Bud Eval Datasets
Collection
OpenCompass-format evaluation datasets mirrored/curated by Bud Ecosystem for the Bud eval platform. • 66 items • Updated
Datasets:
id int64 10 766 | tag stringclasses 6
values | content stringlengths 191 2.91k | solution stringlengths 352 5.62k | answer stringlengths 16 445 |
|---|---|---|---|---|
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 huma... | 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 ... | \[v = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2}\] |
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 $S_1$ an... | 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
$$
E_{1x}=A,E_{2... | $$\gamma = \frac{2}{5}$$ |
285 | MECHANICS | A small bug with a mass of $m$ crawls on a disk with a radius of $2R$. Relative to the disk, its crawling trajectory is a circle of radius $R$ that passes through the center of the disk. The disk rotates with a constant angular velocity $\omega$ about an axis passing through its center and perpendicular to the plane of... | The problem can use the acceleration transformation formula for rotating reference frames, where the actual acceleration of the small insect
$$
\overrightarrow{a}=\overrightarrow{a}+2\overrightarrow{\omega}\times\overrightarrow{v}+\dot{\overrightarrow{\omega}}\times\overrightarrow{r}-\omega^{2}\overrightarrow{r}
$$
... | $$F_{\max} = 5m\omega^2R$$ |
321 | MECHANICS | The civilization of *Three-Body* changes the values of fundamental physical constants, such as the Planck constant, inside the "water drop" to alter the range of strong forces.
What kind of effects would this produce? This problem provides a suitable discussion about this question.
Hideki Yukawa pointed out that t... | $$
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}}
$$
Solving ... | $$\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 subseq... | 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... | $$\frac{3\sqrt{5}}{5}\sqrt{gr}$$ |
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. ... | (1)
Let:
$$
\varphi={\left\{\begin{array}{l l}{A_{1}{\frac{r}{R}}\cos\theta+B_{1}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(r<R)}
{A_{2}{\frac{r}{R}}\cos\theta+B_{2}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(R<r<k R)}\ {A_{3}{\frac{r}{R}}\cos\theta+B_{3}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(r>k R)}\end{array}\right.}
$$
(Tangen... | $$ M = -\frac{27}{31} m B_0 \sin \alpha $$ |
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 smal... | 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 bal... | $$T_2 = 2mg + 4m \frac{v_0^2}{l}$$ |
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 ele... | 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... | $$\frac{q b}{2\pi \varepsilon_0} \left[ ((a+x)^2 + (z-c)^2 + b^2)^{-3/2} - ((x-a)^2 + (z-c)^2 + b^2)^{-3/2} \right]$$ |
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 dista... | 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:
$$
\frac{d S}{2}\sigma T_{2}^{4}\int_{\mathrm{arcsin}\left({\frac{b \sin\theta}{D}}\right)-\mathrm{arcsin}({\frac{r}{D}})}^{\mathrm{arcsin}... | $$
p(\theta) = (\sigma T_2^4 - \sigma T_1^4) \frac{r(R - b \cos \theta)}{R^2 + b^2 - 2Rb \cos \theta}
$$ |
41 | MECHANICS | The rocket ascends vertically from the Earth's surface with a constant acceleration $a = k_1g_0$, where $g_0$ is the gravitational acceleration at the Earth's surface. Inside the rocket, there is a smooth groove containing a testing apparatus. When the rocket reaches a height $h$ above the ground, the pressure exerted ... | At the moment of takeoff, $N_1=(1+k_1)mg_0$, where $m$ is the mass of the tester.
When ascending to a height $h$, the gravitational acceleration changes to $g=g_0(\dfrac{R}{R+h})^2$
The pressure at this time is $N_2=mg_0(\dfrac{R}{R+h})^2+k_1mg_0$
Since $N_2=k_2N_1$, solving the equations together gives
$h=R(\dfr... | $$R\left(\frac{1}{\sqrt{k_2(1+k_1)-k_1}}-1\right)$$ |
453 | ELECTRICITY | Most media exhibit absorption of light, where the intensity of light decreases as it penetrates deeper into the medium.
