt{1-\beta_{1}{}^{2}}}-\frac{E^{\prime}}{c}
$$
Solving, we get:
$$
\beta_{1}=\frac{\left(\sqrt{\displaystyle\frac{1+\beta_{0}}{1-\beta_{0}}}+\frac{2E}{m c^{2}}\right)^{2}-1}{\left(\sqrt{\displaystyle\frac{1+\beta_{0}}{1-\beta_{0}}}+\frac{2E}{m c^{2}}\right)^{2}+1}
$$ | $$\frac{\left(\sqrt{\frac{1+\beta_0}{1-\beta_0}}+\frac{2E}{mc^2}\right)^2 - 1}{\left(\sqrt{\frac{1+\beta_0}{1-\beta_0}}+\frac{2E}{mc^2}\right)^2 + 1}$$ |
110 | MECHANICS | A homogeneous picture frame with a light string, string length $2a$, frame mass $m$, length $2c$, and width $2d$, is hanging on a nail. Ignoring friction, with gravitational acceleration $g$. The mass of the light string is negligible, and it is inextensible, with its ends connected to the two vertices of one long side of the picture frame. Objects other than the picture frame, nail, and light string are not considered. Find the angle $\alpha$ between the long side of the picture frame and the horizontal line when in equilibrium, where $\alpha \in (0, \frac{\pi}{2}], c^4 > a^2d^2 - c^2d^2$. | As shown in the figure, the trajectory of the nail is an ellipse $\begin{array}{r}{\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{b^{2}}=1}\end{array}$ where $b={\sqrt{a^{2}-c^{2}}}$. It is easy to know from geometric relationships that the angle between the line connecting the nail and the center of mass of the picture frame and the $y$-axis is $\alpha$.
Write the parametric equations of the ellipse:
$$
\begin{array}{r}{x=a\sin\theta}\ {}\ {y=b\cos\theta}\end{array}
$$
Obtain the distance between the center of mass of the picture frame and the nail, and $\pmb{\alpha}$:
$$
h={\sqrt{x^{2}+(y+d)^{2}}}={\sqrt{(a\sin\theta)^{2}+(b\cos\theta+d)^{2}}}
$$
$$
\alpha=\arctan({\frac{x}{y+d}})=\arctan({\frac{a\sin\theta}{b\cos\theta+d}})
$$
By the equilibrium of moments, the line connecting the center of mass of the picture frame and the nail is vertical, so the potential energy is:
$$
E_{p}=-m g h=-m g\sqrt{(a\sin\theta)^{2}+(b\cos\theta+d)^{2}}
$$
Condition for extremum of potential energy:
$$
{\frac{d E_{p}}{d\theta}}=-m g{\frac{c^{2}\sin\theta\cos\theta-b d\sin\theta}{\sqrt{(a\sin\theta)^{2}+(b\cos\theta+d)^{2}}}}=0
$$
Solve to obtain:
$$
\theta_{2}=\operatorname{arccos}(\frac{b d}{c^{2}})
$$
The condition for the existence of ${\theta_{2}}$ is:
$$
c^{2}>{\sqrt{a^{2}-c^{2}}}d
$$
Substitute into the previous equation to obtain:
$$
{{\alpha=\arctan({\frac{\sqrt{c^{4}-d^{2}(a^{2}-c^{2})}}{a d}})}}
$$ | $$
\alpha = \arctan\left(\frac{\sqrt{c^4 - d^2(a^2 - c^2)}}{ad}\right)
$$ |
256 | MODERN | Two relativistic particles X, each with rest mass $M$, experience a short-range attractive force $F(r) = \alpha/r^2$ (where $\alpha$ is a positive constant) in the zero momentum reference frame C, and are bound by this short-range attractive force to form a pair $\mathrm{X_{2}}$. The speed of light in a vacuum is $c$, and the reduced Planck constant is $\hbar = h / (2\pi)$.
If a bound pair is stable in the ground state $n=1$, the maximum value of $\alpha$, $\alpha_{c}$, can be calculated.
To bombard $\mathrm{X_{2}}$ and separate $\mathrm{X}$, neglecting the rest mass and utilizing a particle with ultra-high speed and energy $E$, the experimental setup involves bombarding $\mathrm{X_{2}}$ at rest in the laboratory frame $\mathrm{L}$. Set $\alpha = \alpha_{c}/2$. If the ultra-high-speed particle causes the transition of $\mathrm{X_{2}}$ and emit a photon, find the minimum value $E_{1}$ of $E$. | In the center-of-mass system, the momentum of the two particles X is the same, denoted as $p$, and at this time, there exists the angular momentum quantization condition:
$$
p\times{\frac{r}{2}}\times2=p r=n\hbar,n\in\mathbb{N}
$$
According to the dynamics equation, the angular velocity $\omega=2v/r$. Based on Newton's second law:
$$
p\omega={\frac{\alpha}{r^{2}}}\rightarrow\alpha=p\times{\frac{2v}{r}}\times r^{2}=2p r v=2n\hbar v
$$
Note that the attractive potential energy is $V(r)=-\alpha/r$. Introducing the kinetic energy $T$, and noting that the dynamic mass is $T/c^{2}$, it is straightforward to write the system's energy as:
$$
E=-{\frac{\alpha}{r}}+2T=-{\frac{\alpha p}{n\hbar}}+2T=-{\frac{\alpha T v}{n\hbar c^{2}}}+2T=2T\left(1-\left({\frac{\alpha}{2n\hbar c}}\right)^{2}\right)
$$
By calculating the system energy, we obtain:
$$
T={\frac{M c^{2}}{\sqrt{1-\left({\frac{\alpha}{2n\hbar c}}\right)^{2}}}}\to E_{n}=2M c^{2}{\sqrt{1-\left({\frac{\alpha}{2n\hbar c}}\right)^{2}}}
$$
Since the square root must be meaningful, we establish:
$$
\left({\frac{\alpha}{2\hbar c}}\right)^{2}\leq1\rightarrow\alpha_{c}=2\hbar c
$$
Writing out the binding energy for the ground states with $n=1,2$:
$$
\begin{array}{r}{E_{n=1}=2M c^{2}\sqrt{1-\left(\cfrac{\hbar c}{2\hbar c}\right)^{2}}=\sqrt{3}M c^{2}}\ {E_{n=2}=2M c^{2}\sqrt{1-\left(\cfrac{\hbar c}{4\hbar c}\right)^{2}}=\cfrac{\sqrt{15}}{2}M c^{2}}\end{array}
$$
The invariant modulus squared of the initial state is:
$$
\left(\sum E\right)^{2}-\left(\sum P c\right)^{2}=\left(E+{\sqrt{3}}M c^{2}\right)^{2}-E^{2}=3\left(M c^{2}\right)^{2}+2{\sqrt{3}}M c^{2}E
$$
If the system successfully ensures a transition, then in the critical state, the system's total energy in the zero-momentum frame must be no less than the binding energy of the transition to $n=2$:
$$
3\left(M c^{2}\right)^{2}+2{\sqrt{3}}M c^{2}E\geq\left({\sqrt{15}}M c^{2}\right)^{2}/4
$$
Solving this gives:
$$
E_{1}={\frac{15/4-3}{2{\sqrt{3}}}}M c^{2}={\frac{\sqrt{3}}{8}}M c^{2}
$$ | $$
E_1 = \frac{\sqrt{3}}{8}Mc^2
$$ |
112 | ELECTRICITY | Initially, a conductive dielectric sphere with a free charge of 0 is placed in a vacuum. It is known that the radius of the conducting sphere is $R$, its relative permittivity is $\varepsilon_{r}$, and its conductivity is $\sigma$. At the moment $t=0$, a uniform external field ${{\vec{E}}_{0}}$ is applied around the conducting sphere, and free charge begins to accumulate on the surface of the conductor. The permittivity of the vacuum is $\varepsilon_0$. Try to determine the total Joule heat generated by the system from the initial state to the steady state. | Considering the moment when $\ell\to0^{+}$, the polarization properties of the dielectric are prioritized over the conductive properties. At this time, the potential distribution is equivalent to the polarization of a dielectric sphere with a relative dielectric constant of $\varepsilon_{\mathsf{r}}$ in a uniform external field.
$$
\sigma_{i}=P\cos\theta
$$
Further,
$$
\bar{P}=(\varepsilon_{\mathrm{r}}-1)\varepsilon_{0}\left(E_{0}-\frac{P}{3\varepsilon_{0}}\right)
$$
Solving,
$$
P=3\varepsilon_{0}\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}E_{0}
$$
Effective electric dipole outside the sphere,
$$
\mathscr{P}=\frac{4}{3}\pi\mathscr{R}^{3}P=\frac{4\pi\varepsilon_{0}(\varepsilon_{r}-1)}{\varepsilon_{r}+2}E_{0}R^{3}
$$
Solving,
$$
U_{i n.}(t\rightarrow0^{+})=-{\frac{3}{\varepsilon_{r}+2}}E_{0}r\cos\theta
$$
$$
\zeta_{c x}(t\rightarrow0^{+})=-E_{0}r\cos\theta+\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}E_{0}\frac{R^{3}}{r^{2}}\cos\theta
$$
Considering the moment when it reaches a steady state, the conductive dielectric sphere fully exhibits the properties of an ideal conductor $(\varepsilon_{r}\to+\infty)$, solving,
$$
\bar{U}_{i n}(t;\to\infty)=0
$$
$$
U_{\mathrm{cx}}(t\rightarrow\infty)=-E_{0}\tau\cos\theta+E_{0}\frac{R^{3}}{r^{2}}\cos\theta
$$
Considering the moment $t$, when the interior of the conductive dielectric sphere is free from charge, its free charges should be distributed sinusoidally on the surface, assuming the spatial potential distribution satisfies,
$$
U=\left\{\begin{array}{l l}{-E_{0}r\cos\theta+A(t)\frac{\cos\theta}{r^{2}},}&{\mathrm{if~}R<r} \\ {-E_{0}r\cos\theta+B(t)r\cos\theta,}&{\mathrm{if~}0<r<R}\end{array}\right.
