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PHY-016 | Physics | null | {
"en": "The refractive index of the glass plates in the glass plate stack is \\(n = 1.48\\), and it is placed in air. Natural light is incident at the Brewster angle. In order to make the degree of polarization of the polarized light emerging from the glass plate stack reach 98%, how many glass plates are required at least? Disregard the loss of light energy caused by absorption, scattering, etc., and also disregard multiple reflections.",
"zh": "玻片堆玻璃板的折射率\\(n = 1.48\\)，放置在空气中，自然光以布儒斯特角入射。 为了使得从玻片堆射出的偏振光达到 98% 的偏振度，试问至少需要多少块玻璃板？不考虑由吸收和散射等原因引起的光能损失，也不考虑多次反射。"
} | 18 |
PHY-017 | Physics | null | {
"en": "A particle starts from the origin (-a, 0) and is only acted upon by a central potential field V(r). Now, we hope to construct an orbit such that the particle can reach the point (0, b) in the shortest time on this orbit. If the parametric equation of this observed orbit is \\(r = \\frac{p}{1 + e\\cos\\theta}\\), try to find the expression of the potential field V(r) (given \\(V(+\\infty)=V_{1}\\), \\(V(p) = V_{2}\\))",
"zh": "粒子从原点(-a,0)出发，仅受有心势场V(r)作用。现在希望构建轨道使得粒子能在该轨道上到达点(0,b)耗时最少，如果已观测到该轨道的参数方程为r=$\\frac{p}{1+ecosθ}$,试求势场V(r)的表达式（给定V(+∞)=$V_{1}$,V(p)=$V_{2}$)"
} | V(r) =$ V_1 - \frac{(V_1-V_2)*(1-e^2)}{1-e^2+\frac{2*r*e^2}{p}}$ |
PHY-018 | Physics | null | {
"en": "A mixture of \\(n\\)-dimensional ideal gases is in an equilibrium state at temperature \\(T\\). The relative velocity between any two molecules with masses \\(m_1\\) and \\(m_2\\) is defined as \\(\\boldsymbol{u}=\\boldsymbol{v}_1 - \\boldsymbol{v}_2\\), where \\(\\boldsymbol{v}_1\\) and \\(\\boldsymbol{v}_2\\) are the velocities of \\(m_1\\) and \\(m_2\\) respectively. Try to find: the root-mean-square value and the average value of the relative speed, that is, find \\(\\sqrt{\\overline{u^{2}}}\\) and \\(\\overline{u}\\).",
"zh": "混合的n维理想气体处于温度为T的平衡态，其中任意两个质量分别为\\(m_1\\)和\\(m_2\\)的分子之间的相对速度定义为\\(\\boldsymbol{u}=\\boldsymbol{v}_1 - \\boldsymbol{v}_2\\)，式中\\(\\boldsymbol{v}_1\\)和\\(\\boldsymbol{v}_2\\)分别是\\(m_1\\)和\\(m_2\\)的速度 。试求：相对速率的方均根值和平均值，即求\\(\\sqrt{\\overline{u^{2}}}\\)和\\(\\overline{u}\\)。"
} | \sqrt{\overline{u^2}}=\sqrt{\frac{nkT(m_1+m_2)}{m_1m_2}},\overline{u}=\sqrt{\frac{2kT(m_1+m_2)}{m_1m_2}}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}. |
PHY-019 | Physics | null | {
"en": "The \\(H_{\\alpha}\\) line in the Balmer series refers to the spectral line emitted when a hydrogen atom makes a transition from the \\(n = 3\\) energy level to the \\(n = 2\\) energy level, and its wavelength is \\(656.280\\ \\text{nm}\\). The hydrogen isotope deuterium also has a similar Balmer series, and the experimentally measured wavelength of the \\(D_{\\alpha}\\) spectral line is \\(656.101\\ \\text{nm}\\). Let the ratio of the mass of the deuteron to the mass of the hydrogen nucleus obtained in this way be \\(\\alpha\\), find the value of \\(\\beta = 1000(\\alpha - 2)\\), and keep one significant figure.",
"zh": "巴耳末系中的\\(H_{\\alpha}\\)线是指氢原子从\\(n = 3\\)能级跃迁到\\(n = 2\\)能级时发出的谱线，它的波长为\\(656.280\\ \\text{nm}\\)。氢的同位素氘也有类似的巴耳末系，实验测得其中的\\(D_{\\alpha}\\)谱线的波长为\\(656.101\\ \\text{nm}\\)。设以此求得的氘核和氢核的质量比为\\alpha，求\\beta=1000(\\alpha-2)的值，保留一位有效数字。"
} | 3 |
PHY-020 | Physics | null | {
"en": "Consider light passing through a periodic stack of 2N double-layer dielectric films. The thickness of film 1 is \\(d_1\\), and its refractive index is \\(n_{1}\\). The thickness of film 2 is \\(d_2\\), and its refractive index is \\(n_{2}\\). The optical system is arranged periodically as film 1, film 2, film 1, film 2, ... for a total of 2N layers. Assume that at the incident side, the angle of incidence is \\(θ_{0}\\) and the refractive index is \\(n_{0}\\). Inside dielectric film 1, the angle of refraction is \\(θ_{1}\\), inside dielectric film 2, the angle of refraction is \\(θ_{2}\\), and at the exit side, the angle of emergence is \\(θ_{3}\\) with a refractive index of \\(n_{3}\\). Find the reflection coefficient \\(r\\) and the transmission coefficient \\(t\\).",
"zh": "考虑光穿过2N层周期性摆放的双层介质薄膜，薄膜1的厚度为d_1,折射率为$n_{1}$,薄膜2的厚度为d_2,折射率为$n_{2}$,光学系统按照薄膜1，薄膜2，薄膜1，薄膜2.。。。的规律周期性放置一共2N层。假定入射处入射角为$θ_{0}$，折射率为$n_{0}$,介质膜1内折射角为$θ_{1}$，介质膜2内折射角为$θ_{2},出射处出射角为$θ_{3}$,折射率为$n_{3}$，求光的反射系数r和透射系数t"
} | $r= \frac{(a + b*n_3*\cos\theta_3)*n_0*\cos\theta_0-(c + d*n_3*\cos\theta_3)}{(a + b*n_3*\cos\theta_3)*n_0*\cos\theta_0+(c + d*n_3*\cos\theta_3)}$
$t = \frac{2*n_0*\cos\theta_0}{(a + b*n_3*\cos\theta_3)*n_0*\cos\theta_0+(c + d*n_3*\cos\theta_3)}$
where,
$$
\begin{cases}
a = \left[ \cos\beta_1 \cos\beta_2 - \frac{n_2*\cos\theta_2}{n_1*\cos\theta_1} \sin\beta_1 \sin\beta_2 \right] \times \frac{\mathscr{U}_{N - 1}(x)-\mathscr{U}_{N - 2}(x)}{n_1*\cos\theta_1} \\
b=-i \left[ \frac{1}{n_2*\cos\theta_2} \cos\beta_1 \sin\beta_2+\frac{1}{n_1*\cos\theta_1} \sin\beta_1 \cos\beta_2 \right] \times \mathscr{U}_{N - 1}(x) \\
c=-i \left[ n_1*\cos\theta_1 \sin\beta_1 \cos\beta_2 + n_2*\cos\theta_2 \cos\beta_1 \sin\beta_2 \right] \times \mathscr{U}_{N - 1}(x) \\
d = \left[ \cos\beta_1 \cos\beta_2-\frac{n_1*\cos\theta_1}{n_2*\cos\theta_2} \sin\beta_1 \sin\beta_2 \right] \times \left[ \mathscr{U}_{N}(x)-\mathscr{U}_{N - 2}(x) \right]
\end{cases}
$$
where, $\mathscr{U}_{n}(x)=\sin(n*\arccos x)/\sin(\arccos x)$
is the Chebyshev polynomial of the second kind
$x = \cos\beta_1 \cos\beta_2-\frac{1}{2} \left( \frac{n_1*\cos\theta_1}{n_2*\cos\theta_2}+\frac{n_2*\cos\theta_2}{n_1*\cos\theta_1} \right) \sin\beta_1 \sin\beta_2$
$\beta_i = 2*n_i*h_i*\cos\theta_i*\pi/\lambda$
for $i = 0,1,2,3$ |
PHY-021 | Physics | null | {
"en": "Two coherent parallel beams with the same wavelength \\(\\lambda\\) and an amplitude ratio of \\(\\alpha>1\\) are incident at an angle \\(\\theta\\). The observation screen is perpendicular to the angular bisector of the \\(\\theta\\) angle. Try to find: 1. The contrast \\(\\gamma\\) of the interference fringes on the screen. 2. The spacing \\(\\Delta x\\) of the interference fringes.",
"zh": "波长均为\\(\\lambda\\)，振幅比为\\(\\alpha>1\\)的两相干平行光束，以\\(\\theta\\)的夹角射入，观察屏幕与\\(\\theta\\)角的角平分线相垂直。试求：1. 幕上干涉条纹的反衬度\\gamma。2. 干涉条纹的间距\\Delta x。"
} | \[\gamma=\frac{2\alpha}{\alpha^2+1},\Delta x=\frac{\lambda}{2\sin\frac{\theta}{2}}\] |
PHY-022 | Physics | null | {
"en": "There is a thin spherical shell with radius \\(R\\) and uniform surface charge density \\(\\sigma\\), which rotates with a constant angular velocity \\(\\omega\\) about a fixed axis. Try to find the following under the condition that the divergence of the gauge \\(\\mathbf{A}\\) is \\(0\\):\n\n(1) The surface current vector distribution \\(j(\\theta)\\) of the spherical shell, and use this to find the distribution of the magnetic vector potential \\(\\mathbf{A}\\) in the whole space \\(A(r,\\theta)\\), where \\(r\\) is the distance from a certain point in space to the center of the sphere, and \\(\\theta\\) is the angle between the line connecting this point and the center of the sphere and the angular velocity vector.\n\n(2) Use \\(\\mathbf{A}\\) to find the magnetic field distribution \\(\\mathbf{B}(r,\\theta)\\) in space, and find the electric field distribution \\(\\mathbf{E}(r,\\theta)\\) in space.\n\n(3) Use the magnetic field and electric field to find the angular momentum of this electromagnetic field.",
"zh": "现有一半径为R,均匀带电面密度σ的薄球壳，以ω的角速度绕定轴做恒角速度转动，试在规范A的散度为0的情况下求\n(1)球壳的面电流矢量分布j(θ),并以此求全空间的磁矢势A的分布A(r,θ),r为空间某点到球心的距离，θ为该点与球心的连线和角速度矢量的夹角.\n(2)利用A求空间的磁场分布B(r,θ),并求出空间的电场分布E(r,θ)\n(3)利用磁场电场求出该电磁场的角动量"