Let monochromatic parallel light (with angular frequency $\omega$) pass through a uniform medium with a refractive index of $n$. Experimental results show that, over a reasonably wide range of light ... | The dynamics equation for electrons:
$$
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{\... | $$
\alpha=\frac{N}{6\pi n\epsilon_0^2c^4}\frac{(\frac{\omega^2 e^2}{m})^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
|
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... | 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 ... | $$\sqrt{\frac{2\sigma}{\rho g}(1-\sin \theta)}$$ |
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 lig... | 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:
$$
(C_{x},C_{y})=\Big(2m\Omega\dot{y},-2m\Omega(\dot{X}+\dot{x})\Bi... | $$
\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)}
$$ |
101 | MECHANICS | The problem discusses a stick figure model. The stick figure's head is a uniform sphere with a radius of $r$ and a mass of $m$, while the rest of the body consists of uniform rods with negligible thickness and a mass per unit length of $\lambda$. All parts are connected by hinges. Specifically, the torso and both arms ... |
(1) The forces are already marked on the diagram. List Newton's laws and rotational equations, assuming the displacement of the driven object is $\pmb{x}$, positive upwards.
Vertical direction of the leg
$$
T_{3}=1.2\lambda l\times\left(0.6l\times{\frac{d^{2}}{d t^{2}}}\left({\mathrm{cos}}\varphi\right)-{\ddot{x}}\... | $$
A = \frac{25 \sqrt{2}}{204} \theta_0
$$ |
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 tem... | 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 $$ \Delta{{H}_{0}... | $$P_0 \frac{(1+\alpha)(5+10\alpha+2\alpha \frac{\lambda}{R T_0})}{5+8\alpha}$$ |
134 | MECHANICS | A fox is escaping along a straight line $AB$ at a constant speed $v_{1}$. A hound is pursuing it at a constant speed $v_{2}$, always aimed at the fox. At a certain moment, the fox is at $F$ and the hound is at $D$. $FD \bot AB$ and $FD = L$. Find the magnitude of the hound's acceleration at this moment. | To solve $\textcircled{1}$, the hunting dog performs uniform circular motion. Thus, the tangential acceleration is zero, while the normal acceleration changes. Assume that during a very short period of time $\mathrm{d}t$ after the given moment, the curvature radius of the trajectory of the hunting dog is $\rho,$ then t... | $$\frac{v_1 v_2}{L}$$ |
650 | MECHANICS | Two parallel light strings, each with length $L$, are horizontally separated by a distance $d$. Their upper ends are connected to the ceiling, and the lower ends are symmetrically attached to a uniformly distributed smooth semicircular arc. The radius of the semicircle is $R$, and its mass is $m$. The gravitational acc... | Let the angle between the rope and the vertical be $\theta$, and the angle between the line connecting the small ball and the center of the circle and the vertical be $\varphi$. First, consider the kinetic equations. The kinetic energy $$ T={\frac{1}{2}}m L^{2}{\dot{\theta}}^{2}+{\frac{1}{2}}m(L{\dot{\theta}}+R{\dot... | $$\frac{2}{\pi}\sqrt{\frac{E_0 R}{m g}}\sqrt{(\pi-2)^2\frac{R}{L}+2}$$ |
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 o... | 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 $$ \overrightarrow{O B}=(\cos\beta,\sin\beta\sin\varphi,\s... | $$
\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... |
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 co... | 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 G... | $$
i(t)=\frac{U}{r_0\left(1+\frac{2\arccosh(D/R)}{\pi r_0 L(\sigma_1+\sigma_2)}\right)}\left(2+\frac{2\arccosh(D/R)}{\pi r_0 L(\sigma_1+\sigma_2)}-\exp\left[-\left(\frac{\sigma_1+\sigma_2}{\varepsilon_1+\varepsilon_2}+\frac{2\arccosh(D/R)}{\pi r_0 L(\varepsilon_1+\varepsilon_2)}\right)t\right]\right)