$$
Considering the continuity of potential on $\mathcal{R}$,
$$
-E_{0}R\cos{\theta}+A(t)\frac{\cos{\theta}}{R^{2}}=-E_{0}R\cos{\theta}+B(t)R\cos{\theta}
$$
Simplifying to get,
$$
B(t)=A(t)\frac{1}{R^{3}}
$$
Further considering the distribution of space electric field along the $\hat{r}$ direction,
$$
E_{\mathrm{r}}=-\frac{\partial U}{\partial r}=\left\{\begin{array}{l l}{E_{0}\cos\theta+A(t)\frac{2\cos\theta}{r^{3}},}&{\mathrm{if~}R<r} \\ {E_{0}\cos\theta-B(t)\cos\theta,}&{\mathrm{if~}0<r<R}\end{array}\right.
$$
Considering the charging process at the surface,
$$
j_{n}=\sigma\vec{E}_{i n}\cdot\hat{r}=\frac{\mathrm{d}\sigma_{f}}{\mathrm{d}t}
$$
Where,
$$
\boldsymbol{\sigma}_{f}=(\vec{D}_{e x}-\vec{D}_{i n})\cdot\boldsymbol{\hat{r}}=(\varepsilon_{0}\vec{E}_{c x}-\varepsilon_{0}\varepsilon_{r}\vec{E}_{i n})\cdot\boldsymbol{\hat{r}}
$$
Substituting into the previous equation,
$$
\sigma B(t)+\varepsilon_{0}\frac{2}{R^{3}}\frac{\mathrm{d}A(t)}{\mathrm{d}t}+\varepsilon_{0}\varepsilon_{r}\frac{\mathrm{d}B(t)}{\mathrm{d}t}=0
$$
Further considering the previous equation,
$$
\sigma\left(B(t)-E_{0}\right)+\varepsilon_{0}(2+\varepsilon_{r})\frac{\mathrm{d}B(t)}{\mathrm{d}t}=0
$$
Considering the initial condition,
$$
B(t\rightarrow0^{+})=\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}E_{0}
$$
Solving,
$$
B(t)=E_{0}-\frac{3}{\varepsilon_{r}+2}E_{0}e^{-\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}
$$
Further obtaining the spatial potential distribution,
$$
\begin{array}{r}{U=\left\{\begin{array}{l l}{-E_{0}r\cos\theta+\left(1-\frac{3}{\epsilon_{\mathrm{r}}+2}e^{-\frac{\epsilon}{\epsilon_{0}(2+\epsilon_{\mathrm{r}})}t}\right)E_{0}\frac{R^{3}}{r^{2}}\cos\theta,}&{\mathrm{if~}R<r} \\ {-\frac{3}{\epsilon_{\mathrm{r}}+2}e^{-\frac{\sigma}{\epsilon_{0}(2+\epsilon_{\mathrm{r}})}t}E_{0}r\cos\theta,}&{\mathrm{if~}0<r<R}\end{array}\right.}\end{array}
$$
Utilizing the previous equation, it is easy to obtain the radial component distribution of the electric displacement vector inside and outside the surface of the conductive dielectric sphere,
$$
D_{\mathsf{r}}=\left\{\begin{array}{l l}{3\varepsilon_{0}\left[1-\frac{2}{\epsilon_{r}+2}e^{-\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}\right]E_{0}\cos\theta,}&{\mathrm{if~}r\to R^{+}} \\ {-\frac{3\varepsilon_{0}\varepsilon_{\mathrm{r}}}{\varepsilon_{r}+2}e^{-\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}E_{0}\cos\theta,}&{\mathrm{if~}r\to R^{-}}\end{array}\right.
$$
Obtaining the surface free charge distribution,
$$
\sigma_{f}=(\overrightarrow{\vec{D}}_{e x}-\overrightarrow{\vec{D}}_{i n})\cdot\hat{r}=3\epsilon_{0}\left(1-e^{-\frac{\sigma}{\epsilon_{0}(2+\epsilon_{r})}t}\right)E_{0}\cos\theta
$$
Considering the Joule heat theorem, the Joule heat power density inside the conductive dielectric sphere is,
$$
P_{J}=\stackrel{\rightarrow}{\vec{j}}\cdot\vec{E}_{i n}^{\prime}=\frac{9\sigma}{(\varepsilon_{r}+2)^{2}}e^{-2\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}E_{0}^{2}
$$
Total Joule heat power in the entire space,
$$
W_{J}=\underbrace{P_{J}\mathrm{d}V}_{V_{\Theta\:\sharp\wedge\Theta}}=\frac{12\pi\sigma}{(\varepsilon_{r}+2)^{2}}e^{-2\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}R^{3}E_{0}^{2}
$$
Further obtaining the total Joule heat,
$$
Q_{t o t}=\int_{0}^{\infty}W_{J}\mathrm{d}t=\int_{0}^{\infty}\frac{12\pi\sigma}{(\varepsilon_{r}+2)^{2}}e^{-2\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}R^{3}E_{0}^{2}\mathrm{d}t=\frac{6\pi\varepsilon_{0}}{(\varepsilon_{r}+2)}R^{3}E_{0}^{2}
$$ | $$
Q_{tot} = \frac{6 \pi \varepsilon_0}{\varepsilon_r + 2} R^3 E_0^2
$$ |
505 | MECHANICS | It is often observed that when a puppy gets wet, it will vigorously shake its body, and with just a few quick shakes, it can rid itself of most of the water. In fact, the loose skin and longer fur of dogs provide a biological rationale for this behavior. However, this might drive you crazy when you are giving the dog a bath!
To address this, consider the following kinematics problem: treat the puppy as a sphere with radius $R$ in a uniform gravitational field with gravity $g$. The center of the sphere is suspended at a height $H>R$ above the ground. Establish a Cartesian coordinate system where the $z$-axis is vertical, the ground is at $z=0$, and the sphere's center is located at $(0,0,H)$. The sphere rotates uniformly with an angular velocity vector along the $x$ direction, and the magnitude is $\omega$. Consider a droplet of water being flung off the surface of the sphere, and then follows a trajectory under the influence of gravity.
If the angular velocity is too small, water droplets cannot be flung off the top, so there's a minimum angular velocity $\omega_{1}$.
When the angular velocity $\omega>\omega_{1}$, there is still a region on the sphere where droplets cannot be normally flung off. Find the equation of the curve projecting the boundary of this region onto the $xy$-plane (express this as an equation involving $y^2$ in terms of $x$). | Take the $x$-axis as the polar axis, and the $z$-axis as the reference axis for the azimuthal angle, i.e.:
$$
\begin{aligned}{x}&={R\cos{\theta}}\\ {y}&={R\sin{\theta}\sin{\varphi}}\\{z}&={H+R\sin{\theta}\cos{\varphi}}\end{aligned}
$$
Then the velocity of the circular motion at the point:
$$
v=\omega r=\omega R\sin\theta
$$
The radius of curvature of the parabolic path of the water droplet follows:
$$
g\cos\varphi={\frac{v^{2}}{\rho}}
$$
In order to eject the water drop, it needs to satisfy:
$$
\rho>r=R\sin\theta
$$
Setting the equality gives the boundary condition:
$$
\omega^{2}R\sin\theta=g\cos\varphi
$$
Expressing $\theta, \varphi$ in terms of $x, y$ and simplifying:
$$
\boxed{y^{2}=\frac{\omega^{4}}{g^{2}}(R^{2}-x^{2})\left(x^{2}-R^{2}+\frac{g^{2}}{\omega^{4}}\right)}
$$ | $$
\boxed{y^{2}=\frac{\omega^{4}}{g^{2}}(R^{2}-x^{2})\left(x^{2}-R^{2}+\frac{g^{2}}{\omega^{4}}\right)}
$$ |
331 | OPTICS | Building the experimental setup for diffraction with a steel ruler: Establish a spatial Cartesian coordinate system, with the positive $x$ direction pointing vertically downward and the positive $z$ direction pointing perpendicularly towards the wall. The angle between the steel ruler and the $z$-axis in the horizontal plane is $\phi$, and the angle between the steel ruler and the horizontal plane is $\xi$. A laser is directed along the positive $z$ direction onto the surface of the steel ruler, with the horizontal distance from the point of contact to the wall being $L$. The graduations on the steel ruler are evenly distributed with a spacing of $d$, oriented perpendicular to the direction of the ruler.
By considering the wave vector continuous along the direction of the scale markings, one can derive the equation of the curve where the diffraction pattern appears (general property); by considering the strong interference between the graduations, one can determine the locations of the primary maxima in the diffraction pattern (specific property).