} | (1)j(θ) = σωRsinθ\hat{φ}
If r>R then A(r,θ) = $\frac{μ_0*\pi*σωR^4*sinθ}{3r^2}\hat{φ}$
If r<R then A(r,θ) = $\frac{μ_0*\pi*σω*R*rsinθ}{3}\hat{φ}$
(2) If r>R, then B = $\frac{μ_0*\pi*σωR^4}{3r^2}(2cosθ\hat{r}+sinθ\hat{θ})$
E = $\frac{σR^2}{ε_0*r^2}\hat{r}$
If r<R, then B = $\frac{μ_0*\pi*σωR^4}{3r^2}(cosθ\hat{r}-sinθ\hat{θ})$
E = 0
(3) L = \frac{4(\pi)^2*σ^2*R^5*ω}{9} |
PHY-023 | Physics | null | {
"en": "A ring with radius \\(R_{1}\\) is uniformly charged with a total charge of \\(Q\\). To its right, there is a grounded conducting sphere with radius \\(R_{2}\\). The distance between the center of the sphere and the center of the ring is \\(L\\), and the line connecting the two centers is perpendicular to the plane of the ring. Find the total induced charge \\(Q'\\) on the sphere.",
"zh": "半径为\\(R_{1}\\)的圆环均匀带电，总电量为\\(Q\\)。在其右侧有一接地的半径为\\(R_{2}\\)的导体球面，球心与环心的间距为\\(L\\)，且球心与环心的连线垂直于圆环平面。求球面上的总感应电荷量\\(Q'\\)。"
} | \(Q'=-\frac{R_2}{\sqrt{R_1^{2}+L^{2}}}Q\) |
PHY-024 | Physics | null | {
"en": "A stepped grating is composed of $N$ stepped optical media on an opaque screen. The refractive index of the optical medium is $n$, the horizontal spacing between adjacent steps is $d$, and the vertical spacing is $h$. Given that the wavelength of parallel light incident perpendicularly to the plane of the stepped grating from infinity is $\\lambda$, and assuming the refractive index of air is 1, find the intensity distribution of the outgoing light $I(\\theta)$, where $\\theta$ is the diffraction angle.",
"zh": "某阶梯光栅由不透明屏上N条阶梯状光介质构成，光介质折射率为n，相邻两个阶梯的水平间距为d,纵向间距为h。给定无穷远处以与阶梯光栅的平面垂直的入射的平行光波长为λ，再假设空气折射率为1，求出射光强分布I(θ),其中θ为衍射角"
} | $I = I_0 \times \left(\frac{\sin\left(N\pi(nd - h\sin\theta - d\cos\theta)/\lambda\right)}{\sin\left(\pi(nd - h\sin\theta - d\cos\theta)/\lambda\right)}\right)^2\times\left(\frac{\sin\left(\pi\times h\sin\theta/\lambda\right)}{\pi\times h\sin\theta/\lambda}\right)^2$ |
PHY-025 | Physics | null | {
"en": "In a $k$-dimensional closed container, there is stored a $k$-dimensional ideal gas with molecular mass $m$. Suppose at equilibrium the temperature is $T$, and both the molecular velocity and speed follow the Maxwell distribution. The Boltzmann constant is $k_{B}$.\n\\subsection*{(1)}\nDenote the number density of molecules (that is, the number of molecules per $k$-dimensional space volume) as $n_k$. Try to find the number of molecules $\\Gamma$ that collide with a unit \"$k - 1$ dimensional area\" of the container wall per unit time.\n\\subsection*{(2)}\nDerive the relationship between the average speed $\\overline{v}$ of the $k$-dimensional ideal gas and the temperature $T$, and give the specific expression of $\\overline{v^2}\\sim T$.\n\\subsection*{(3)}\nA $k$-dimensional ideal gas with molecular mass $m$ is enclosed by a \"wall\" formed by a closed plane curve, and the gas is already in an equilibrium state at temperature $T$.\n\\subsubsection*{(3.1)}\nOpen a small slit on the \"wall\" of the closed container, then gas molecules can escape through the slit. Try to find the speed distribution $f_k(v)$ of the molecules in the ejected molecular beam, and then find the root-mean-square of the relative speed between pairs of particles.\n\\subsubsection*{(3.2)}\nDerive the average speed $\\overline{v_k}$ of the molecules in the ejected molecular beam $\\Gamma_k$. From this, deduce the heat capacity $C$ of the ejected gas.",
"zh": "$k$ 维空间闭合容器中贮存着分子质量为 $m$ 的 $k$ 维理想气体，设处于平衡态时温度为 $T$，分子的速度和速率都服从麦克斯韦分布。玻尔兹曼常数为k_{B}\n\\subsection*{(1)}\n将分子数密度（即单位 $k$ 维空间体积中的分子数）记为 $n_k$，试求单位时间内碰到单位\"面积\"容器壁上的分子数 $\\Gamma$。\n\\subsection*{(2)}\n导出 $k$ 维理想气体速率平均值 $\\overline{v}$ 与温度 $T$ 的关系式，并给出 $\\overline{v^2} \\sim T$ 具体表述式。\n\\subsection*{(3)}\n由平面闭合曲线\"壁\"围住的k维理想气体的分子质量为 $m$，气体已处于温度为 $T$ 的平衡态。\n\\subsubsection*{(3.1)}\n在闭合容器\"壁\"上开一条小缝，气体分子便可从小缝射出，试求出射分子束中分子的速率分布 $f_k(v)$,再求两两粒子之间的相对速率的均方根\n\\subsubsection*{(3.2)}\n导出 $\\Gamma_k$ 的出射分子束中分子的平均速率 $\\overline{v_k}$。由此推导出出射气体的热容C"
} | (1)n_k*\sqrt{\frac{k_B*T}{2*\pi*m}}
(2)$\overline{v}$ = $\sqrt{\frac{2k_{B}T}{\pi m}} \cdot \frac{\Gamma\left(\frac{k + 1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)}
$
$\overline{v^2}$ = k*k_{B}*T/m
(3)
(3.1)
f_k(v) = $v^k*exp(\frac{-m v^2}{2 k_{B} T}*\frac{2*m^((k+1)/2)}{\Gamma(\frac{k+1}{2})*(2*k_{B}*T)^((k+1)/2)}$
$\overline{v_r^2}$ = (2*(k+1)-\pi)*k_{B}*T/m
(3.2)
$\overline{v_k}$ = $\sqrt{\frac{2k_{B}T}{\pi m}} \cdot \frac{\Gamma\left(\frac{k + 2}{2}\right)}{\Gamma\left(\frac{k+1}{2}\right)}
C = (k+1)*k_{B}*T |
PHY-026 | Physics | null | {
"en": "Moist air flows adiabatically and continuously over a hill, going up the slope, crossing the peak, and then flowing down the slope. Meteorological stations H0 and H3 measure the atmospheric pressure as both being \\(p_0\\). At H0, the air temperature is \\(T_0\\); as the air rises to H1, the measured pressure is \\(p_1\\), and clouds start to form. The air continues to rise, and after time \\(t\\) it reaches the ridge at H2, where the pressure is \\(p_2\\). During this process, water vapor condenses into raindrops and falls. Assume that the mass of moist air over each square meter is \\((m_{wet})\\ kg\\), and for every kilogram of moist air, \\(m_{rain}\\) of rainwater condenses out.\n\nKnown conditions:\n- Air is an ideal gas, and the influence of water vapor on specific heat capacity and density can be ignored.\n- The specific heat capacity at constant pressure of air is \\(c_p\\).\n- The latent heat of vaporization of water is \\(l\\).\n- The adiabatic index is \\(\\gamma\\), and the acceleration due to gravity is \\(g\\).\n- The air density at H0 is \\(\\rho_0\\).\n\nProblems:\n(1) Find the air temperature \\(T_1\\) at the height of the bottom of the cloud layer, H1.\n(2) Assuming that the air density decreases linearly with height, find the height H1 of the bottom of the cloud layer.\n(3) Find the air temperature \\(T_2\\) at the ridge H2.\n(4) Find the air temperature \\(T_3\\) at H3.",
"zh": "潮湿空气绝热地持续流过山坡，翻越山峰后沿坡而下。气象站H0和H3测得大气压强均p_0。在H0处，空气温度为T_0；空气上升至H1处，测得压强为p_1，开始有云形成。空气继续上升，经过t后到达山脊H2处，压强为p_2，此过程中水蒸气凝结成雨滴落下。设每平方米上空潮湿空气质量为(m_wet)kg，每千克潮湿空气凝结出m_rain雨水。\n已知条件：\n空气为理想气体，水蒸气对比热容和密度的影响可忽略\n空气比定压热容cp\n水的汽化热l\n绝热指数γ，重力加速度g\nH0处空气密度ρ0\n问题：\n(1)求云层底部高度H1处的空气温度T1；\n(2)假设空气密度随高度线性减少，求云层底部高度H1；\n(3)求山脊H2处的空气温度T2；\n(4)求H3处的空气温度T3，"
} | (1)$T_{1} = T_{0} \left( \frac{p_{1}}{p_{0}} \right)^{1 - \frac{1}{\gamma}}$
(2)$h_{1} = \frac{2(\rho_{0} - \rho_{1})}{\rho_{0}g\left(1 + \frac{p_{1}T_{0}}{p_{0}T_{1}}\right)}$
(3)$T_{2} = T_{1} \left( \frac{p_{2}}{p_{1}} \right)^{1 - \frac{1}{\gamma}}+\frac{L*m_rain}{C_p}$
(4)$T_3 = T_{2} \left( \frac{p_{3}}{p_{2}} \right)^{1 - \frac{1}{\gamma}}$ |
PHY-027 | Physics | null | {
"en": "Two long straight wires, carrying opposite uniform line free charges ±λ, are situated on either side of a long dielectric cylinder . The cylinder (which carries no free net charge) has radius R, and the wires are a distance a (a>R) from the axis.The permittivity inside the cylinder is $ε_{1}$ and the permittivity outside the cylinder is $ε_{2}$.Find the potential.",
"zh": "Two long straight wires, carrying opposite uniform line free charges ±λ, are situated on either side of a long dielectric cylinder . The cylinder (which carries no free net charge) has radius R, and the wires are a distance a (a>R) from the axis.The permittivity inside the cylinder is $ε_{1}$ and the permittivity outside the cylinder is $ε_{2}$.Find the potential."