$$ |
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 e... | 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 for... | $$
\boxed{\sigma=\frac{2\varepsilon_{0}E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$ |
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,
$$
v_{x} = v \cos\theta,
$$... | $$\frac{v^3}{R v_x}$$ |
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 cy... | 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 ellipti... | $$
f=\frac{1}{2\pi}\sqrt{\frac{4g(b+R-\frac{b^2R}{a^2})}{(a^2+5b^2)\left(\frac{bR}{a^2}-1\right)}}
$$ |
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... | 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 b... | $$
w=\frac{\lambda N_0 E_0}{4\pi}\frac{\left(1-\left(\frac{v}{c}\right)^2\right)^2}{\left(1-\frac{v}{c}\cos\theta\right)^3}e^{-\frac{4\lambda}{7}}
$$ |
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
$$
\nabla\cdot{\pmb{E}}=\frac{\rho}{\varepsilon_{0}}
$$
$$
\nabla\times{\pmb{{E}}}=-\frac{\partial{\pmb{{B}}}}{\partial t}
$$
$$
\nabla\cdot\pmb{B}=0
$$
... | $$
\pmb{\varepsilon}\rightarrow\pmb{\varepsilon}
$$
It can be obtained:
$$
{\pmb B}\rightarrow\mu_{0}{\pmb B}c
$$
$$
\nabla\cdot\pmb{\cal E}=\rho
$$
$$
\nabla\times\pmb{\mathcal{E}}=-\frac{1}{c}\frac{\partial\pmb{\mathcal{B}}}{\partial t}
$$
$$
\nabla\cdot\pmb{B}=0
$$
$$
\nabla\times\pmb{B}=\frac{1}{c}\pmb{j}+\fr... | $$\lambda = -c$$ |
617 | MECHANICS | Consider each domino as a uniform, smooth slender rod with height $h$ and mass $m$ (neglecting thickness and width), connected to the ground with a smooth hinge. All rods are initially vertical to the ground and arranged evenly in a straight line, with the distance between adjacent rods being $l$, where $l \ll h$. The ... | Suppose the initial angular velocity of the $n$th rod when it falls is $\omega _{n}$. From the beginning of the fall until it collides with the next rod, the angle through which the rod rotates is denoted as $\theta$. Since $l\ll h$, we have $$\theta\approx\tan\theta\approx\sin\theta=\frac lh$$ Rigid body dynamics: ... | $$\omega_\infty = \frac{l}{h} \sqrt{\frac{g}{2h}}$$ |
702 | ELECTRICITY | To establish a rectangular coordinate system in an infinitely large three-dimensional space, there are two uniformly positively charged infinite lines at $(a,0)$ and $(-a,0)$, parallel to the $z$-axis, with a charge line density of $\lambda$; and two uniformly negatively charged lines at $(0,a)$ and $(0,-a)$, also para... | Near the origin, the electric potential can be expressed as $$ \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\varepsil... | $$\sqrt{\frac{2q\lambda}{\pi\varepsilon_0 a^2 m}}$$ |
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 ori... | 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 objec... | $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4\pi \epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} $$ |
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... | 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}\... | \[
\Delta A = \frac{m A_0}{4M}\left[\frac{1}{3}(\gamma_1 - 1) + \gamma_2\right]
\] |
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 ... | 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:
$$
I=\frac{m l^{2}}{2}(\frac{1}{3}\sin^{2}\frac{\p... | $$
{\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})}}
$$ |
432 | ELECTRICITY | A homogeneous sphere with a mass of $m$ and a radius of $R$ carries a uniform charge of $Q$ and rotates around the $z$-axis (passing through the center of the sphere) with a constant angular velocity $\omega$. The formula for calculating the magnetic moment is $\sum_{i} I_{i} S_{i}$, where $I_{i}$ represents the curren... | Using the method of electric imaging and magnetic imaging, the electric force is given by