For simplicity, assume the steel ruler is completely horizontal, setting $\xi=0$. Find: the spacing of the primary maxima near the zeroth-order maximum $\Delta y$. | Let the coordinates of the diffraction point be $(x,y)$, we can derive the incident wave vector and the reflected wave vector
$$
\left\{\begin{array}{c}{\displaystyle\boldsymbol{k}=\frac{2\pi}{\lambda}\hat{\boldsymbol{z}}}\ {\displaystyle\boldsymbol{k}^{\prime}=\frac{2\pi}{\lambda}\frac{x\hat{\boldsymbol{x}}+y\hat{\boldsymbol{y}}+L\hat{\boldsymbol{z}}}{\sqrt{x^{2}+y^{2}+L^{2}}}}\end{array}\right.
$$
The wave vector is continuous along the direction of the scale mark
$$
({\pmb k}^{\prime}-{\pmb k})\cdot{\widehat{\pmb x}}=0
$$
This yields
$$
x=0
$$
The interference between scales is extremely strong (assume the order is $n$)
$$
(k^{\prime}-k)\cdot d(\sin\phi\widehat{\pmb{y}}+\cos\phi\widehat{\pmb{z}})=2n\pi
$$
Solving, we obtain
$$
(\frac{n\lambda}{d}+\cos\phi)^{2}x^{2}+\left[(\frac{n\lambda}{d}+\cos\phi)^{2}-\sin\phi^{2}\right]y^{2}-2\sin\phi\cos\phi L y+\left[(\frac{n\lambda}{d})^{2}+2\frac{n\lambda}{d}\cos\phi\right]L^{2}=0
$$
The solution for $y$ is
$$
y_{n}=\frac{\sin\phi\cos\phi+\sqrt{\sin\phi^{2}\cos\phi^{2}-\left[(\frac{n\lambda}{d}+\cos\phi)^{2}-\sin\phi^{2}\right]\left[(\frac{n\lambda}{d})^{2}+2\frac{n\lambda}{d}\cos\phi\right]}}{(\frac{n\lambda}{d}+\cos\phi)^{2}-\sin\phi^{2}}L
$$
According to the problem statement
$$
\frac{n\lambda}{d}\ll1
$$
Then
$$
y_{n}\approx\tan2\phi L-\frac{1+\tan2\phi^{2}}{\sin\phi}\frac{n\lambda}{d}L
$$
The difference in $y$ is
$$
\Delta y=\frac{1+\tan2\phi^{2}}{\sin\phi}\frac{\lambda L}{d}
$$ | $$
\Delta y = \frac{1 + \tan^2(2\phi)}{\sin\phi} \frac{\lambda L}{d}
$$ |
498 | OPTICS | The interference phenomenon in variable refractive index systems is a new issue of concern in the field of optics in recent years. There is a thin film of non-dispersive variable refractive index medium with a constant thickness \(d\), where the refractive index within the film changes linearly with the distance from the film's surface. The refractive index at the lower surface is \(n_a\), and at the upper surface is \(n_b > n_a\). A beam of parallel light enters the film from the lower surface at an incident angle \(i\). Taking the incident point as the origin, the positive direction of the \(x\)-axis is along the lower surface of the film to the right, and the positive direction of the \(y\)-axis is perpendicular to the film surface upward.
If the incident light is monochromatic light with wavelength \(\lambda\), and the refractive index of air for this wavelength of light is \(n_0 = 1\), the distance between the incident points of two adjacent beams of incident light is \(2S\). The incident point of the first beam on the lower surface is denoted as Q, and the point from which it exits from the lower surface again after its first reflection from the upper surface is denoted as P, with OP = \(2S\). It is important to note that \(S\) is unknown and can be determined through other physical quantities; the final answer should not contain \(S\). Ignoring the phase change caused by reflection, find the phase difference \(\varphi\) between the two beams of light. | The phase difference is \(\varphi = \frac{2\pi}{\lambda} \Delta L\), where \(\Delta L = L - L_0\).
\(L\) is the total optical path from the incident point Q through reflection to the exit point P in the medium.
\(L_0 = 2S \sin i\) is the optical path difference in air for two adjacent incident beams, which can also be understood as: if the light travels a horizontal distance \(2S\) in air (with a refractive index of 1), it is the projection of the optical path along the direction parallel to the incident light.
This definition of \(\Delta L\) describes the additional optical path difference caused by light traveling in the medium relative to traveling the same equivalent distance in air, which is often used to describe the formation of interference patterns.
1. **Calculate the optical path \(L\) in the medium**:
We have already calculated:
\(L = 2 \int_0^d \frac{n(y)^2}{\sqrt{n(y)^2 - \sin^2 i}} dy\)
Substituting variables \(u = n(y) = n_a + ky\), \(dy = \frac{d}{n_b - n_a} du\).
\(L = 2 \int_{n_a}^{n_b} \frac{u^2}{\sqrt{u^2 - \sin^2 i}} \frac{d}{n_b - n_a} du = \frac{2d}{n_b - n_a} \int_{n_a}^{n_b} \frac{u^2}{\sqrt{u^2 - \sin^2 i}} du\)
Using the integration formula \(\int \frac{x^2 dx}{\sqrt{x^2 - A^2}} = \frac{x}{2} \sqrt{x^2 - A^2} + \frac{A^2}{2} \ln|x + \sqrt{x^2 - A^2}| + C\), where \(A = \sin i\).
\(L = \frac{2d}{n_b - n_a} \left[ \frac{u}{2} \sqrt{u^2 - \sin^2 i} + \frac{\sin^2 i}{2} \ln(u + \sqrt{u^2 - \sin^2 i}) \right]_{n_a}^{n_b}\)
\(L = \frac{d}{n_b - n_a} \left[ u \sqrt{u^2 - \sin^2 i} + \sin^2 i \ln(u + \sqrt{u^2 - \sin^2 i}) \right]_{n_a}^{n_b}\)
\(L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} + \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right]\)
2. **Calculate the horizontal displacement \(2S\)**:
We have already calculated:
\(2S = 2 \int_0^d \frac{\sin i}{\sqrt{n(y)^2 - \sin^2 i}} dy\)
\(2S = 2 \sin i \int_0^d \frac{dy}{\sqrt{n(y)^2 - \sin^2 i}}\)
\(2S = 2 \sin i \left[ \frac{d}{n_b - n_a} \int_{n_a}^{n_b} \frac{du}{\sqrt{u^2 - \sin^2 i}} \right]\)
\(2S = \frac{2d \sin i}{n_b - n_a} \left[ \ln(u + \sqrt{u^2 - \sin^2 i}) \right]_{n_a}^{n_b}\)
\(2S = \frac{2d \sin i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right)\)
3. **Calculate the optical path difference \(L_0\) in air**:
\(L_0 = 2S \sin i\)
\(L_0 = \left[ \frac{2d \sin i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \sin i\)
\(L_0 = \frac{2d \sin^2 i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right)\)
4. **Calculate the total optical path difference \(\Delta L\)**:
\(\Delta L = L - L_0\)
\(\Delta L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} + \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] - \frac{2d \sin^2 i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right)\)
\(\Delta L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} + (\sin^2 i - 2\sin^2 i) \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right]\)
\(\Delta L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right]\)
5. **Calculate the phase difference \(\varphi\)**:
\(\varphi = \frac{2\pi}{\lambda} \Delta L\)
\[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \]
Or written in the form of arccosh:
\[ \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) = \text{arccosh}\left(\frac{n_b}{\sin i}\right) - \text{arccosh}\left(\frac{n_a}{\sin i}\right) \]
Therefore:
\[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \left( \text{arccosh}\left(\frac{n_b}{\sin i}\right) - \text{arccosh}\left(\frac{n_a}{\sin i}\right) \right) \right] \]
Final answer:
The phase difference \(\varphi\) between the two beams is:
\[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \] | \[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \] |