} | When r<R,
V(r,θ) = $\frac{λ)}{2\pi*(\varepsilon_{1}+\varepsilon_{2})}*ln(\frac{r^2+a^2-2racosθ}{r^2+a^2+2racosθ})$
When r>R
V(r,θ) = $\sum_{n = 1}^{\infty}(\frac{λ(\varepsilon_{1}-\varepsilon_{2})}{\varepsilon_{2}*\pi*(\varepsilon_{1}+\varepsilon_{2})}*cosnθ*(\frac{R^2}{a*r})^n*(1-(-1)^n)/n)+\frac{λ)}{4\pi*\varepsilon_{2}}*ln(\frac{r^2+a^2-2racosθ}{r^2+a^2+2racosθ})$ = $\frac{λ (ε_1 - ε_2)}{2 ε_2 \pi (ε_1 + ε_2)} \ln \frac{a^2 r^2 + 2 a r R^2 \cos θ + R^4}{a^2 r^2 - 2 a r R^2 \cos θ + R^4}+\frac{λ)}{4\pi*ε_{2}}*ln(\frac{r^2+a^2-2racosθ}{r^2+a^2+2racosθ})$ |
PHY-028 | Physics | null | {
"en": "An ideal electric dipole is situated at the origin, and points in the z direction. An electric charge is released from rest at a point in the x y plane. suppose the mass of electric charge is m,the charge is q,the dipole is p,the distance is R at the begin.\n(1)Show that it swings back and forth in a semi-circular arc, as though it were a pendulum supported at the origin.Find the period T of this system(use gamma function).\n(2)If the electric charge is given a start speed $v_{0}$ which points in the line between dipole and charge,,find the time function of the distance r(t) between them,where r(0)=R is given,also find the time cost for electric charge from θ=$\\frac{\\pi}{2}$ at the beginning to $-\\frac{\\pi}{2} $",
"zh": "An ideal electric dipole is situated at the origin, and points in the z direction. An electric charge is released from rest at a point in the x y plane. suppose the mass of electric charge is m,the charge is q,the dipole is p,the distance is R at the begin.\n(1)Show that it swings back and forth in a semi-circular arc, as though it were a pendulum supported at the origin.Find the period T of this system(use gamma function).\n(2)If the electric charge is given a start speed $v_{0}$ which points in the line between dipole and charge,,find the time function of the distance r(t) between them,where r(0)=R is given,also find the time cost for electric charge from θ=$\\frac{\\pi}{2}$ at the beginning to $-\\frac{\\pi}{2} $"
} | (1)Use energy is constant and force analyze we can get:
$\frac{(d^2)r}{dt^2}+\frac{(dr/dt)^2}{r}=0$
solve this differential equation we get r=R is the only solution which satisfies both differential equation and beginning state.
And use the energy function V = $\frac{pq}{4\pi*ε_{0}*r^2}$>0 we get θ∈[3\pi/2,\pi/2] is a semi_circular arc.
Use integration we get:
T = $2\pi*\Gamma{1/4}\sqrt{2ε_{0}mR^4/(pq)}/\Gamma{3/4}$
(2)r(t) = R+$v_{0}t$
t =$ 1/(1/R-\frac{\sqrt{pq}*\Gamma(3/4)}{\pi*\sqrt{2ε_{0}m}*\Gamma(1/4)})-R/v_{0}$ |
PHY-029 | Physics | null | {
"en": "A hemispherical shell-shaped bowl with mass \\(M\\) and inner radius \\(R\\) can slide and roll without friction on a horizontal plane. Inside the bowl, a uniform solid sphere with mass \\(m\\) and radius \\(r\\) undergoes pure rolling. Taking the position where the sphere is at the bottom of the bowl when the bowl is stationary as the equilibrium position, now a tangential small perturbation is given to the sphere such that the velocity vectors of both the shell and the sphere are confined within the plane of the perturbation. If the period of small vibrations of the system is \\(T\\), introduce the dimensionless parameter \\(\\chi = \\frac{T}{2\\pi\\sqrt{\\frac{r}{g}}}\\). If \\(\\alpha=\\frac{M}{m}=1\\), \\(\\beta=\\frac{R}{r}=2\\), please give the value of \\(\\chi\\), retaining 5 significant figures.",
"zh": "质量为 \\( M \\)、内半径为 \\( R \\) 的半球壳形碗可无摩擦地在水平面上滑动和滚动，碗内有一质量为 \\( m \\)、半径为 \\( r \\) 的均匀实心小球作纯滚动。以碗静止时小球位于碗底为平衡位置，现给小球一个切向微扰，使得球壳和小球的速度矢量都被限制在微扰平面内，若系统微小振动的周期为T，引入无量纲参数$\\chi=\\frac{T}{2\\pi\\sqrt{r}{g}}$，若取$\\alpha=\\frac{M}{m}=1, \\beta=\\frac{R}{r}=2$，请给出$\\chi$的值，保留5位有效数字。"
} | \[\chi=0.90696\] |
PHY-030 | Physics | null | {
"en": "In a vacuum diode, electrons are “boiled” off a hot cathode, at potential zero, and accelerated across a gap to the anode, which is held at positive\npotential $V_{0}$. The cloud of moving electrons within the gap (called space charge)\nquickly builds up to the point where it reduces the field at the surface of the cathode\nto zero. From then on, a steady current I flows between the plates.\nSuppose the plates are large relative to the separation (A>>d where A is the square of both plates and d is the distance of the two plates), so\nthat edge effects can be neglected. Then V, ρ, and v (the speed of the electrons) are\nall functions of x alone.\n(a) Write Poisson’s equation for the region between the plates.\n(b) Assuming the electrons start from rest at the cathode, what is their speed at point\nx, where the potential is V(x)?\n(c) In the steady state, I is independent of x. What, then, is the relation between\nρ and v?\n(d) Use these three results to obtain a differential equation for V, by eliminating\nρ and v.\n(e) Solve this equation for V as a function of x, $V_{0}$, and d. Plot V(x), and compare\nit to the potential without space-charge. Also, find ρ and v as functions of x.\n(f) Show that\nI = K*${V_{0}}^{\\frac{3}{2}}$\n, (2.56)\nand find the constant K. (Equation 2.56 is called the Child-Langmuir law.\nIt holds for other geometries as well, whenever space-charge limits the current.\nNotice that the space-charge limited diode is nonlinear—it does not obey Ohm’s\nlaw.)\n(g)Considering first-order relativistic corrections(eV/mc^2<<1), recalculate the above voltage-current relationship.",
"zh": "In a vacuum diode, electrons are “boiled” off a hot cathode, at potential zero, and accelerated across a gap to the anode, which is held at positive\npotential $V_{0}$. The cloud of moving electrons within the gap (called space charge)\nquickly builds up to the point where it reduces the field at the surface of the cathode\nto zero. From then on, a steady current I flows between the plates.\nSuppose the plates are large relative to the separation (A>>d where A is the square of both plates and d is the distance of the two plates), so\nthat edge effects can be neglected. Then V, ρ, and v (the speed of the electrons) are\nall functions of x alone.\n(a) Write Poisson’s equation for the region between the plates.\n(b) Assuming the electrons start from rest at the cathode, what is their speed at point\nx, where the potential is V(x)?\n(c) In the steady state, I is independent of x. What, then, is the relation between\nρ and v?\n(d) Use these three results to obtain a differential equation for V, by eliminating\nρ and v.\n(e) Solve this equation for V as a function of x, $V_{0}$, and d. Plot V(x), and compare\nit to the potential without space-charge. Also, find ρ and v as functions of x.\n(f) Show that\nI = K*${V_{0}}^{\\frac{3}{2}}$\n, (2.56)\nand find the constant K. (Equation 2.56 is called the Child-Langmuir law.\nIt holds for other geometries as well, whenever space-charge limits the current.\nNotice that the space-charge limited diode is nonlinear—it does not obey Ohm’s\nlaw.)\n(g)Considering first-order relativistic corrections(eV/mc^2<<1), recalculate the above voltage-current relationship."
} | (a)$\laplacian{V} = -\frac{\rho(x)}{\epsilon_0} \quad \Rightarrow \quad \frac{d^2 V}{dx^2} = -\frac{\rho(x)}{\epsilon_0}
$
(b)$ v(x) = \sqrt{\frac{-2eV(x)}{m}}$
(c)$I = JA = \rho(x)v(x)A \Rightarrow \rho(x) = \frac{I}{Av(x)}
$
(d)$\frac{d^2V}{dx^2} = \frac{I}{A\epsilon_0}\sqrt{\frac{m}{2e}}*V^{-1/2}
$
(e)$V(x) = V_0\left(\frac{x}{d}\right)^{4/3}$
(f)$K = \frac{4\epsilon_0 A}{9d^2}\sqrt{\frac{2e}{m}}$
(g) $V_0 = (\frac{9I*d^2*\sqrt{\frac{m}{2e}}}{4\epsilon_0 A})^{(2/3)}+\frac{e*(\frac{9I*d^2*\sqrt{\frac{m}{2e}}}{4\epsilon_0 A})^{(4/3)}}{14mc^2}$ |
PHY-031 | Physics | null | {
"en": "A spherical hot air balloon has a total mass (including a well-insulated balloon skin, a basket, and other loads) of 350 kg. After heating, the balloon expands to its maximum volume with a diameter of 20 m. Assume the gas composition inside and outside the balloon is the same, and the gas pressure inside the balloon is slightly higher than atmospheric pressure. Given that the atmospheric temperature is 27 °C, the pressure is 1 atm, and the density of air under standard conditions is 1.3 kg/m³. What should the temperature of the air inside the balloon be when the hot air balloon is just able to rise?",
"zh": "一球形热气球，总质量（包括隔热很好的球皮以及吊篮等装载）为 350 kg。经加热后，气球膨胀到最大体积，其直径为 20m。设球内外气体成份相同，球内气体压强稍高于大气压 。已知大气温度为 27 ℃，压强为 1 atm，标准状态下空气的密度为 1.3 kg/m³。试问热气球刚好能上升时，球内空气的温度应为多少？"
} | 322.8K |
PHY-032 | Physics | null | {
"en": "An infinitely long charged straight line with linear charge density $\\lambda$ is located at $(-a,0)$, and an infinitely long grounded cylindrical conductor with radius $r < a$ whose axis is parallel to the line charge is located at $(a,0)$. Find the electric potential distribution $V(x,y)$ in space.",
"zh": "线电荷密度为$\\lambda$的无限长带电直线位于$(-a,0)$，轴线与其平行的半径为$r<a$的无限长圆柱形接地导体位于$(a,0)$，求空间中的电势分布$V(x,y)$."