$$
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... | $$
h={\sqrt{\frac{-Q^{2}+{\sqrt{Q^{4}+{\frac{96\pi\varepsilon_{0}^{2}m g\mu_{0}Q^{2}R^{4}\omega^{2}}{25}}}}}{32\pi\varepsilon_{0}m g}}}
$$
|
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 $x$ d... | 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 $\widetil... | $$\varphi=2\arctan\left(\frac{\sqrt{\sin^2\theta-\left(1+\frac{\delta n}{n}\right)^2}}{\cos\theta}\right)$$ |
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 resistan... | 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 mot... | $$
T = 2\pi \sqrt{\frac{\rho_1 H}{\rho_2 g}}
$$ |
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 fricti... | 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:
$$
\begin{array}
{r l}{m a_{x}+M A_{x}=0.}&{(1)}\\
\ \frac{a_{x}-A_{x}}{a_{y}}=\cot \theta&{(2)}\\
... | $$
T=2\pi\sqrt{\frac{m(M+m\sin^2\theta)}{k(M+m)}}
$$ |
55 | ELECTRICITY | In a vacuum, there is an infinitely long, uniformly charged straight line fixed in place, with a charge line density of $\lambda$. Additionally, there is a dust particle with mass $m$, which can be considered as an isotropic, uniform dielectric sphere with a volume $V$, and a relative permittivity $\varepsilon_{r}$. It... | Since the volume of the medium sphere $V$ is very small, we can approximate the external electric field $E$ at the medium sphere as a uniform field. Therefore, the medium sphere is uniformly polarized, and let the polarization intensity inside the sphere be $P$. The polarization surface charge density follows the cosin... | $$
-\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... | 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... | \[
R_{s}=\left(1+\left(\frac{2\mathrm{e}^{-\frac{2\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}{1 - \mathrm{e}^{-\frac{4\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}\frac{2\sqrt{(1 - \sin^{2}i_{2})(\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\right)^{2}\right)^{-1}
\] |
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 th... | 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 i... | $ 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 t... | **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 spher... | $$
\frac{V_0}{\ln\left(\frac{\tan\left(\frac{\alpha_2}{2}\right)}{\tan\left(\frac{\alpha_1}{2}\right)}\right)}\ln\left(\frac{\tan\left(\frac{\alpha}{2}\right)}{\tan\left(\frac{\alpha_1}{2}\right)}\right)
$$ |
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 plan... | Analyzing the electric field at a small displacement $z$ from the equilibrium position along the $z$ axis
we obtain
$$
\vec{E_{z}}=\frac{Q}{4\pi\varepsilon_{0}}\left(\frac{l+z}{[R^{2}+(l+z)^{2}]^{\frac{3}{2}}}-\frac{l-z}{[R^{2}+(l-z)^{2}]^{\frac{3}{2}}}\right)
$$
$$
\vec{E_{z}}=\frac{Q}{2\pi\varepsilon_{0}}\frac{\... | $$
\sqrt{\frac{Q q}{2 \pi \varepsilon_0 m} \frac{R^2 - 2l^2}{(R^2+l^2)^{5/2}}}
$$ |
478 | ELECTRICITY | In an infinitely large, isotropic, linear dielectric medium, there exists a uniform external electric field $\vec{E}_{0}$. The vacuum permittivity is given as $\varepsilon_{0}$.
The dielectric medium is a liquid dielectric with a relative permittivity of $\varepsilon_{r}$. A solid, ideal conducting sphere with a radi... | Spatial Electric Field Distribution
$$
\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
$$
Total Surface Charge Density on the Conductor Sphere (including free charge and polarization charge):... | $$
F=\frac{1}{\varepsilon_{r}}QE_{0}
$$ |
71 | ELECTRICITY | Given a particle with charge $q$ and mass $m$ moving in an electric field $\pmb{E}=E_{x}\pmb{x}+E_{z}\pmb{z}$ and a magnetic field $\pmb{B}=B\pmb{z}$. The initial conditions are: position $(x_{0},y_{0},z_{0})$ and velocity $(v_{\perp}\cos\delta,v_{\perp}\sin\delta,v_{z})$.