737 | ELECTRICITY | There is a centrally symmetric magnetic field in space that is directed inward perpendicular to the paper. The magnitude of the magnetic field varies with distance $r$ from the center O, and is given by the formula ${\mathbf{B}(\mathbf{r})=\mathbf{B}_{0}\left(\frac{r}{R}\right)^{n}}$. A charged particle with charge $q$ and mass $m$ moves in uniform circular motion of radius $R$ around O in the plane perpendicular to the magnetic field. If the particle is given a small radial disturbance, it will oscillate slightly along the radial direction. Find the period of these small oscillations. | (1) According to Newton's second law: $$ {\mathfrak{q B}}_{0}v_{0}={\frac{m v_{0}^{2}}{R}} $$ We obtain: $$ \mathrm{v}_{0}={\frac{q B_{0}R}{m}} $$ (2) The principle of angular momentum: $$ \frac{d\mathrm{L}}{d t}=\mathrm{B}_{0}\Big(\frac{r}{R}\Big)^{n}r q\dot{r} $$ Rearrange terms: $$ dL-\frac{B_0q}{R^n}r^{n+1}dr=0 $$ Integrate and substitute the result from question (1): $$ \mathrm{L}-\frac{\mathrm{B}_{0}q}{(n+2)R^{n}}r^{n+2}=q B_{0}R^{2}(1-\frac{1}{n+2}) $$ (3) Conservation of energy: $$ {\frac{{\mathbf{L}}^{2}}{2m r^{2}}}+{\frac{1}{2}}m{\dot{r}}^{2}=c $$ That is: $$ \frac{(q B_{0}R^{2}\Big(1-\frac{1}{n+2}\Big)+\frac{{\bf B}_{0}q}{(n+2)R^{n}}r^{n+2})^{2}}{2m r^{2}}+\frac{1}{2}m{\dot{r}}^{2}=c $$ Differentiate the above equation with respect to time: $$ \mathrm{m}\ddot{r}+\frac{(n+1)B_{0}^{2}q^{2}r^{2n+1}}{m(n+2)^{2}R^{2n}}+\frac{q^{2}B_{0}^{2}n(n+1)r^{n-1}}{(n+2)^{2}R^{n-2}m}-\frac{q^{2}B_{0}^{2}R^{4}(n+1)^{2}}{m(n+2)^{2}r^{3}}=0 $$ Let: $$ {\bf{r}}={\bf{R}}+\delta{\bf{r}} $$ Expand the above equation for small quantities and get: $$ \mathrm{m}\ddot{\delta{r}}+\frac{(n+1)B_{0}^{2}q^{2}(2n+1)\delta r}{m(n+2)^{2}}+\frac{q^{2}B_{0}^{2}n(n+1)(n-1)\delta r}{(n+2)^{2}m}+\frac{3 q^{2}B_{0}^{2}R^{4}(n+1)^{2}\delta r}{m(n+2)^{2}}=0 $$ That is: $$ \mathrm{m}\ddot{\delta{r}}+(n+1)\frac{q^2B_{0}^2\delta r}{m}=0 $$ The above equation is the standard formula for simple harmonic motion, yielding: $$ \displaystyle\mathbb{T}=\frac{2\pi}{\sqrt{n+1}}\frac{m}{q B_{0}} $$ | $$
\frac{2\pi}{\sqrt{n+1}} \frac{m}{q B_0}
$$ |
636 | ELECTRICITY | Establish a Cartesian coordinate system Oxyz, with a hypothetical sphere of radius $R$ at the origin. Place $n$ rings of radius $R$ along the meridional circles, all passing through the points (0, 0, R) and (0, 0, -R). The angle between any two adjacent rings is $\frac{\pi}{n}$, and each ring is uniformly charged with positive charge $Q$. The setup is stationary, and a negative charge $βq$ with mass $m$ is performing circular motion on the equatorial plane at a distance $r_{0}$ from the center of the sphere ($r_0\gg R$). Now, give the charge a radial disturbance, and attempt to find the difference between its radial oscillation period and angular revolution period. The vacuum permittivity is known as $\epsilon_0$. | Consider the case of a single circular loop, establishing a spherical coordinate system $(r, \alpha)$ with the loop axis as the polar axis. Then: $$ V = {\frac{1}{4\pi\varepsilon_{0}}}\int_{0}^{2\pi}{\frac{Q d\theta}{2\pi}}{\frac{1}{\sqrt{r^{2}+R^{2}-2R r \sin\alpha \cos\Theta}}}={\frac{Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}+{\frac{R^{2}}{r^{3}}}{\frac{1-3\cos^{2}\alpha}{2}}\right)\#\left({\frac{1}{r}}+{\frac{R^{2}}{r^{2}}}\right). $$ Next, consider the superposition of multiple circular loops. Based on geometric relations: $$ V_{i}={\frac{Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}+{\frac{R^{2}}{r^{3}}}{\frac{1-3\sin^{2}\theta\cos^{2}\left(\varphi-{\frac{\mathrm{i}\pi}{\mathrm{n}}}\right)}{2}}\right) $$ Summing results in: $$ V={\frac{n Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{r^{3}}}{\frac{1-3\cos^{2}\theta}{4}}\right) $$ On the equatorial plane, $\textstyle{\theta={\frac{\pi}{2}}}$ $$ V={\frac{n Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{4r^{3}}}\right) $$ Assuming the angular momentum is $L=m\omega_{\theta}r_{0}^{2}$, the effective potential energy is $$ \mathrm{V}_{\mathrm{eff}}=-{\frac{n Q q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{4r^{3}}}\right)+{\frac{L^{2}}{2m r^{2}}} $$ The first derivative set to zero: $$ -{\frac{n Q q}{4\pi\varepsilon_{0}}}{\left({\frac{-1}{r^{2}}}+{\frac{3R^{2}}{4r^{4}}}\right)}-{\frac{L^{2}}{m r^{3}}}=0 $$ Solving gives: $$ \omega_{\theta}^{2}=\frac{n Q q}{4\pi\varepsilon_{0}m r_{0}}\left(\frac{1}{r_{0}^{2}}-\frac{3R^{2}}{4r_{0}^{4}}\right) $$ The second derivative: $$ -{\frac{n Q q}{4\pi\varepsilon_{0}}}\left({\frac{2}{r^{3}}}-{\frac{3R^{2}}{r^{5}}}\right)+{\frac{3L^{2}}{m r^{4}}}=m\omega_{r}^{2} $$ Solving gives: $$ \omega_{r}^{2}=\frac{n Q q}{4\pi\varepsilon_{0}m r_{0}}\left(\frac{1}{r_{0}^{2}}+\frac{3R^{2}}{4r_{0}^{4}}\right) $$ $$ T_{r}-T_{\oplus}=\frac{2\pi}{\omega^{2}}(\omega_{\theta}-\omega_{r})=-\frac{3\pi R^{2}}{2r_{0}^{2}}\sqrt{\frac{4\pi\varepsilon_{0}m r_{0}^{3}}{n Q q}} $$ | $$-\frac{3\pi R^2}{2r_0^2}\sqrt{\frac{4\pi\varepsilon_0 m r_0^3}{n Q q}}$$ |
420 | MECHANICS | B and C are two smooth fixed pulleys with negligible size, positioned on the same horizontal line. A and D are two objects both with mass $m$, connected by a light and thin rope that passes over the fixed pulleys. Initially, the system is stationary, and the distances between AB and CD are both in the direction of gravity, where the distance between AB is $x_{0}$ and the distance between CD is $L$, which is sufficiently large such that D will not touch C within the time frame discussed in the problem. At this moment, ball A is pulled so that the line AB deviates from the vertical line by a small angle $\theta_{0}$ (without changing the length of AB), and the system begins to move. Taking gravitational acceleration as $\pmb{g}$, and assuming that A descends very slowly so that we can approximately consider the length of AB remains unchanged, A moves around B with a pendulum-like motion. Try to solve for the amplitude of the oscillation angle $\theta$ of A when $AB=x$.
| Let the tension in the rope be $\intercal$. According to Newton's second law for block D, $T-mg=m{\ddot{x}}$ (1). Using the given assumptions, write the relation of the pendulum's oscillation angle with time: $$ \theta_{t} = \theta \cos(\sqrt{\frac{g}{x}}t + \phi) $$ For $\mathsf{A}$, write down the expression of Newton's second law: $$ mg-T+m\dot{\theta}_{t}^{2}x = m\ddot{x} $$ Combining (2) and (3), we get: $$ mg-T+{\textstyle{\frac{1}{2}}}mg\theta^{2}=m\ddot{x} $$ Since the work done on A by the tension in the rope and the gravity of A is equal to the increment in A's kinetic energy, and A's kinetic energy can be separately written as the kinetic energy of vertical motion and the kinetic energy of oscillation (as the oscillation angle is small, the two can be considered independent): $$ (mg-T)\dot{x}=\frac{d}{dt}(\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}mg x\theta^{2}) $$ Combining (1), (4), and (5), we get: $$ g\theta^{2}=4\ddot{x} \quad -\ddot{x}\dot{x}=\frac{d}{dt}(\frac{1}{2}\dot{x}^{2}+2x\ddot{x}) $$ $\textstyle{\bar{\mathcal{F}}}\psi/{\bar{\mathcal{H}}}{\ddot{x}}dx={\frac{1}{2}}d{\dot{x}}^{2}$, simplifying the above equation, we get: $$ \dot{x}^{2}+2x\ddot{x}=const $$ Using the initial conditions at $\mathtt{t}{=}0$: $$ x=x_{0} \quad \ddot{x}=g\theta_{0}^{2}/4 $$ We can rewrite equation (8) as: $$ \dot{x}^{2}+2x\ddot{x}=g\theta_{0}^{2}x_{0}/2 $$ Multiplying both sides of the above equation by ${\mathsf{dx}}$, and using $\ddot{\bf x}dx=\frac{1}{2}d\dot{x}^{2}$, we get: $$ \dot{x}^{2}dx+xd\dot{x}^{2}=d(x\dot{x}^{2})=\frac{1}{2}g\theta_{0}^{2}x_{0}dx $$ Integrating, we obtain: $$ \dot{x}^{2}=\frac{1}{2}g\theta_{0}^{2}x_{0}(x-x_{0})/x $$ Differentiating both sides with respect to time, eliminating the first-order derivative of $\mathsf{x}$ with respect to $\mathtt{t}$, we get: $$ \ddot{x}=\frac{1}{4}g\theta_{0}^{2}\frac{x_{0}^{2}}{x^{2}} $$ Combining with equation (6), we can solve for the amplitude of the oscillation angle $\theta$: $$ \theta=\theta_{0}\frac{x_{0}}{x} $$
| $$
\theta=\theta_{0}\frac{x_{0}}{x}
$$ |
228 | MECHANICS | AB is a uniform thin rod with mass $m$ and length $l_{2}$. The upper end B of the rod is suspended from a fixed point O by an inextensible soft and light string, which has a length of $l_{1}$. Initially, both the string and the rod are hanging vertically and at rest. Subsequently, all motion occurs in the same vertical plane, with all angles rotating counterclockwise being positive.