} | \frac{-\lambda}{4\pi\varepsilon_0}\ln[\frac{r^2}{4a^2}\frac{(x+a)^2+y^2}{[x-(1-\frac{r^2}{2a^2})a]^2+y^2}] |
PHY-033 | Physics | null | {
"en": "When the distance between two atoms is small, the interaction between them manifests as a strong repulsion; when the distance is large, the interaction manifests as a weak attraction. A large number of atoms can combine into a crystal through this interaction. When the temperature approaches 0K, atoms are arranged in a periodic spatial lattice and are in a state of force equilibrium; when the temperature is higher than 0K, atoms will perform small-amplitude vibrations near their equilibrium positions. Consider a cubic crystal composed of atoms of the same mass $M$, and use the following model to discuss how atoms combine into a crystal and the influence of atomic vibrations in the crystal on its thermal properties.\n\n(1) The interaction potential energy $V(R)$ between two atoms at a distance $R$ can be approximately expressed as the Lennard-Jones potential\n$$\nV(R)=\\epsilon\\left[\\left(\\frac{r_0}{R}\\right)^{2u}-\\left(\\frac{r_0}{R}\\right)^{2v}\\right]\n$$\nwhere $\\epsilon > 0$ characterizes the interaction strength, $r_{0}$ characterizes the range of force, and both $u$ and $v$ are positive integers. Assume that at equilibrium, atoms are arranged at the vertices of a cubic lattice, that is, the equilibrium position of an atom is \n$$R_{u^{u}}=\\left(v_{1} a_{1} a_{1} \\right{a} \\left( \\frac{\\pi}{R} ）^{2} - \\right) \\right]$$\nwhere $a$ is the lattice constant. For simplicity, assume that the form of the Lennard-Jones potential remains valid until the crystal melts, and the parameters $\\epsilon$ and $r_{o}$ are also considered unchanged. Try to derive an expression for the lattice constant $a$. The calculation result can contain the following constant:\n$A_{n}=$ $$ A_n=\\sum_{x,y,z}(x^2 + y^2+z^2)^{-n/2} $$\nwhere $n$ is an arbitrary positive integer, and $\\sum$ represents the summation over all $x,y,z$ not all zero.\n\n(2) Try to calculate the contributions of nearest-neighbor atoms, next-nearest-neighbor atoms, and next-next-nearest-neighbor atoms to the value of $A_n$ respectively.\n\n(3) It is very complicated to strictly solve the small-amplitude vibrations of atoms near the equilibrium position $$R_{u|n|_{x|}}$$ in the crystal. To simplify, Einstein assumed that the motion of atoms does not interfere with each other, that is, when considering the vibration of any one atom, assume that all other atoms are stationary at their respective equilibrium positions. Under this model, find the angular frequency $\\omega$ of the small-amplitude vibration of each atom in the above crystal (the result can contain $a$ and $A_{n}$).\n(4) On the basis of (3), if the potential parameter $r_0$ quasi-statically and slowly changes to $2r_0$, try to find the rate of change of the micro-vibration amplitude.",
"zh": "当两个原子之间的距离较小时，两者之间的相互作用表现为强的排斥；距离较\n大时，其相互作用表现为弱的吸引。大量原子可通过此相互作用结合成晶体。温度趋于0K\n时，原子排列为周期性的空间点阵，处于力平衡状态；温度高于0K时，原子将在平衡位置\n附近做小幅振动。考虑由质量为$M$的同种原子组成的立方晶体，试用如下模型讨论原子如\n何结合成晶体以及晶体中原子振动对其热学性质的影响。\n\n(1)距离为$R$的两个原子之间的相互作用势能$V(R)$可以近似表示为伦纳德-琼斯势\n$$\nV(R) = \\epsilon \\left[ \\left(\\frac{r_0}{R}\\right)^{2u} - \\left(\\frac{r_0}{R}\\right)^{2v} \\right]\n$$\n其中$\\epsilon>0$表征相互作用强度，$r_{0}$表征力程,u,v均为正整数，假设平衡时原子排列在立方点阵的顶点上，即\n原子的平衡位置为\n$$R_{u^{u}} = \\left(v_{1} a_{1} a_{1} \\right{a} \\left( \\frac{\\pi}{R} ）^{2} - \\right) \\right]$$\n其中$a$是晶格常数。为简单起见，假设伦纳德琼斯势的形式在晶体熔化前一直成立，参量$\\epsilon$\n和$r_{o}$也视为不变。试导出晶格常数$a$的表达式。计算结果中可包含如下常数：\n$A_{n}$ = $$ A_n = \\sum_{x,y,z} (x^2 + y^2 + z^2)^{-n/2} $$\n其中$n$是任意正整数， 义...表示对所有不全为零的 $$I_{x} = 1, y_{1},l_{2}$$ 求和。\n\n(2)试计算近邻原子、次近邻原子和次次近邻原子分别对$A_n$值的贡献。\n\n(3)严格求解晶体中原在平衡位置$$R_{u|n|_{x|}$$ 附近的小幅振动是非常复杂的。为简化起见，爱\n因斯坦假设原子的运动互不干扰，即考虑任意一个原子的振动时，假设其它原子都静止于各\n自的平衡位置处。在此模型下，求上述晶体中每个原子的小幅振动的圆频率$\\mathrm{CB}$（结果可包\n含$a$和$A_{n}$）。\n(4)在(3)的基础上，如果势能参数r_0准静态缓慢变化到2*r_0,试求微振动振幅的变化率"
} | (1)a=(\frac{uA_2u}{vA_2v})^(1/(2u-2V))r_0
(2)(A_n)^1=6,(A_n)^2=12*2^(-n/2)
(A_n)^3 = 8*3^(-n/2)
(3)\sqrt{\frac{2ε*(\frac{u(2u-1)A_(2u+2)(r_0)^2u}{3a^(2u+2)}-\frac{v(2v-1)A_(2v+2)(r_0)^2v}{3a^(2v+2)})}{M}}
(4)\sqrt(2)-1 |
PHY-034 | Physics | null | {
"en": "There are two points in space on the same horizontal plane $y = 0$ with a distance of $l$ between them. A homogeneous catenary with a length of $L>l$ and a linear mass density of $\\lambda$ is hung with these two points as endpoints. Given that the positive direction of the $y$ - axis in space is vertically downward, and there is a non-uniform gravitational field $g = g_0\\exp(-y/L)$ in the downward direction in space, find the second-order differential equation satisfied by the shape $y(x)$ of the catenary.",
"zh": "空间中有两点在同一水平面y=0上相距$l$，以这两点为端点悬挂一长度为$L>l$，线密度为$\\lambda$的匀质悬链线，已知空间中y轴正方向为竖直向下，空间中存在向下的非均匀重力场g=g_0exp(-y/L)，求悬链线形状y(x)所满足的二阶微分方程。"
} | [\frac{y'^2-1}{L(e^{y/L}-1)}+y''](1+y'^2)-y'y''=0 |
PHY-035 | Physics | null | {
"en": "If an ideal gas with a constant volume heat capacity $C_v$ of $\\frac{1}{\\sqrt{3}-1}R$ undergoes a thermodynamic process that appears as a regular $n$-gon on a $P - V$ diagram, and we denote the vertices of the regular $n$-gon as $\\{A_{i},i = 1,2,\\ldots,n\\}$, then the $P - V$ coordinates of the $k$-th point can be written as $(a*(N+\\cos\\frac{2k\\pi}{n}),b*(N + \\sin\\frac{2k\\pi}{n}))$, where $N$ satisfies: $N>2*n/\\pi$. Find the thermal efficiency $\\eta_{n}$ of this heat engine cycle.",
"zh": "如果某定容热容C_v为$\\frac{1}{\\sqrt(3)-1}R$的理想气体经历的热力学过程在P-V图上显示为正n边形，记正n边形的各个顶点为{A_{i},i=1,2,...n},则第k个点的PV坐标可以记作(a*(N+cos$\\frac{2k\\pi}{n}$,b*（N+sin$\\frac{2k\\pi}{n}）$)，其中N满足：N>2*n/\\pi,求该热机循环的热力效率$η_{n}$"
} | $η_{n}$ = \frac{nsin2*\pi/n}{(\sqrt{3}+1)((N+cos$\frac{2(p+1)\pi}{n})*(N+sin$\frac{2(p+1)\pi}{n})-(N+cos$\frac{2q\pi}{n})*(N+sin$\frac{2q\pi}{n}))+2*(N \sin \left( \frac{2(p+1)\pi}{n} \right) + \frac{p + 1}{2} \sin \left( \frac{2\pi}{n} \right) + \frac{1}{4} \sin \left( \frac{4(p+1)\pi}{n} \right))-2*(N \sin \left( \frac{2q\pi}{n} \right) + \frac{q}{2} \sin \left( \frac{2\pi}{n} \right) + \frac{1}{4} \sin \left( \frac{4q\pi}{n} \right))}
where, \(p = \lfloor n/6 - 1/2 \rfloor\)
\(q=\lceil -n/3 - 1/2 \rceil\) |
PHY-036 | Physics | null | {
"en": "On the slope, a row of fixed inclined fences is formed by small thin rods that are parallel to each other, with an adjacent horizontal spacing of \\(l\\) and a vertical spacing of \\(h\\). Above the fence, there is a homogeneous circular plate with a mass of \\(m\\) and a radius \\(r\\gg l\\), and the circular plate will not come into contact with the ground.\n\nA slender soft rope passes through a small hole \\(C\\) at the center of the plate, with half on the back of the circular plate and half on the front of the circular plate. The two ends of the rope are joined together and denoted as the \\(P\\) end. A horizontal rightward pulling force \\(T\\) is applied at the \\(P\\) end, causing the circular plate to roll diagonally upward along the fence without jumping or relative sliding.\n\nThe average velocity of the circular plate moving diagonally upward to the right can be approximately treated as the magnitude \\(v\\) of the velocity of the center \\(C\\) of the circular plate when it is perpendicular to the direction of the line connecting the thin rods. Let \\(v\\) be a constant. Ignoring the friction at all contact parts between the rope and the plate, try to find the average pulling force \\(T\\) applied to the \\(P\\) end.",
"zh": "在坡面上用彼此平行、相邻水平间距为 l ，竖直间距为h的小细杆构成一排固定的斜栅栏。栅栏上方有一个质量为 m、半径为 r>>l 的匀质圆板，圆板不会与地面接触。\n一根细长的软绳穿过板的中央小孔 C，一半在圆板的背面，一半在圆板的正面，绳的两端合在一起记为 P 端。在 P 端施加水平向右的拉力 T，使圆板沿栅栏无跳动、无相对滑动地向右上滚动。\n圆板水平向右上的平均速度可近似处理为圆板中心 C 在相与细杆连线方向垂直x时的速度大小 v，设 v 为不变量。忽略绳与板间所有接触部位的摩擦，试求施加于 P 端的平均拉力 T。"
} | $mv^2*(h^2+l^2)/(2r^2*l)+mgh/l$ |
PHY-037 | Physics | null | {
"en": "Suppose we have a particle with potential\n\\(V = V_{0}-i\\Gamma,\\)\nwhere \\(V_{0}\\) is the true potential energy and \\(\\Gamma\\) is a positive real constant? Find the lifetime of the particle in terms of \\(\\Gamma\\), you can use $\\hbar$ to simplify your answer.",
"zh": "Suppose we have a particle with potential\n\\(V = V_{0}-i\\Gamma,\\)\nwhere \\(V_{0}\\) is the true potential energy and \\(\\Gamma\\) is a positive real constant? Find the lifetime of the particle in terms of \\(\\Gamma\\), you can use $\\hbar$ to simplify your answer."