We know that a particle in a uniform magnetic... | This problem has been modified; the original problem had four questions.
If there is an electric field present, we find that the motion of the particle will be a combination of two movements: the normal circular Larmor gyration and a drift towards the center of guidance. We can choose the $\pmb{x}$ axis along the dire... | $$
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 mom... | 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 ... | $$\frac{2 v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}$$ |
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 i... | 【Solution】Let 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 f... | $$
T = \frac{5 + \cos^2 \theta}{(1 + \cos^2 \theta)^2} m g \sin \theta
$$ |
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 tu... | 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 ha... | $$\omega_B = 4 \sqrt{\frac{13 k q Q}{23 m L^3}}$$ |
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.... | 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 ... | $$
R_{04} = (3 - \sqrt{3})R
$$ |
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 b... | 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{\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}$$ |
110 | MECHANICS | A homogeneous picture frame with a light string, string length $2a$, frame mass $m$, length $2c$, and width $2d$, is hanging on a nail. Ignoring friction, with gravitational acceleration $g$. The mass of the light string is negligible, and it is inextensible, with its ends connected to the two vertices of one long side... | As shown in the figure, the trajectory of the nail is an ellipse $\begin{array}{r}{\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{b^{2}}=1}\end{array}$ where $b={\sqrt{a^{2}-c^{2}}}$. It is easy to know from geometric relationships that the angle between the line connecting the nail and the center of mass of the picture frame a... | $$
\alpha = \arctan\left(\frac{\sqrt{c^4 - d^2(a^2 - c^2)}}{ad}\right)
$$ |
256 | MODERN | Two relativistic particles X, each with rest mass $M$, experience a short-range attractive force $F(r) = \alpha/r^2$ (where $\alpha$ is a positive constant) in the zero momentum reference frame C, and are bound by this short-range attractive force to form a pair $\mathrm{X_{2}}$. The speed of light in a vacuum is $c$, ... | In the center-of-mass system, the momentum of the two particles X is the same, denoted as $p$, and at this time, there exists the angular momentum quantization condition:
$$
p\times{\frac{r}{2}}\times2=p r=n\hbar,n\in\mathbb{N}
$$
According to the dynamics equation, the angular velocity $\omega=2v/r$. Based on Ne... | $$
E_1 = \frac{\sqrt{3}}{8}Mc^2
$$ |
112 | ELECTRICITY | Initially, a conductive dielectric sphere with a free charge of 0 is placed in a vacuum. It is known that the radius of the conducting sphere is $R$, its relative permittivity is $\varepsilon_{r}$, and its conductivity is $\sigma$. At the moment $t=0$, a uniform external field ${{\vec{E}}_{0}}$ is applied around the co... | Considering the moment when $\ell\to0^{+}$, the polarization properties of the dielectric are prioritized over the conductive properties. At this time, the potential distribution is equivalent to the polarization of a dielectric sphere with a relative dielectric constant of $\varepsilon_{\mathsf{r}}$ in a uniform exter... | $$
Q_{tot} = \frac{6 \pi \varepsilon_0}{\varepsilon_r + 2} R^3 E_0^2
$$ |
505 | MECHANICS | It is often observed that when a puppy gets wet, it will vigorously shake its body, and with just a few quick shakes, it can rid itself of most of the water. In fact, the loose skin and longer fur of dogs provide a biological rationale for this behavior. However, this might drive you crazy when you are giving the dog a... | Take the $x$-axis as the polar axis, and the $z$-axis as the reference axis for the azimuthal angle, i.e.:
$$
\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 ... | $$
\boxed{y^{2}=\frac{\omega^{4}}{g^{2}}(R^{2}-x^{2})\left(x^{2}-R^{2}+\frac{g^{2}}{\omega^{4}}\right)}
$$ |
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... | Let the coordinates of the diffraction point be $(x,y)$, we can derive the incident wave vector and the reflected wave vector