Assume at a certain moment, the angles between the string, the rod, and the vertical direction are $\theta_{1}$ and $\theta_{2}$ respectively. The angular velocities of the string around the fixed point O and the rod around its center of mass are $\omega_{1}$ and $\omega_{2}$ respectively. Determine the angular acceleration $\alpha_{2}$ of the rod around its center of mass, expressing the answer only in terms of the physical quantities given in the problem and the gravitational acceleration $g$. | In the vertical plane where points O, B, and A are located, establish a plane coordinate system with point O as the origin, the horizontal ray to the right as the $\mathbf{X}$ axis, and the vertical ray upward as the $\mathbf{y}$ axis. The acceleration of the pole's center of mass C, denoted as $(\boldsymbol{a}_{\mathrm{G}x},\boldsymbol{a}_{\mathrm{G}y})$, satisfies the center of mass motion theorem
$$
m a_{\mathrm{c}x}=-T\sin\theta_{1},m a_{\mathrm{c}y}=-m g+T\cos\theta_{1}\textcircled{7}
$$
In this equation, $T$ is the magnitude of the tension in the rope. At the same time, the pole rotates around its center of mass under the moment caused by the rope tension $T$ relative to the center of mass. By the rotational theorem, we have
$$
\frac{1}{12}m l_{2}^{2}\alpha_{2}=-T\frac{1}{2}l_{2}\sin(\theta_{2}-\theta_{1})\textcircled{8}
$$
From the geometric relationship, we have
$$
x_{\mathrm{{B}}}(t)=x_{\mathrm{{C}}}(t)-{\frac{1}{2}}l_{2}\sin\theta_{2}(t),y_{\mathrm{{B}}}(t)=y_{\mathrm{{C}}}(t)+{\frac{1}{2}}l_{2}\cos\theta_{2}(t)
$$
Differentiating both sides of the above equations with respect to time $t$, the velocity of point B satisfies the conditions
$$
{\boldsymbol{v}}_{\mathrm{{B}}x}(t)={\boldsymbol{v}}_{\mathrm{{C}}x}(t)-{\frac{1}{2}}{\boldsymbol{\omega}}_{2}(t){\boldsymbol{l}}_{2}\cos\theta_{2}(t),\quad{\boldsymbol{v}}_{\mathrm{{B}}y}(t)={\boldsymbol{v}}_{\mathrm{{C}}y}(t)-{\frac{1}{2}}{\boldsymbol{\omega}}_{2}(t){\boldsymbol{l}}_{2}\sin\theta_{2}(t),
$$
Differentiating both sides of the above equations with respect to time $t$ again, the acceleration of point B satisfies the conditions
$$
a_{\mathrm{B}x}=a_{\mathrm{C}x}-\frac{1}{2}\alpha_{2}l_{2}\cos\theta_{2}+\frac{1}{2}\omega_{2}^{2}l_{2}\sin\theta_{2},a_{\mathrm{B}y}=a_{\mathrm{C}y}-\frac{1}{2}\alpha_{2}l_{2}\sin\theta_{2}-\frac{1}{2}\omega_{2}^{2}l_{2}\cos\theta_{2}
$$
Meanwhile, point B rotates around point O with a fixed axis under the condition of an inextensible rope, thus
$$
x_{\mathrm{{B}}}(t)=l_{\mathrm{{1}}}\sin\theta_{\mathrm{{1}}}(t),y_{\mathrm{{B}}}(t)=-l_{\mathrm{{1}}}\cos\theta_{\mathrm{{1}}}(t)
$$
Differentiating both sides of the above equations with respect to time $t$, the velocity of point B also satisfies the conditions
$$
V_{_{\mathrm{B}x}}(t)=\omega_{1}(t)l_{1}\cos\theta_{1}(t),~V_{_{\mathrm{B}y}}(t)=\omega_{1}(t)l_{1}\sin\theta_{1}(t)
$$
Differentiating both sides of the above equations with respect to time $t$ again, the acceleration of point B also satisfies the conditions
$$
a_{\scriptscriptstyle{\mathrm{B}x}}=\alpha_{1}l_{1}\cos\theta_{1}-\alpha_{1}^{2}l_{1}\sin\theta_{1},a_{{\scriptscriptstyle{\mathrm{B}y}}}=\alpha_{1}l_{1}\sin\theta_{1}+\omega_{1}^{2}l_{1}\cos\theta_{1}
$$
The angular accelerations of the rope around the suspension point and the pole around the center of mass can be solved as
$$
\begin{array}{r l}&{\alpha_{2}=\cfrac{3\sin(\theta_{1}-\theta_{2})\left[2g\cos\theta_{1}+2l_{1}\omega_{1}^{2}+l_{2}\omega_{2}^{2}\cos(\theta_{1}-\theta_{2})\right]}{l_{2}\left[1+3\sin^{2}(\theta_{1}-\theta_{2})\right]}}\end{array}
$$ | $$
\frac{3\sin(\theta_1-\theta_2)\left[2g\cos\theta_1+2l_1\omega_1^2+l_2\omega_2^2\cos(\theta_1-\theta_2)\right]}{l_2\left[1+3\sin^2(\theta_1-\theta_2)\right]}
$$ |
450 | MECHANICS | Xiao Ming discovered an elliptical plate at home with semi-major and semi-minor axes of \( A \) and \( B \), respectively. Using one focus \( F \) as the origin, a polar coordinate system was established such that the line connecting the focus and the vertex closest to the focus defines the polar axis direction. Through extremely precise measurements, it was found that the mass surface density satisfies the equation \(\sigma = \sigma_{0}(1 + e\cos\varphi)^3\), where \( e \) is the eccentricity of the ellipse. When the plate's major axis is positioned vertically and released from rest on a sufficiently rough tabletop, the plate moves only within the plane it lies in. Find the angular acceleration \(\beta\) of the plate when the major axis becomes horizontal. | At this time, the angular velocity is $\omega$, the angular acceleration is $\beta$, and the acceleration of the center of mass is $a_{x}, a_{y}$.
The moment of inertia about the instantaneous center $\mathrm{P}$ is $I_{P}=I+m A^{2}$.
From the conservation of energy, the contact condition at point P gives the acceleration
$a = \omega^2 \rho = \omega^2 \frac{A^2}{B}$.
Considering the rotational theorem about the instantaneous center:
$I_{P} \vec{\beta} = \vec{M} - m \vec{r}_{c} \times \vec{a}$
We obtain:
$$
I_{P} \beta = m \sqrt{A^2 - B^2} \bigg(g + \omega^{2} \frac{A^{2}}{B} \bigg)
$$
$$
\beta = 4A \sqrt{A^2 - B^2} g \frac{2A^{4} + 2A^{3} \sqrt{A^2 - B^2} - A^{3}B + B^{4}}{(2A^{3} + B^{3})^{2}B}
$$ | $$
\beta=4A \sqrt{A^2 - B^2} g\frac{2A^{4}+2A^{3}\sqrt{A^2 - B^2}-A^{3}B+B^{4}}{(2A^{3}+B^{3})^{2}B}
$$ |
708 | OPTICS | \"Choose a sodium lamp for Young's double-slit interference experiment. The wavelengths of the sodium lamp's double yellow lines are $\lambda_{1}$ and $\lambda_{2}$ respectively. Due to a very small wavelength difference, the higher-order interference fringes on the screen will become blurred. Neglecting the width of the slits themselves, introduce the contrast function $\gamma \equiv\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$ to measure the clarity of the interference. When $\gamma = 0$, the fringes will become blurred. Additionally, if the background light intensity becomes $1/e$ of the system's maximum possible light intensity, it will also cause the observer to subjectively perceive the interference fringes as too dim, leading to blurriness. If the width of the double slits is $d$ and the horizontal distance from the screen to the double slits is $L$. Use a filter to filter out the light with a wavelength of $\lambda_{1}$. Due to the broadening of the spectral line caused by the thermal motion of molecules in the sodium lamp, if the temperature of the sodium lamp is $T$ and the mass of the sodium atom is $m$, try to calculate the position where the fringes become blurred for the first time. Assume that the sodium atoms follow the Maxwell velocity distribution law (and that the problem can be solved using the far-field condition). | Examine the effect of monochromatic light on the screen: $$ U = U_0 \left( e^{i k \cdot \frac{d x}{L}} + 1 \right) $$ The resulting light intensity is: $$ I = U \cdot U^* = U_0^2 \left(2 + 2\cos\left(k \cdot \frac{d x}{L}\right)\right) = I_0 \left(1 + \cos\left(k \cdot \frac{d x}{L}\right)\right) $$ For Doppler frequency shift due to thermal motion, examine the Doppler effect of electromagnetic waves: $$ \frac{k}{k_0} = \frac{\sqrt{1 - \beta^2}}{1 - \beta \cos\theta} \approx 1 + \beta \cos\theta = 1 + \frac{v_x}{c} \quad $$ For particles with horizontal velocity in the interval $v - v + \mathrm{d}v$ (in terms of emitted light intensity), we have: $$ \frac{\mathrm{d}I}{I_0} = f_x(v) \, \mathrm{d}v \quad $$ For light intensity on the screen: $$ I \propto \int f_x(v)\,\mathrm{d}v \cos\frac{k_0 \left(1 + \frac{v_x}{c}\right) d x}{L}\ = \int f_x(v)\,\mathrm{d}v \left( \cos\frac{k_0 d x}{L}\cos\frac{v k_0 d x}{L c}\ - \sin\frac{k_0 d x}{L}\sin\frac{v k_0 d x}{L c} \right) \quad $$ Notice that the second term is an odd function, and its integral equals 0. The integral of the first term is calculated as: $$ s_1 = \operatorname{Re} \int_{-\infty}^{\infty} \sqrt{\frac{m}{2\pi k T}} e^{- \frac{m v^2}{2k T} + i \frac{v k_0 d x}{L c}} \, \mathrm{d}v = e^{\frac{\frac{k_0 d x^2}{L c}}{\frac{2m}{k T}}} \quad $$ When it decreases to $\frac{1}{e}$, the background light will become blurred, thus: $$ \frac{\frac{k_0 d x^2}{L c}}{\frac{2m}{k T}} = 1 \quad $$ Solve for: $$ x = \sqrt{\frac{2m}{k T}} \cdot \frac{L c \lambda_2}{2\pi d} \quad $$ | $$