} | \frac{\hbar}{2\Gamma} |
PHY-038 | Physics | null | {
"en": "In space, there is an experimental device consisting of a concentric inner sphere and a spherical shell. The space between the inner sphere and the inner surface of the spherical shell is a vacuum. The radius of the inner sphere is \\(r\\), its temperature is kept constant, and its emissivity is \\(e\\). The thermal conductivity of the spherical shell is \\(\\kappa\\), with inner and outer radii \\(R_1\\), \\(R_2\\) respectively. The thermal radiation of each surface is isotropic, and the outer surface can be regarded as a blackbody. The experimental device has reached a thermal stable state, and at this time the emissivity of the inner surface of the spherical shell is \\(E\\). The Stefan constant is \\(\\sigma\\), and the temperature of the cosmic microwave background radiation is \\(T\\). If the heat transferred from the inner surface of the spherical shell to the outer surface per unit time is \\(Q\\), find: \n(1) The temperature \\(T_2\\) of the outer surface of the spherical shell; \n(2) The temperature \\(T_1\\) of the inner surface of the spherical shell; \n(3) The temperature \\(T_0\\) of the inner sphere. \n\n\n**Known:** \n- The ratio of the radiation power per unit area of an object's surface to that of a blackbody at the same temperature on the same surface is called the emissivity. \n- When radiation strikes an object's surface, the ratio of the radiation power absorbed per unit area by the object's surface to the radiation power incident per unit area is called the absorptivity. \n- The absorptivity of an object at a certain temperature is equal to its emissivity at the same temperature. \n- When the temperature increment per unit distance in the \\(s\\) direction (heat direction) at a certain point inside an object is \\(\\frac{dT}{ds}\\), the heat flowing per unit area per unit time in the \\(s\\) direction at that point inside the object is \\(q = -\\kappa\\frac{dT}{ds}\\), which is Fourier's law of heat conduction.",
"zh": "太空中有一由同心的内球和球壳构成的实验装置，内球和球壳内表面之间为真空。内球半径为r，温度保持恒定，比辐射率为e；球壳的导热系数为κ ，内、外半径分别为R₁，R₂，各表面的热辐射都是各向同性的，而外表面可视为黑体；该实验装置已处于热稳定状态，此时球壳内表面比辐射率为E。斯特藩常量为σ，宇宙微波背景辐射温度为T。若单位时间内由球壳内表面传递到球壳外表面的热量为Q，求：\n（1）球壳外表面温度T_2；\n（2）球壳内表面温度T_1；\n（3）内球温度T_0。\n\n已知：物体表面单位面积上的辐射功率与同温度下的黑体在该表面单位面积上的辐射功率之比称为比辐射率。当辐射照射到物体表面时，物体表面单位面积吸收的辐射功率与照射到物体单位面积上的辐射功率之比称为吸收比。物体在某一温度下的吸收比等于其在同一温度下的比辐射率。当物体内某处在s方向（热方向）每单位距离温度的增量为dT/ds时，物体内该处单位时间在s方向每单位面积流过的热量为q=-κdT/ds，此即傅里叶热传导定律。"
} | (1)$
T_{2} = \left( \frac{Q}{4\pi R_2^{2}\sigma} + T^{4} \right)^{\frac{1}{4}}
$
(2)$
T_{1} = \frac{(R_{2}-R_{1})) Q}{4\pi κ R_{2} R_{1}} + \left( \frac{Q}{4\pi R_2^{2}\sigma} + T^{4} \right)^{\frac{1}{4}}
$
(3)$
T_{0} = \sqrt[4]{\frac{1 - (1 - e)(1 - E)\dfrac{r^{2}}{R_{1}^{2}}}{4\pi E e \sigma r^{2}} Q + (\frac{(R_{2}-R_{1})) Q}{4\pi κ R_{2} R_{1}} + \left( \frac{Q}{4\pi R_2^{2}\sigma} + T^{4} \right)^{\frac{1}{4}})^{4}}}
$
|
PHY-039 | Physics | null | {
"en": "There is an isolated metal sphere with a radius of \\(R\\) and a total charge of \\(0\\). Along two mutually perpendicular diameter directions, at a distance of \\(\\lambda R\\) from the center of the sphere on its upper, lower, left and right sides, there is a point charge \\(q\\) respectively.\n(1) If the point charges are exactly in electrostatic equilibrium, find the value of \\(\\lambda\\) (retain five significant figures).\n(2) Calculate the electrostatic interaction energy \\(W\\) of this system. Based on the previous question, give the specific value of \\(4\\pi\\epsilon_0R^{2}W/q^{2}\\) (five significant figures).\n(3) Discuss the perturbation of a certain point charge in the direction of its line connecting to the center of the sphere (with other point charges fixed). Assume that the perturbation is stable. In the case where the mass of the point charge is \\(m\\), find the perturbation frequency \\(\\omega\\) in terms of \\(\\lambda\\). Substitute the value of \\(\\lambda\\) obtained in (1) to get \\(\\omega_0\\) and determine whether the perturbation is stable.",
"zh": "有一半径为R的孤立金属球，总电量为0，在其上下左右两侧沿两相互垂直的直径方向距离球心各位λ*R位置，各有一个点电荷q。\n(1)如果点电荷恰好在静电力下平衡，求出λ数值(保留五位有效数字）。\n(2)计算这个体系的静电相互作用能W。在上一题的基础上给出4*\\pi*ε_0*R^2*W/q^2具体数值(5位有效数字)\n(3)讨论对某一点电荷在其与球心的连线方向的微扰下(其他点电荷固定），假设其该微扰稳定则在点电荷质量为m的情况下求含λ微扰频率ω带入(1)中的λ数值解得到ω_0并判断该扰动是否稳定"
} | (1)λ = 1.4727
(2)W = $\frac{q^2((1+2\sqrt{2})/λ+16/λ^2-4/(λ^2-1)-4/(λ^2+1)-8λ/\sqrt{λ^4+1})}{4\pi*ε_0*R}$
4*π*ε₀*R²*W/q²=0.35923
(3)
ω = $\sqrt{\frac{q^2*((\sqrt{2}/4+1/4)/λ^3+9/λ^4-(λ^2+1)/(λ^2-1)^3-(2λ^2)/(λ^2+1)^3-(4λ^6-2λ^2)/(λ^4+1)^(5/2)}{4\pi*ε_0*R^3m}}$
ω₀ = $\sqrt{\frac{-q^2*0.4514}{4\pi*ε_0*R^3m}}$
Not a stable perturbation |
PHY-040 | Physics | null | {
"en": "A non-uniformly charged spherical shell of radius R has surface charge density σ(θ) = σ₀(1 + cosθ)，where θ is the polar angle. The shell rotates about its symmetry axis with angular velocity ω. Find the dipole form of A(z) for z > R, and z is lying on the symmetry axis.",
"zh": "A non-uniformly charged spherical shell of radius R has surface charge density σ(θ) = σ₀(1 + cosθ)，where θ is the polar angle. The shell rotates about its symmetry axis with angular velocity ω. Find the dipole form of A(z) for z > R, and z is lying on the symmetry axis."