$$
\left\{\begin{array}{c}{\displaystyle\boldsymbol{k}=\frac{2\pi}{\lambda}\hat{\boldsymbol{z}}}\ {\displaystyle\boldsymbol{k}^{\prime}=\frac{2\pi}{\lambda}\frac{x\hat{\boldsymbol{x}}+y\hat{\... | $$
\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 t... | 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 ... | \[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \] |
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 $q$... | (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: $$ dL-\frac{B_0q}{R^n}r^{n+1}d... | $$
\frac{2\pi}{\sqrt{n+1}} \frac{m}{q B_0}
$$ |
636 | ELECTRICITY | Establish a Cartesian coordinate system Oxyz, with a hypothetical sphere of radius $R$ at the origin. Place $n$ rings of radius $R$ along the meridional circles, all passing through the points (0, 0, R) and (0, 0, -R). The angle between any two adjacent rings is $\frac{\pi}{n}$, and each ring is uniformly charged with ... | Consider the case of a single circular loop, establishing a spherical coordinate system $(r, \alpha)$ with the loop axis as the polar axis. Then: $$ 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({... | $$-\frac{3\pi R^2}{2r_0^2}\sqrt{\frac{4\pi\varepsilon_0 m r_0^3}{n Q q}}$$ |
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 grav... | 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 N... | $$
\theta=\theta_{0}\frac{x_{0}}{x}
$$ |
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... | 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}_{\mathr... | $$
\frac{3\sin(\theta_1-\theta_2)\left[2g\cos\theta_1+2l_1\omega_1^2+l_2\omega_2^2\cos(\theta_1-\theta_2)\right]}{l_2\left[1+3\sin^2(\theta_1-\theta_2)\right]}
$$ |
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. Throug... | 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 acceler... | $$
\beta=4A \sqrt{A^2 - B^2} g\frac{2A^{4}+2A^{3}\sqrt{A^2 - B^2}-A^{3}B+B^{4}}{(2A^{3}+B^{3})^{2}B}
$$ |
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 t... | 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 f... | $$
x = \sqrt{\frac{2m}{k T}} \cdot \frac{L c \lambda_2}{2\pi d}
$$ |
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 t... | 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 $$ \vec{B_{z}}=\frac{\mu_{0}Q\Omega}{2\pi}\fra... | $$
\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}}
$$ |
688 | ELECTRICITY | In space, there is an axisymmetric magnetic field, with the direction of the magnetic field pointing outward perpendicular to the plane, and its magnitude depends only on the distance from the center of symmetry, $B(r) = B_{0} \left(\frac{r}{r_{0}}\right)^{n}$. A particle with mass $m$ and charge $q$ moves in a circula... | According to symmetry, the system in this problem has conserved canonical angular momentum with respect to the magnetic field's center of symmetry. To find the canonical angular momentum, we list the corresponding rate of change equation: $$ \frac{d L}{d t} = q v_{r} B_{0} \Big(\frac{r}{r_{0}}\Big)^{n} r $$ Rearrangi... | $$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}
$$
After extracting $-2r$, the remaining resistance value can be constructed using the forced current method. Inject $19I_{1}$ into node $B$, and each of the other n... | $$
R_{BC} = \frac{38}{41}r
$$ |
318 | MECHANICS | Consider a small cylindrical object with radius $R$ and height $h$. Determine the expression for the force $F$ acting on the cylinder when a sound wave passes through it. The axial direction of the cylinder is the direction of wave propagation. In the sound wave, the displacement of a particle from its equilibrium posi... | The force on the cylinder will be
$$
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... | $$
F = -\pi R^2 h \gamma P_0 k^2 A \cos(kx) \cos(2\pi ft)
$$ |
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 ... | 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... | $$\gamma B(2JS + \gamma B)$$ |
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 par... | 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 th... | $$2\left(\pi+\frac{2a}{b}\right)\sqrt{\frac{m}{k}}$$ |
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