x = \sqrt{\frac{2m}{k T}} \cdot \frac{L c \lambda_2}{2\pi d}
$$ |
649 | ELECTRICITY | In modern plasma physics experiments, negative particles are often constrained in two ways. In the following discussion, we do not consider relativistic effects or retarded potentials. Uniformly charged rings with radius $R$ are placed on planes $z=l$ and $z=-l$ in space, respectively. The rings are perpendicular to the $z$ axis, and each carries a charge of $Q_{0}$. A particle with charge $-q$ and mass $m$ is placed at the origin of the coordinate system. Given that $Q_0, q > 0$, the vacuum permittivity is $\epsilon_{0}$ and the vacuum permeability is $\mu_{0}$. To keep the point charge stable in both the $\hat{z}$ and $\hat{r}$ directions, we consider rotating the two charged rings in the $\hat{z}$ direction with a constant angular velocity $\Omega$. Given a small disturbance to the point charge at the origin, provide the minimum $\Omega$ required for the particle to remain stable in all directions. | Analyzing the magnetic field at a small displacement $z$ along the $z$-axis away from the coordinate origin $$ \vec{B_{z}}={\frac{\mu_{0}Q\Omega R}{4\pi}}\left({\frac{R}{[R^{2}+(l+z)^{2}]^{\frac{3}{2}}}}+{\frac{R}{[R^{2}+(l-z)^{2}]^{\frac{3}{2}}}}\right) $$ yields $$ \vec{B_{z}}=\frac{\mu_{0}Q\Omega}{2\pi}\frac{R^{2}}{(R^{2}+l^{2})^{\frac{3}{2}}}\hat{z} $$ Observing that near the origin, $\vec{B}$ can be considered as a uniform field along the $\hat{z}$ direction, denoted as $B_{0}$, consider a particle starting from the origin. By the angular momentum theorem $$ \frac{\mathrm{d}\vec{L}}{\mathrm{d}t}=-q B_{0}r\dot{r}\vec{\varphi} $$ $$ L+\frac{1}{2}q B_{0}r^{2}=0 $$ $$ v_{\varphi}=-\frac{q B_{0}r}{2m} $$ Substitute into the conservation of energy equation to obtain the system's effective potential energy $$ V_{e f f}=[{\frac{\mu_{0}^{2}q^{2}Q^{2}\Omega^{2}}{32m\pi^{2}}}{\frac{R^{4}}{(R^{2}+l^{2})^{3}}}-{\frac{q Q}{8\pi\varepsilon_{0}}}{\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}]r^{2}+{\frac{q Q}{4\pi\varepsilon_{0}}}{\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}z^{2} $$ For the system to be stable along $\hat{r},\hat{z}$, the coefficients should both be greater than 0, solving gives $$ R^{2}>2l^{2} $$ $$ \Omega>\frac{2\pi}{\mu_{0}R^{2}}\sqrt{\frac{m(R^{2}-2l^{2})(R^{2}+l^{2})^{\frac{1}{2}}}{\pi\varepsilon_{0}Q q}} $$ That is, the minimum value is $$ \frac{2\pi}{\mu_{0}R^{2}}\sqrt{\frac{m(R^{2}-2l^{2})(R^{2}+l^{2})^{\frac{1}{2}}}{\pi\varepsilon_{0}Q q}} $$ | $$
\frac{2\pi}{\mu_0 R^2} \sqrt{\frac{m(R^2 - 2l^2)(R^2 + l^2)^{1/2}}{\pi \varepsilon_0 Q q}}
$$ |
688 | ELECTRICITY | In space, there is an axisymmetric magnetic field, with the direction of the magnetic field pointing outward perpendicular to the plane, and its magnitude depends only on the distance from the center of symmetry, $B(r) = B_{0} \left(\frac{r}{r_{0}}\right)^{n}$. A particle with mass $m$ and charge $q$ moves in a circular motion with radius $r_{0}$ around the center of symmetry under the influence of the magnetic field. Find: if the charged particle is given a small radial disturbance, determine the period $T$ of the small radial oscillations. | According to symmetry, the system in this problem has conserved canonical angular momentum with respect to the magnetic field's center of symmetry. To find the canonical angular momentum, we list the corresponding rate of change equation: $$ \frac{d L}{d t} = q v_{r} B_{0} \Big(\frac{r}{r_{0}}\Big)^{n} r $$ Rearranging terms and integrating yields: $$ d L - \frac{q B_{0}}{r_{0}^{n}} r^{n+1} d r = 0 \to L - \frac{q B_{0}}{(n+2) r_{0}^{n}} r^{n+2} = C = \frac{n+1}{n+2} q B_{0} r_{0}^{2} $$ (Conservation quantity 6 points) Thus, we obtain the function of angular momentum and $r$: $$ L = L(r) = \frac{q B_{0}}{n+2} \biggl(\frac{r^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\biggr) $$ The energy conservation of the charged particle gives: $$ \frac{1}{2} m \dot{r}^{2} + \frac{1}{2} m v_{\theta}^{2} = \frac{1}{2} m \dot{r}^{2} + \frac{L(r)^{2}}{2m r^{2}} = E $$ (Energy conservation 6 points) Give the particle a radial perturbation, let $r = r_{0} + \delta r$, $\dot{r} = (\dot{\delta r})$, resulting in: $$ \frac{1}{2} m (\dot{\delta r})^{2} + \frac{q^{2} B_{0}^{2}}{2m(n+2)^{2}} \frac{\left(\frac{(r_{0}+\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\right)^{2}}{(r_{0}+\delta r)^{2}} = E $$ To analyze the oscillation, we need to expand the potential energy and retain terms up to the second order, so we first look at the denominator: $$ \begin{array}{r l} &\left(\frac{(r_{0}+\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\right)^{2} = \left(r_{0}^{2}\Bigl(1+\frac{\delta r}{r_{0}}\Bigr)^{n+2} + (n+1) r_{0}^{2}\right)^{2} \\ &\qquad = r_{0}^{4} \Biggl(1 + (n+2) \frac{\delta r}{r_{0}} + \frac{(n+2)(n+1)}{2} \Bigl(\frac{\delta r}{r_{0}}\Bigr)^{2} + (n+1) \Biggr)^{2} \\ &\qquad = r_{0}^{4} (n+2)^{2} \Biggl(1 + \frac{\delta r}{r_{0}} + \frac{n+1}{2} \Bigl(\frac{\delta r}{r_{0}}\Bigr)^{2} \Biggr)^{2} = r_{0}^{4} (n+2)^{2} \left(1 + 2 \frac{\delta r}{r_{0}} + (n+2) \frac{\delta r}{r_{0}}\right)^{2} \end{array} $$ In the above equation, we repeatedly used the small quantity approximation formula $\begin{array}{r} (1+x)^{n} = 1 + n x + \frac{n(n-1)}{2}x^{2} + \ldots, \end{array}$ so the potential energy term can be written as: $$ \frac{\left(\displaystyle\frac{(r_{0}+\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\right)^{2}}{(r_{0}+\delta r)^{2}} = r_{0}^{2} (n+2)^{2} \left(1 + 2 \frac{\delta r}{r_{0}} + (n+2) \Big(\frac{\delta r}{r_{0}}\Big)^{2}\right) \left(1 + \frac{\delta r}{r_{0}}\right)^{-2} $$ $$ = r_{0}^{2} (n+2)^{2} \left(1 + 2 \frac{\delta r}{r_{0}} + (n+2) \Big(\frac{\delta r}{r_{0}}\Big)^{2}\right) \left(1 - 2 \frac{\delta r}{r_{0}} + 3 \Big(\frac{\delta r}{r_{0}}\Big)^{2}\right) $$ Thus the second-order terms of potential energy are: $$ \frac{q^{2} B_{0}^{2}}{2m(n+2)^{2}} r_{0}^{2} (n+2)^{2} (n+1) \bigg(\frac{\delta r}{r_{0}}\bigg)^{2} = \frac{q^{2} B_{0}^{2}}{2m} (n+1) \delta r^{2} $$ (Correctly expanding small terms 12 points) Thus, the energy conservation equation is: $$ \frac{1}{2} m (\dot{\delta r})^{2} + \frac{q^{2} B_{0}^{2}}{2m} (n+1) \delta r^{2} = C $$ According to the formula for simple harmonic oscillation, we know this represents an oscillation period of: $$ T = \frac{2 \pi m}{q B_{0} \sqrt{n+1}} $$ | $$T = \frac{2\pi m}{q B_0 \sqrt{n+1}}$$ |
118 | ELECTRICITY | A regular dodecahedron resistor network is given. Except for $R_{BC} = 2r$, the resistance between all other adjacent vertices is $r$. Points $B$ and $C$ are the two endpoints of one edge of the dodecahedron. Find the resistance between points $B$ and $C$. | Boldly introducing negative resistance, $BC$ is equivalent to a parallel combination of $\pmb{r}$ and $-2r$:
$$
\frac{1}{r}+\frac{1}{-2r}=\frac{1}{2r}
$$
After extracting $-2r$, the remaining resistance value can be constructed using the forced current method. Inject $19I_{1}$ into node $B$, and each of the other nodes emits a current of 1. Then consider injecting $I$ into each of the other nodes while node $C$ emits $191$. Thus, the current flowing through the branch:
$$
I_{BC}=20I
$$
Voltage:
$$
U_{BC}=\frac{19Ir}{3}\cdot2
$$
Thus:
$$
U_{BC}=I_{BC}R_{BC}^{\prime}
$$
Resulting in:
$$
R_{BC}^{\prime}=\frac{19}{30}r
$$
Then, parallel it with $-2r$:
$$
\frac{1}{R_{BC}}=\frac{1}{R_{BC}^{\prime}}+\frac{1}{-2r}
$$
Finally obtained:
$$
R_{BC}=\frac{38}{41}r
$$ | $$
R_{BC} = \frac{38}{41}r
$$ |
318 | MECHANICS | Consider a small cylindrical object with radius $R$ and height $h$. Determine the expression for the force $F$ acting on the cylinder when a sound wave passes through it. The axial direction of the cylinder is the direction of wave propagation. In the sound wave, the displacement of a particle from its equilibrium position is $\psi = A\cos kx\cos 2\pi ft$, the ambient pressure is $P_0$, and the adiabatic index of air is $\gamma$. | The force on the cylinder will be