} | \frac{\mu_0\omega R^2\sigma_0}{4(z/R-1)}B(3/2, 7/2) _2F_1(1/2, 3/2; 5; -\frac{4z/R}{(z/R-1)^2}) |
PHY-041 | Physics | null | {
"en": "On a $k$-dimensional finite hypercube grid plane, denote the coordinates of the lower-left corner of the grid as the origin of coordinates $(0,0,\\ldots,0)$. For the $i$-th coordinate, there are resistors with resistance value $R_{i}$ connected between adjacent points on the corresponding axis, and there are a total of $m_{i}$ resistors in each row on this axis. Find the resistance value between the point $(0,0,\\ldots,0)$ and the point $(x_{1},x_{2},\\ldots,x_{k})$ ($0\\leq x_{i}\\leq m_{i}$)",
"zh": "k维有限大超方体网格平面上，记网格左下角坐标为坐标原点(0,0,……,0),其中第i个坐标对应坐标轴上相邻两点连接有阻值为$R_{i}$的电阻，该坐标轴上每排总计均有$m_{i}$个电阻，求点(0,0,……,0)到点$(x_{1},x_{2},……,x_{k})$两点间的电阻值$（0=<x_{i}<=m_{i})$"
} | $\boxed{R_{eq}(\mathbf{0}, \mathbf{x}) = \sum_{q_1=0}^{m_1} \dots \sum_{q_k=0}^{m_k} {}^{'} \frac{\left[1 - \prod_{i=1}^k \cos\left(\frac{q_i \pi x_i}{m_i}\right)\right]^2}{N_{\mathbf{q}} \lambda_{\mathbf{q}}} \text{ where } \lambda_{\mathbf{q}} = 4 \sum_{i=1}^k \frac{1}{R_i} \sin^2\left(\frac{q_i \pi}{2m_i}\right), N_{\mathbf{q}} = \prod_{i=1}^k S(q_i, m_i), S(q, m) = \begin{cases} m+1 & \text{if } q=0 \text{ or } q=m \\ (m+2)/2 & \text{if } 0 < q < m \end{cases} \text{ and } \sum^{'} \text{ excludes } \mathbf{q}=0}$ |
PHY-042 | Physics | null | {
"en": "There is a simple pendulum with mass $m$ and string length $l$. Its suspension point undergoes high-frequency vibration with angular frequency $\\gamma\\gg\\sqrt{\\frac{g}{l}}$, an amplitude of $a$, and the direction makes an angle $\\varphi$ with the horizontal direction. Try to find: (1) The new “equilibrium” position of the simple pendulum; (2) The angular frequency of small perturbations near this equilibrium position.",
"zh": "现有单摆质量为m绳长为l,将其悬挂点以角频率γ>>$\\sqrt{\\frac{g}{l}}$做振幅为a方向与水平方向夹φ角的高频振动，试求：(1)单摆新的“平衡”位置（2）在该平衡位置附近的微扰角频率"
} | (1) $\theta = 0$, or $\theta=\pi$,
if $\frac{2gl}{(a\gamma)^2}<\cos2\varphi$, then there is also $\cos^{-1}\left(\frac{2gl}{(a\gamma)^2\cos2\varphi}\right)$
(2) $\theta = 0$ is stable when $\frac{2gl}{(a\gamma)^2}>\cos2\varphi$,
at this time $\omega=\sqrt{\frac{g}{l}-\cos2\varphi*\frac{a^{2}\gamma^{2}}{2l^{2}}}$
$\theta=\pi$ is stable when $-\frac{2gl}{(a\gamma)^2}>\cos2\varphi$,
at this time $\omega=\sqrt{-\frac{g}{l}-\cos2\varphi*\frac{a^{2}\gamma^{2}}{2l^{2}}}$
$\theta=\cos^{-1}\left(\frac{2gl}{(a\gamma)^2\cos2\varphi}\right)$
$\omega=\sqrt{\frac{a^{2}\gamma^{2}\cos2\varphi}{2l}-\frac{g^{2}(4 - 2(\cos2\varphi)^{2})}{a^{2}\gamma^{2}\cos2\varphi}}$
It is required that the content in the parentheses is greater than 0 |
PHY-043 | Physics | null | {
"en": "A homogeneous thin-walled cylinder (mass \\(m\\), radius \\(r\\)) starts to roll down from rest at the top of a semi-cylindrical surface with radius \\(R\\). It is known that the axis of the cylinder is always parallel to the axis of the semi-cylindrical surface, and the acceleration due to gravity is \\(g\\).\n\n1. When the cylinder undergoes pure rolling within the range of \\(\\varphi\\leq45^{\\circ}\\), derive the expression for the minimum value of the coefficient of static friction \\(\\mu\\).\n2. If \\(\\mu = 0.5\\), find the tangential velocity \\(v_1\\) of the cylinder's center of mass when pure rolling ends.",
"zh": "一均质薄壁圆筒（质量\\( m \\)、半径\\( r \\)）自半径\\( R \\)的半圆柱面顶端静止开始下滚。已知圆筒轴线始终与半圆柱面轴线平行，重力加速度为\\( g \\)。\n\n1. 当圆筒在\\( \\varphi \\leq 45^\\circ \\)范围内作纯滚动时，试推导静摩擦系数\\( \\mu \\)的最小值表达式。\n2. 若\\( \\mu = 0.5 \\)，求纯滚动结束时圆筒质心的切向速度\\( v_1 \\)。"
} | (1)\mu_{\min}=\frac{\sqrt{2+\sqrt{2}}}{4}
(2)v_1=\sqrt{\frac{1}{5}g(R+r)} |
PHY-044 | Physics | null | {
"en": "In an infinitely long straight cylindrical conductor of radius \\(a\\), when there is no current, the volume charge density of free electrons is a constant \\(\\rho_{0}\\), and that of positive ions is \\(-\\rho_{0}\\), so the conductor is electrically neutral. When a direct current flows along the length of the conductor, the volume current generates an axisymmetric ring-shaped magnetic field distribution in the conductor. Free electrons moving along the axial direction will move towards the axis under the action of the magnetic force, thus disrupting the original \\(\\rho_{0}\\) distribution. When the radial electric field generated by the accumulated net charge exactly cancels out the magnetic force acting on the moving free electrons, a steady current only along the length of the conductor is formed. Given that the drift velocity of free electrons at this time is \\(\\overline{u}\\), find the distribution of the net charge density in the conductor and the magnetic field distribution outside the conductor.",
"zh": "无限长直半径为a的圆柱形导体在没有电流时，自由电子的体电荷密度为常量\\(\\rho_{0}\\)，正离子的体电荷密度为\\(-\\rho_{0}\\)，导体内处于电中性。当沿导体长度方向通以直流电时，体电流会在导体中产生轴对称的环状分布磁场，沿轴向运动的自由电子便会在磁场力的作用下向中轴移动，从而破坏了原来的\\(\\rho_{0}\\)分布。当积聚的净电荷产生的径向电场与运动自由电子所受的磁场力恰好抵消时，便形成只沿导体长度方向的稳恒电流。设此时自由电子的漂移速度为\\(\\overline{u}\\)，试求导体中净电荷密度的分布和导体外的磁场分布。"
} | \rho=\frac{\varepsilon_0\mu_0u^2}{1-\varepsilon_0\mu_0u^2}\rho_0, B(r)=\frac{\mu_0a^2u\rho_0}{2(1-\varepsilon_0\mu_0u^2)r} |
PHY-045 | Physics | null | {
"en": "A mass of \\(2.0\\ \\text{kg}\\), temperature of \\(- 13^{\\circ}C\\), and volume of \\(0.19\\ \\text{m}^3\\) of Freon (molecular weight is 121, van der Waals constants \\(a = 1.06\\ \\text{Pa}\\cdot\\text{m}^6/\\text{mol}^2\\), \\(b = 5.42\\times10^{-5}\\ \\text{m}^3/\\text{mol}\\)) is compressed under isothermal conditions, with its volume reduced to \\(0.10\\ \\text{m}^3\\). How many kilograms of Freon are liquefied in this process? It is known that at \\(-13^{\\circ}C\\), the density of liquid Freon is \\(\\rho = 1.44\\times10^{3}\\ \\text{kg/m}^3\\), the saturated vapor pressure is \\(p_{sat}=2.08\\times10^{5}\\ \\text{Pa}\\), and the saturated vapor of Freon can be approximately regarded as a van der Waals gas.",
"zh": "质量为\\(2.0\\ \\text{kg}\\)，温度为\\(-13^{\\circ}C\\)，体积为\\(0.19\\ \\text{m}^3\\)的氟利昂（分子量为 121，范德瓦尔斯常数 \\(a = 1.06\\ \\text{Pa·m}^6/\\text{mol}^2\\)，\\(b = 5.42 \\times 10^{-5}\\ \\text{m}^3/\\text{mol}\\)），在保持温度不变的条件下被压缩，体积减小为\\(0.10\\ \\text{m}^3\\)。试问在此过程中有多少千克的氟利昂被液化 。已知在\\(-13^{\\circ}C\\)时，液态氟利昂的密度为\\(\\rho = 1.44×10^{3}\\ \\text{kg/m}^3\\)，饱和蒸汽压为\\(p_{饱}= 2.08×10^{5}\\ \\text{Pa}\\)，又氟利昂的饱和蒸汽可近似看作范德瓦尔斯气体 。"
} | 0.79kg |
PHY-046 | Physics | null | {
"en": "Consider following quantum Hamiltonian:\n$$ H_0 = \\frac{p_1^2}{2m} + \\frac{1}{2}m\\omega^2 x_1^2 + \\frac{p_2^2}{2m} + \\frac{1}{2}m\\omega^2 x_2^2 $$\nThis is the Hamiltonian for two decoupled harmonic oscillators.\n\n(a) Calculate the eigenstates and eigenvalues for $H_0$ (an energy eigenstate could be labeled as $|n_1, n_2\\rangle$).\n(b) Assume the creation and annihilation operators for the two harmonic oscillators are $a_i^\\dagger$, $a_i$ ($i = 1, 2$). Define the following operators:\n$$ J_+ = a_1^\\dagger a_2, \\quad J_- = a_2^\\dagger a_1, \\quad J_z = \\frac{1}{2}(a_1^\\dagger a_1 - a_2^\\dagger a_2) $$\ni. Prove that:\n$$ [J_z, J_\\pm] = \\pm J_\\pm, \\quad [J_+, J_-] = 2J_z $$\nii. Consider one eigenvalue $E_n$ of $H_0$ (here $n_1 + n_2 = n$). Prove that all eigenstates of $E_n$ form an irreducible representation of the $\\mathfrak{su}(2)$ Lie algebra, and compute the spin.\n(c) Consider the following perturbed Hamiltonian ($\\lambda$ is small):\n$$ H = H_0 + \\lambda x_1^2 p_2^2 $$\nCompute the first-order correction to the energy level for the state with $n_1 + n_2 = 2$.\n(d) Consider the following perturbed Hamiltonian:$$ H = H_0 + ρ*mω^2*x_{1}*x_{2} $$ where ρ is between (-1,1),resolve eigenvalues for $H$",
"zh": "Consider following quantum Hamiltonian:\n$$ H_0 = \\frac{p_1^2}{2m} + \\frac{1}{2}m\\omega^2 x_1^2 + \\frac{p_2^2}{2m} + \\frac{1}{2}m\\omega^2 x_2^2 $$\nThis is the Hamiltonian for two decoupled harmonic oscillators.\n\n(a) Calculate the eigenstates and eigenvalues for $H_0$ (an energy eigenstate could be labeled as $|n_1, n_2\\rangle$).\n(b) Assume the creation and annihilation operators for the two harmonic oscillators are $a_i^\\dagger$, $a_i$ ($i = 1, 2$). Define the following operators:\n$$ J_+ = a_1^\\dagger a_2, \\quad J_- = a_2^\\dagger a_1, \\quad J_z = \\frac{1}{2}(a_1^\\dagger a_1 - a_2^\\dagger a_2) $$\ni. Prove that:\n$$ [J_z, J_\\pm] = \\pm J_\\pm, \\quad [J_+, J_-] = 2J_z $$\nii. Consider one eigenvalue $E_n$ of $H_0$ (here $n_1 + n_2 = n$). Prove that all eigenstates of $E_n$ form an irreducible representation of the $\\mathfrak{su}(2)$ Lie algebra, and compute the spin.\n(c) Consider the following perturbed Hamiltonian ($\\lambda$ is small):\n$$ H = H_0 + \\lambda x_1^2 p_2^2 $$\nCompute the first-order correction to the energy level for the state with $n_1 + n_2 = 2$.\n(d) Consider the following perturbed Hamiltonian:$$ H = H_0 + ρ*mω^2*x_{1}*x_{2} $$ where ρ is between (-1,1),resolve eigenvalues for $H$"
} | (a)$|n_1, n_2\rangle$=($a_1^\dagger$)^($n_1$)/($n_1$!)^0.5($a_2^\dagger$)^($n_2$)/($n_2$!)^0.5$|0, 0\rangle$,
E = (n_1+n_2+1)hω/(2\pi)
(b)i.use [a_i,$a_i^\dagger$]=1 for i =1,2 to directly verify it
ii.J_z$|n_1, n_2\rangle$=\frac{1}{2}(a_1^\dagger a_1 - a_2^\dagger a_2)($a_1^\dagger$)^($n_1$)/($n_1$!)^0.5($a_2^\dagger$)^($n_2$)/($n_2$!)^0.5$|0, 0\rangle$=\frac{1}{2}(n_1-n_2)$|n_1, n_2\rangle$,
so the operator J_z has eigenvalue \frac{1}{2}(n_1-n_2),so it forms a spin n/2 representation.