$$
F = -\pi R^{2}(p(y+h)-p(y)) = -\pi R^{2}h\frac{d p}{d y}
$$
For a traveling wave, substitute
$$
\Delta p(y,t) = -\gamma p_{0}A k\cos(k y-2\pi f t)
$$
$$
\frac{d p}{d y} = \gamma p_{0}A k^{2}\sin(k y-2\pi f t)
$$
Therefore, the force is
$$
F = -\pi R^2 h\gamma p_0 A k^2\sin(ky-2\pi ft)
$$
However, for a standing wave, we have
$$
\psi = A\cos{k x}\cos{2\pi}f t
$$
Thus,
$$
\Delta p = \gamma P_{0}k A\sin k x\cos2\pi f t
$$
Therefore,
$$
\frac{d p}{d x} = \gamma P_{0}k^{2}A\cos k x\cos2\pi f t
$$
Thus, the force is
$$
F = -\pi R^2 h\gamma P_0 k^2 A\cos kx \cos 2\pi ft
$$ | $$
F = -\pi R^2 h \gamma P_0 k^2 A \cos(kx) \cos(2\pi ft)
$$ |
62 | ADVANCED | In certain solids, ions have spin angular momentum and can be regarded as a three-dimensional real vector $\vec{S}$ with a fixed length under the semiclassical approximation, where the length $\vert\vec{S}\vert = S$ is a constant. The magnetic moment $\overrightarrow{M}$ of the ions is usually proportional to the spin vector, ${\overrightarrow{M}} = \gamma{\overrightarrow{S}}$, where $\gamma(>0)$ is a constant. The spin-spin interaction between ions $i$ and $j$ is usually the Heisenberg interaction, $-J_{i,j}\vec{S}_{i} \cdot \vec{S}_{j}$. Under a static uniform external magnetic field $\overrightarrow{B}$, the total energy (Hamiltonian) of the system is typically given by $\begin{array}{r}{H = -\sum_{(i,j)}J_{i,j}\vec{S}_{i}\cdot\vec{S}_{j} - }\end{array}$ $\begin{array}{r}{\gamma\overrightarrow{B}\cdot\sum_{i}\overrightarrow{S}_{i}}\end{array}$. Here, $(i,j)$ indicates a sum over all distinct pairs without order, meaning $(i,j)$ and $(j,i)$ represent the same pair and are not counted twice. The semiclassical time evolution of the spin vector satisfies the Heisenberg equation of motion (here $\times{}$ represents the vector cross product):
$$
\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{i}=\frac{\partial H}{\partial \vec{S}_{i}}\times\vec{S}_{i}
$$
Consider two spins, $H = -J\vec{S}_{1}\cdot\vec{S}_{2} - \gamma\vec{B}\cdot\left(\vec{S}_{1}+\vec{S}_{2}\right)$, where $J>0$. Without loss of generality, let $\overrightarrow{B}$ be along the $\mathbf{Z}$ direction, $\vec{B}=B\hat{\bf z}$, $B \geq 0$. The ground state (lowest energy state) of the system is $\vec{S}_{1}=\vec{S}_{2}=S\hat{\mathbf{z}}$, and this state does not evolve over time. Consider states close to the ground state, $\vec{S}_{1}=S\big(x_{1},y_{1},\sqrt{1-{x_{1}}^{2}-{y_{1}}^{2}}\big)$, $\vec{S}_{2}=S\big(x_{2},y_{2},\sqrt{1-{x_{2}}^{2}-{y_{2}}^{2}}\big)$, where $\left|x_{1,2}\right|\ll1, \left|y_{1,2}\right|\ll1$. Find the product of the final normal mode frequencies. | Solution: The Heisenberg equation of motion is
$$
\begin{array}{r}{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{1}=-\left(J\vec{S}_{2}+\gamma\vec{B}\right)\times\vec{S}_{1}}\ {\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{2}=-\left(J\vec{S}_{1}+\gamma\vec{B}\right)\times\vec{S}_{2}}\end{array}
$$
Note that $\vec{S}_{\pm}=\vec{S}_{1}\pm\vec{S}_{2}$ can be defined, then $\begin{array}{r}{H=\frac{J}{2}\big|\overrightarrow{S}_{+}\big|^{2}-J S^{2}-\gamma\overrightarrow{B}\cdot\overrightarrow{S}_{+}}\end{array}$,
$$
\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{+}=-\gamma\vec{B}\times\vec{S}_{+}
$$
$$
\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{-}=-\left(\vec{J S}_{+}+\gamma\vec{B}\right)\times\vec{S}_{-}
$$
The equations of motion for $x_{1,2},y_{1,2}$ in the lowest order approximation are
$$
\begin{array}{c}{{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(x_{1}+x_{2})=\gamma B\cdot(y_{1}+y_{2})}}\ {{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(y_{1}+y_{2})=-\gamma B\cdot(x_{1}+x_{2})}}\ {{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(x_{1}-x_{2})\approx(2J S+\gamma B)\cdot(y_{1}-y_{2})}}\ {{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(y_{1}-y_{2})\approx-(2J S+\gamma B)\cdot(x_{1}-x_{2})}}\end{array}
$$
The product of the characteristic frequencies is
$\gamma B(2J S+\gamma B)$ | $$\gamma B(2JS + \gamma B)$$ |
143 | MECHANICS | Consider an elastic soft rope with original length $a$, and elastic coefficient $k$. With one end fixed, the other end is attached to a particle with a mass of $m$. And the rope remains horizontal. The particle moves on a smooth horizontal surface. Initially, the rope is stretched to a length of $a+b$, and then the particle is released. Determine how much time it takes for the particle to return to the original release position. | First consider \( x > a \). At this time, the particle only experiences the elastic force \( -k(x-a) \) in the horizontal direction. According to Newton's second law, the equation of motion for the particle is
$$
m{\ddot{x}} = -k(x-a),
$$
which is the equation of simple harmonic motion. To solve equation (1), make the substitution \(\xi = x - a\), then equation (1) becomes
$$
m\ddot{\xi} = -k\xi.
$$
Further, with the transformation \(\ddot{\xi} = \frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}} = \frac{\mathrm{d}\xi}{\mathrm{d}t} \frac{\mathrm{d}}{\mathrm{d}\xi}\left(\frac{\mathrm{d}\xi}{\mathrm{d}t}\right) = \frac{1}{2} \frac{\mathrm{d}\dot{\xi}^{2}}{\mathrm{d}\xi}\), equation (2) becomes
$$
\mathrm{d}\dot{\xi}^{2} = -2\omega^{2}\xi\mathrm{d}\xi,
$$
where \(\omega = \sqrt{k/m}\). Integrating the above, and setting the initial conditions \(\xi = \xi_{0}, \dot{\xi} = v_{0} \) at \(t=0\), yields
$$
\dot{\xi}^{2} = v_{0}^{2} - \omega^{2}(\xi^{2} - \xi_{0}^{2}) = (v_{0}^{2} + \omega^{2}\xi_{0}^{2}) - \omega^{2}\xi^{2},
$$
which implies
$$
\frac{\mathrm{d}\xi}{\mathrm{d}t} = \pm\sqrt{(v_{0}^{2} + \omega^{2}\xi_{0}^{2}) - \omega^{2}\xi^{2}} = \pm\omega\sqrt{A^{2} - \xi^{2}},
$$
where \({\cal A} = \sqrt{\frac{v_{0}^{2}}{\omega^{2}} + \xi_{0}^{2}}\). Further integration and considering the initial conditions gives
$$
\xi = A\sin(\pm\omega t + \arcsin \frac{\xi_{0}}{A}) = A\cos(\omega t + \alpha),
$$
where \(\alpha = \pm\arcsin{\frac{\xi_{0}}{A}} \mp {\frac{\pi}{2}}\). Thus, the general solution of equation (1) is
$$
x = a + A\cos(\omega t + \alpha).
$$
According to the problem, at \(t=0\), \(x=a+b, \dot{x}=0\). Substituting into equation (3) determines the constants of integration to be \(A=b, \alpha=0\). Substituting back into equation (3), we obtain
$$
x = a + b\cos\omega t.
$$
This shows that the motion of the particle in the domain \(x > a\) is periodic. The time for the particle to move from \(x = a + b\) to \(x = a\) and back is the period, which is
$$
t_{0} = {\frac{1}{2}}\frac{2\pi}{\omega} = \pi\sqrt{\frac{m}{k}}.
$$
When \(x < -a\), the particle experiences an elastic force of \(-k(x+a)\) in the horizontal direction. Note that because \(x < -a\), \( -(x+a) > 0\), so the elastic force points in the positive \(x\) direction, which is consistent with the actual situation. The equation of motion is
$$
m{\ddot{x}} = -k(x+a).
$$
Using a similar method to the above, the general solution of the equation is
$$
x = -a + B\cos(\omega t + \beta),
$$
where \(B, \beta\) are constants of integration. According to analysis in \(\textcircled{2}\), \(\dot{x}|_{x=a} = \dot{x}|_{x=-a}\), we determine \(B=b\). Equation (5) indicates that the motion of the particle in the region \(x < -a\) is also simple harmonic, with the same frequency as in the region \(x > a\). The time for the particle to move from \(x=-a\) to \(x=-(a+b)\) and back is also \(t_{0}\).
In summary, the total time taken for the particle to move from \(x = a + b \) to \(x = a\) and from \(x = -a\) to \(x = -(a + b)\) and back is
$$
t_{1} = 2t_{0} = 2\pi\sqrt{\frac{m}{k}}.
$$
The previous discussion also shows that considering just \(x > a\) is sufficient to determine the characteristics of the particle's motion in the interval \(|x| > a\), especially concerning the motion's time.