(c)the correction matrix is
$
\begin{pmatrix}
-5 & 0 & 2 \\
0 & -9 & 0 \\
2 & 0 & -5 \\
\end{pmatrix}
$
(d)use orthogonal transformation Q,where
y= Qx,Q=
$
\begin{pmatrix}
1 & -1 \\
1 & 1 \\
\end{pmatrix}
$/\sqrt{2},the new Hamiltonian is :
$$ H_1 = \frac{p_1^2}{2m} + \frac{1+ρ}{2}m\omega^2 x_1^2 + \frac{p_2^2}{2m} + \frac{1-ρ}{2}m\omega^2 x_2^2 $$
then:
E = ((n_1+1/2)(1+ρ)^0.5+(n_2+1/2)(1-ρ)^0.5)hω/(2\pi) |
PHY-047 | Physics | null | {
"en": "At the origin \\((0,0)\\) of the polar coordinate system, there is a mass point with mass \\(M\\). In space, there is a homogeneous soft spring with an original length of \\(0\\), linear mass density \\(\\lambda\\), and a uniform spring constant \\(k\\). Its two ends are fixed at \\((r_{1},\\theta_{1})\\) and \\((r_{2},\\theta_{2})\\) respectively. Considering the gravitational field of \\(M\\) on the soft spring without taking into account the self-gravitation of the soft rope, (1) Prove that the equation of the soft rope is an ellipse with precession; (2) If it is measured that \\(r_{1}=r_{2} = R\\), find the shape equation \\(f(r,\\theta)\\) of the soft spring.",
"zh": "在极坐标系(0,0)处有一质点质量为M，空间中有一根原长为0的匀质软弹簧，质量线密度为λ，劲度系数均匀为常数k,两端分别被固定在($r_{1}$,$θ_{1}$),($r_{2}$,$θ_{2}$)处，考察M对软弹簧的万有引力场而不考虑软绳的自引力，(1)证明软绳的方程为圆带有进动的椭圆，(2)如果测得r_{1} = r_{2} = R,求软弹簧的形状方程f(r,θ)"
} | (1) According to the law of conservation of generalized angular momentum \(Tr\sin\beta=\text{Const}\) (where \(\beta\) is defined as \(\tan\beta = r\frac{d\theta}{dr}\)) and the radial force equation \(\cos\beta dT-\sin\beta T(d\beta + d\theta)=\frac{GM\lambda dr}{r^{2}\cos\beta}\), we can obtain through two integrations:
\(\frac{1}{r}=C_{2}\left(1 + \left(\frac{C_{1}}{GM\lambda}\right)\cos\left(\sqrt{1-\left(\frac{GM\lambda}{C_{1}}\right)^{2}}(\theta - \theta_{0})\right)\right)\)
where \(C_{1}\), \(C_{2}\), \(\theta_{0}\) are all undetermined constants. It can be seen that this is an ellipse with precession.
(2)
\(r(\theta)=\frac{R\left(1+\frac{\cos\left(e(\theta_{1}-\theta_{2})/2\right)}{\sqrt{1 - e^{2}}}\right)}{\left(1+\frac{\cos\left(e(\theta-(\theta_{1}+\theta_{2})/2\right)}{\sqrt{1 - e^{2}}}\right)}\)
where the parameter \(e\) satisfies:
\(kR\left(\sqrt{1 - e^{2}}+\cos\left(e(\theta_{1}-\theta_{2})/2\right)\right)\times\ln\left(\frac{1+\sqrt{1 - e^{2}}\cos\left(e(\theta_{1}-\theta_{2})/2\right)+e\sin\left(e(\theta_{1}-\theta_{2})/2\right)}{1+\sqrt{1 - e^{2}}\cos\left(e(\theta_{1}-\theta_{2})/2\right)-e\sin\left(e(\theta_{1}-\theta_{2})/2\right)}\right)=GM\lambda\times\frac{1+\sqrt{1 - e^{2}}\cos\left(e(\theta_{1}-\theta_{2})/2\right)}{R\sqrt{1 - e^{2}}\left(\sqrt{1 - e^{2}}+\cos\left(e(\theta_{1}-\theta_{2})/2\right)\right)}\) |
PHY-048 | Physics | null | {
"en": "Set up a fixed $Oxyz$ coordinate system. There is a glass ellipsoid with the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}+\\frac{z^{2}}{c^{2}} = 1$, which is placed flat on the lower half of a hyperboloid of two sheets - like surface with the equation $- 2\\rho\\sinh(\\frac{xy}{ab})+(\\frac{z}{c})^{2}=1$ (i.e., the part where $z < 0$). Monochromatic parallel light with a wavelength of $\\lambda$ is incident against the positive $z$ - axis. An interference pattern similar to Newton's rings is formed on the air film between the spherical surface and the cylindrical surface. The pattern appears as a series of interference rings in the $Oxy$ plane.\n\nFind the shape and the equation $f(x,y)=0$ of the $k$-th order interference ring.",
"zh": "设置固定的$Oxyz$坐标系，有一方程为$x^2/a^2+y^2/b^2+z^2/c^2=1$的玻璃椭球，平放在方程为$-2ρsinh(xy/ab)+(z/c)^2=1$的类双叶双曲面的下半扇面上(即z<0的部分）的波长为$\\lambda$的单色平行光逆着$z$轴正入射，在球面与圆柱面之间的空气膜上形成类似于牛顿环的干涉图样，图样在$Oxy$平面上表现为一系列的干涉环。\n求第k级的干涉环的形状以及方程f(x,y)=0"
} | $f(x,y) = \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{2\rho xy}{ab}-(\ 2k+\frac{1}{2})\frac{\lambda}{c}, 0 < |\rho| < 1$ is an ellipse, $\rho=\pm1$ is a parabola, $|\rho| > 1$ is a hyperbola |
PHY-049 | Physics | null | {
"en": "An infinitely long paraboloid of revolution has the equation \\(z = \\frac{r^{2}}{4a}\\) (where \\(r=\\sqrt{x^{2}+y^{2}}\\)). There is a current uniformly distributed along the latitudinal direction on its surface, with a current linear density of \\(\\sigma\\). Try to find the magnetic induction intensity \\(\\boldsymbol{B}\\) at the focus.",
"zh": "一个无限长的旋转抛物面，其方程为\\( z = \\frac{r^2}{4a} \\)（其中\\( r = \\sqrt{x^2 + y^2} \\)），表面均匀分布着沿纬线方向的电流，电流线密度为$\\sigma$。试求焦点处的磁感应强度\\(\\boldsymbol{B}\\)。"
} | \frac{4}{3}\mu_0\sigma |
PHY-050 | Physics | null | {
"en": "A simple pendulum at a latitude of $\\theta$ has a string length of $l$ and the acceleration due to gravity is $g$, and it undergoes small-angle motion. Assume the angular velocity of the Earth's rotation is $\\omega$, and the air resistance coefficient is $\\eta$ (not necessarily a small quantity). Find the equation of motion of the simple pendulum (initially, the simple pendulum deviates from the vertical plane by a small angle $\\alpha\\ll1$ and is released from rest. Take the vertical direction as the $z$ - axis, the direction of the initial deviation of the simple pendulum as the $x$ - axis, and the direction perpendicular to the above two axes as the $y$ - axis. Let $r = x + iy$, and give the relationship of $r$ with respect to $t$).",
"zh": "现有位于纬度为θ处的单摆绳长为l,重力加速度为g做小角运动。假设地球自转角速度为ω,空气阻力系数为η(不一定是小量),求单摆的运动方程(初态单摆与垂直面偏离小角度α<<1并静止释放,以垂直方向为z轴，初态单摆偏移方向为x轴，垂直于上述二轴为y轴，令r = x+iy,给出r关于t的关系)"
} | r(t) = $αl*\frac{ω_2*exp{iω_1t}-ω_1*exp{iω_2t}}{ω_2-ω_1}$
where, $ω_{1,2}=-η/(2m)-iωsinθ±i\sqrt{g/l+(iωsinθ+η/(2m))^2}$ |
PHY-051 | Physics | null | {
"en": "Inside a vacuum chamber, there are two identical liquid bubbles, each filled with a monatomic ideal gas at a temperature of \\(T_{1}\\). The radius of each liquid bubble is \\(R_{1}\\), and the relationship between the surface tension coefficient and temperature is \\(\\sigma(T)=\\sigma_0\\left(1 - \\frac{T}{T_0}\\right)^n\\). The following processes are considered quasistatic and adiabatic. Each liquid bubble has a heat capacity of \\(c\\).\n\nTake \\(n = 2\\)\n\n(1) If the two liquid bubbles stick together without the middle being punctured, with the contact surface being circular while the non-contact parts of the liquid bubbles themselves still remain spherical (that is, the two liquid bubbles become two mutually - contacting spherical caps). Determine the respective radii \\(R_{2}\\) and temperature \\(T_{2}\\) of the two contacting liquid bubbles at this time.\n\n(2) If the two liquid bubbles merge into one, find the radius \\(R_{3}\\) and temperature \\(T_{3}\\) of the liquid bubble at this time.",