\(\textcircled{2}\) When \(|x|<a\), i.e., \(-a<x<a\), the particle is not subject to any external force in the horizontal direction, and thus will move uniformly with the velocity it has at \(x = \pm a\). The speed of the particle at \(x = a\) is
$$
\dot{x}|_{x=a} = \dot{x}|_{t=t_{1}} = -b\omega = -b\sqrt{\frac{k}{m}}.
$$
The time for the particle to travel from \(x=a\) to \(x=-a\) and back to \(x=a\) is
$$
t_{2} = 4\frac{a}{|\dot{x}|_{x=a}|} = \frac{4a}{b}\sqrt{\frac{m}{k}}.
$$
The process for the particle to return to its release point is \(x = a+b \rightarrow x = a \rightarrow x = 0 \rightarrow x = -a \rightarrow x = -(a+b) \rightarrow x = -a \rightarrow x = 0 \rightarrow x = a \rightarrow x = a+b\), so the total time experienced is
$$
t = t_{1} + t_{2} = 2\left(\pi + \frac{2a}{b}\right)\sqrt{\frac{m}{k}}.
$$ | $$2\left(\pi+\frac{2a}{b}\right)\sqrt{\frac{m}{k}}$$ |
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PHYBench: Holistic Evaluation of Physical Perception and Reasoning in Large Language Models
[π Project] [π Paper] [π» Code] [π Overview] [π§ Data Details] [π© Citation]
New Updates
- 2025.4.25: We release our code of EED Score. View and star on our github page!
- Recently: The leaderboard is still under progress, we'll release it as soon as possible.
π Acknowledgement and Progress
We're excited to announce the initial release of our PHYBench dataset!
- 100 fully-detailed examples including handwritten solutions, questions, tags, and reference answers.
- 400 additional examples containing questions and tags.
π Dataset Access
You can access the datasets directly via Hugging Face:
- PHYBench-fullques.json: 100 examples with complete solutions.
- PHYBench-onlyques.json: 400 examples (questions and tags only).
- PHYBench-questions.json: Comprehensive set of all 500 questions.
π Full-Dataset Evaluation & Leaderboard
We are actively finalizing the full-dataset evaluation pipeline and the real-time leaderboard. Stay tuned for their upcoming release!
Thank you for your patience and ongoing support! π
For further details or collaboration inquiries, please contact us at [email protected].
π Overview
PHYBench is the first large-scale benchmark specifically designed to evaluate physical perception and robust reasoning capabilities in Large Language Models (LLMs). With 500 meticulously curated physics problems spanning mechanics, electromagnetism, thermodynamics, optics, modern physics, and advanced physics, it challenges models to demonstrate:
- Real-world grounding: Problems based on tangible physical scenarios (e.g., ball inside a bowl, pendulum dynamics)
- Multi-step reasoning: Average solution length of 3,000 characters requiring 10+ intermediate steps
- Symbolic precision: Strict evaluation of LaTeX-formulated expressions through novel Expression Edit Distance (EED) Score
Key innovations:
- π― EED Metric: Smoother measurement based on the edit-distance on expression tree
- ποΈ Difficulty Spectrum: High school, undergraduate, Olympiad-level physics problems
- π Error Taxonomy: Explicit evaluation of Physical Perception (PP) vs Robust Reasoning (RR) failures
π§ Example Problems
Put some problem cards here
Answer Types:
πΉ Strict symbolic expressions (e.g., \sqrt{\frac{2g}{3R}})
πΉ Multiple equivalent forms accepted
πΉ No numerical approximations or equation chains
π οΈ Data Curation
3-Stage Rigorous Validation Pipeline
Expert Creation & Strict Screening
- 178 PKU physics students contributed problems that are:
- Almost entirely original/custom-created
- None easily found through direct internet searches or standard reference materials
- Strict requirements:
- Single unambiguous symbolic answer (e.g.,
T=2mg+4mvβΒ²/l) - Text-only solvability (no diagrams/multimodal inputs)
- Rigorously precise statements to avoid ambiguity
- Solvable using only basic physics principles (no complex specialized knowledge required)
- Single unambiguous symbolic answer (e.g.,
- No requirements on AI test to avoid filtering for AI weaknesses
- 178 PKU physics students contributed problems that are:
Multi-Round Academic Review
- 3-tier verification process:
- Initial filtering: Reviewers assessed format validity and appropriateness (not filtering for AI weaknesses)
- Ambiguity detection and revision: Reviewers analyzed LLM-generated solutions to identify potential ambiguities in problem statements
- Iterative improvement cycle: Questions refined repeatedly until all LLMs can understand the question and follow the instructions to produce the expressions it believes to be right.
- 3-tier verification process:
Human Expert Finalization
- 81 PKU students participated:
- Each student independently solved 8 problems from the dataset
- Evaluate question clarity, statement rigor, and answer correctness
- Establish of human baseline performance meanwhile
π Evaluation Protocol
Machine Evaluation
Dual Metrics:
- Accuracy: Binary correctness (expression equivalence via SymPy simplification)
- EED Score: Continuous assessment of expression tree similarity
The EED Score evaluates the similarity between the model-generated answer and the ground truth by leveraging the concept of expression tree edit distance. The process involves the following steps:
Simplification of Expressions:Both the ground truth (
gt) and the model-generated answer (gen) are first converted into simplified symbolic expressions using thesympy.simplify()function. This step ensures that equivalent forms of the same expression are recognized as identical.Equivalence Check:If the simplified expressions of
gtandgenare identical, the EED Score is assigned a perfect score of 100, indicating complete correctness.Tree Conversion and Edit Distance Calculation:If the expressions are not identical, they are converted into tree structures. The edit distance between these trees is then calculated using an extended version of the Zhang-Shasha algorithm. This distance represents the minimum number of node-level operations (insertions, deletions, and updates) required to transform one tree into the other.
Relative Edit Distance and Scoring:The relative edit distance ( r ) is computed as the ratio of the edit distance to the size of the ground truth tree. The EED Score is then determined based on this relative distance:
- If ( r = 0 ) (i.e., the expressions are identical), the score is 100.
- If ( 0 < r < 0.6 ), the score is calculated as ( 60 - 100r ).
- If ( r \geq 0.6 ), the score is 0, indicating a significant discrepancy between the model-generated answer and the ground truth.
This scoring mechanism provides a continuous measure of similarity, allowing for a nuanced evaluation of the model's reasoning capabilities beyond binary correctness.
Key Advantages:
- 204% higher sample efficiency vs binary metrics
- Distinguishes coefficient errors (30<EED score<60) vs structural errors (EED score<30)
Human Baseline
- Participants: 81 PKU physics students
- Protocol:
- 8 problems per student: Each student solved a set of 8 problems from PHYBench dataset
- Time-constrained solving: 3 hours
- Performance metrics:
- 61.9Β±2.1% average accuracy
- 70.4Β±1.8 average EED Score
- Top quartile reached 71.4% accuracy and 80.4 EED Score
- Significant outperformance vs LLMs: Human experts outperformed all evaluated LLMs at 99% confidence level
- Human experts significantly outperformed all evaluated LLMs (99.99% confidence level)
π Main Results
The results of the evaluation are shown in the following figure:
Significant Performance Gap: Even state-of-the-art LLMs significantly lag behind human experts in physical reasoning. The highest-performing model, Gemini 2.5 Pro, achieved only a 36.9% accuracy, compared to the human baseline of 61.9%.
EED Score Advantages: The EED Score provides a more nuanced evaluation of model performance compared to traditional binary scoring methods.
Domain-Specific Strengths: Different models exhibit varying strengths in different domains of physics:
- Gemini 2.5 Pro shows strong performance across most domains
- DeepSeek-R1 and o3-mini (high) shows comparable performance in mechanics and electricity
- Most models struggle with advanced physics and modern physics
Difficulty Handling: Comparing the advantage across problem difficulties, Gemini 2.5 Pro gains a pronounced edge on harder problems, followed by o3 (high).
π΅βπ« Error Analysis
We categorize the capabilities assessed by the PHYBench benchmark into two key dimensions: Physical Perception (PP) and Robust Reasoning (RR):
- Physical Perception (PP) Errors: During this phase, models engage in intensive semantic reasoning, expending significant cognitive effort to identify relevant physical objects, variables, and dynamics. Models make qualitative judgments about which physical effects are significant and which can be safely ignored. PP manifests as critical decision nodes in the reasoning chain. An example of a PP error is shown in Example Problem 1.
- Robust Reasoning (RR) Errors: In this phase, models produce numerous lines of equations and perform symbolic reasoning. This process forms the connecting chains between perception nodes. RR involves consistent mathematical derivation, equation solving, and proper application of established conditions. An example of a RR error is shown in Example Problem 2.
π© Citation
@misc{qiu2025phybenchholisticevaluationphysical,
title={PHYBench: Holistic Evaluation of Physical Perception and Reasoning in Large Language Models},
author={Shi Qiu and Shaoyang Guo and Zhuo-Yang Song and Yunbo Sun and Zeyu Cai and Jiashen Wei and Tianyu Luo and Yixuan Yin and Haoxu Zhang and Yi Hu and Chenyang Wang and Chencheng Tang and Haoling Chang and Qi Liu and Ziheng Zhou and Tianyu Zhang and Jingtian Zhang and Zhangyi Liu and Minghao Li and Yuku Zhang and Boxuan Jing and Xianqi Yin and Yutong Ren and Zizhuo Fu and Weike Wang and Xudong Tian and Anqi Lv and Laifu Man and Jianxiang Li and Feiyu Tao and Qihua Sun and Zhou Liang and Yushu Mu and Zhongxuan Li and Jing-Jun Zhang and Shutao Zhang and Xiaotian Li and Xingqi Xia and Jiawei Lin and Zheyu Shen and Jiahang Chen and Qiuhao Xiong and Binran Wang and Fengyuan Wang and Ziyang Ni and Bohan Zhang and Fan Cui and Changkun Shao and Qing-Hong Cao and Ming-xing Luo and Muhan Zhang and Hua Xing Zhu},
year={2025},
eprint={2504.16074},
archivePrefix={arXiv},
primaryClass={cs.CL},
url={https://arxiv.org/abs/2504.16074},
}
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