"zh": "真空室内有两个完全一样的液泡，内部装有温度为$T_{1}$的单原子理想气体，液泡半径为$R_{1}$，表面张力系数与温度的关系为σ(T)=σ_0(1-\\frac{T}{T_0})^n。以下过程认为是准静态过程，且绝热。每个液泡的热容为c.\n取n=2\n(1)如果两个液泡粘在一起但中间不被戳破，接触面为圆形而液泡本身非接触部分依然保持球形（即两个液泡变成了相互接触的球冠）。求出两接触液泡此时各自的半径$R_{2}$和温度$T_{2}$\n(2)如果两个液泡合为一体，求液泡此时的半径$R_{3}$和温度$T_{3}"
} | (1)$T_2$=$\frac{T_0}{2}+\frac{c*T_1*T_0}{32*\pi*(R_1)^2*σ_0}+\sqrt{(\frac{T_0}{2}+\frac{c*T_1*T_0}{32*\pi*(R_1)^2*σ_0})^2-\frac{c*(T_1)^2*T_0}{16*\pi*(R_1)^2*σ_0}-T_1*(T_0-T_1)}$
$R_2$ = $\sqrt{\frac{32*T_2}{27*T_1}}*R_1$
(2) T_3 = T_2
R_3 = $\sqrt{\frac{2*T_3}{T_1}}*R_1$ |
PHY-052 | Physics | null | {
"en": "Use the uncertainty relation to estimate the ground-state energy $E_0$ of a quantum harmonic oscillator.",
"zh": "试用不确定性关系，估算基态量子谐振子的能量E_0"
} | hω\(4\pi) |
PHY-053 | Physics | null | {
"en": "There is a horizontal uniform magnetic field with a very large magnetic induction intensity \\(B\\). A metal concentric ring with an inner diameter of \\(R_1\\) and an outer diameter of \\(R_2\\) (\\(R_2 > R_1\\)) and a thickness of \\(D\\) (\\(D\\ll R_1, R_2\\)) falls vertically in this magnetic field. The plane of the ring is always in the vertical plane and parallel to the direction of the magnetic field. Assume the resistance of the metal is zero and the density is \\(\\rho = 9\\times 10^3\\,\\mathrm{kg/m^3}\\), and air resistance can be ignored. In order to make the acceleration of the ring during its fall in the magnetic field decrease by one-thousandth compared to when there is no magnetic field, find the magnitude of the magnetic induction intensity \\(B\\).",
"zh": "有一水平方向的匀强磁场，磁感应强度\\( B \\)很大。一个内径为\\( R_1 \\)、外径为\\( R_2 \\)（\\( R_2 > R_1 \\)）的金属同心圆环，厚度为\\( D \\)（\\( D \\ll R_1, R_2 \\)），在此磁场中竖直下落，环面始终在竖直平面内并与磁场方向平行。设金属的电阻为零，密度为\\( \\rho = 9 \\times 10^3 \\, \\mathrm{kg/m^3} \\)，空气阻力可忽略。为使圆环在磁场中下落的加速度比没有磁场时减小千分之一，试求磁感应强度\\( B \\)的大小。"
} | B=10^6T |
PHY-054 | Physics | null | {
"en": "At the six vertices of a regular hexagon with side length \\(a\\), there are fixed point charges each with charge \\(Q\\). At the center of the regular hexagon, a free point charge with mass \\(m\\) and charge \\(q\\) (where \\(q\\) has the same sign as \\(Q\\)) is placed. Now, \\(q\\) is moved a very small distance along a long diagonal and then released. Find the period \\(T\\) of the free point charge's motion.",
"zh": "在边长为a的正六边形的六个顶点分别有电量均为Q的固定的点电荷。在正六边形中心上放置一个质量为m、电量为q（q与Q同号）的自由点电荷。今将q沿某一长对角线移动一个很小的距离后释放。求自由点电荷的运动周期T；"
} | $2\pi\sqrt{\frac{ma^3}{3kQq}}$ |
PHY-055 | Physics | null | {
"en": "An initially macroscopically uncharged long metal cylinder (radius \\(a\\), length \\(L\\gg a\\)) contains an equal number density \\(n\\) of uniformly distributed electron gas and positive ion gas inside. Now the cylinder rotates at a high angular velocity \\(\\omega\\) about its central axis, and the system temperature remains constant at \\(T\\). After reaching a steady state, find: \nThe charge distribution \\(\\rho(r)\\) inside the body \nThe distributions of the electric field \\(\\mathbf{E}(r)\\) and magnetic field \\(\\mathbf{B}(r)\\) inside the body. \nIt is known that electrons can move freely in the metal, the electron charge is \\(e\\), the mass is \\(m\\), and positive ions are fixed.",
"zh": "一个初态宏观无电荷的长金属圆柱体（半径a，长度L≫a）圆柱体内部有等量的数密度为n的均匀分布的电子气体和正离子气体。现在圆筒以角速度ω绕其中心轴高速旋转，系统温度恒为T,达到稳态后，求：\n体内的电荷分布 ρ(r)\n体内电场 E(r) 和磁场 B(r) 的分布。\n已知在金属中电子可自由移动，电子电荷为e，质量为m,正离子固定。"
} | ρ(r) = $\frac{-e*n*m*ω^2*R^2*\exp{mω^2*r^2/(2KT)}}{mω^2*R^2*\exp{mω^2*R^2/(2KT)}+1-\exp{mω^2*R^2/(2KT)}}+ne$
E =$ n*e*r/(2ε_0)-\frac{n*e*R^2*(mω^2*r^2*\exp{mω^2*r^2/(2KT)}+1-\exp{mω^2*r^2/(2KT)})}{2*r*ε_{0}mω^2*R^2*\exp{mω^2*R^2/(2KT)}+1-\exp{mω^2*R^2/(2KT)}}$$\hat{r}$
B = $μ_{0}*n*e*ω*r^2/2-\frac{μ_{0}*ω*n*e*R^2*(mω^2*r^2*\exp{mω^2*r^2/(2KT)}+1-\exp{mω^2*r^2/(2KT)})}{2*mω^2*R^2*\exp{mω^2*R^2/(2KT)}+1-\exp{mω^2*R^2/(2KT)}}$ |
PHY-056 | Physics | null | {
"en": "There are two homogeneous rods with masses $m_{1}$, $m_{2}$ and lengths $l_{1}$, $l_{2}$ respectively. The first rod is hinged to the wall, and one end of the second rod is hinged to the other end of the first rod, with the other end free. (1) Find all possible angular frequencies of the system's vibration. (2) Assume $m_1 = m_2 = m$ and $l_1 = l_2 = l$. If during the periodic motion of the system, a bug with mass $M$ crawls along the two rods at a speed much smaller than the speed of the rods' motion, try to find the relationship between the amplitude of the system's motion and the crawling distance $x$ ($x < l$) (written in the form of a conserved quantity).",
"zh": "现有匀质的两条杆质量分别为$m_{1}$,$m_{2}$,长度分别为$l_{1}$,$l_{2}$,第一根杆铰接于墙面，第二根杆一端铰接于第一根杆的另一端，另一端自由，（1）求系统的振动的所有可能的角频率（2）下设m_1=m_2=m,l_1=l_2=l,假如在系统周期运动的过程中，有一质量为M的小虫沿着两条杆以速度远远小于杆的运动速度爬行，试求系统运动振幅与爬行距离x(x<l)的关系(写成守恒量的形式)"
} | (1)ω=$\sqrt{\frac{B±\sqrt{B^2 - 4AC}}{2C}}$
where
A = m_2*g_2*l_2*(m_1*g*l_1/2 + m_2*g*l_1)
B = m_2*m_1*g*l_1*l_2*(l_1/3 + l_2/6)+(m_2)^2*g*l_1*l_2*(l_1 + l_2/3)
C = (m_1*m_2/9+(m_2)^2/12)*(l_1*l_2)^2
(2)
Let ω_{1,2} = $\sqrt{\frac{B'±\sqrt{B'^2 - 4C'}}{2C'}}$
where
$B' = 2g/3l+\frac{4ml^2/3+Mx^2}{3mgl/2+Mgx}$
$C' = \frac{2Mx^2/3+7ml^2/18}{3mgl/2+Mgx}$
Then the following quantity is conserved:
Let a be the amplitude of the vibration with the ω_1 angular frequency component of rod 1, and b be the amplitude of the vibration with the ω_2 angular frequency component of rod 2, then:
a^2*ω_1 + b^2*ω_2*(l/(g(ω_2)^2)-2/3)^2 = const
a^2*ω_1*1/((l/(g(ω_1)^2)-2/3)^2)+b^2*ω_2 = const |
PHY-057 | Physics | null | {
"en": "A stationary electric dipole **p** = p**ẑ** is situated at the origin. A negative point charge -q (mass m) moves in a uniform magnetic field **B** = B**ẑ**, superimposed with the dipole’s electric field. The charge executes a circular orbit of radius s above x-y plane, and its angular momentum has positive projection along $\\hat{a}$. Find: The speed of the charge, expressed in terms of p, q, m, s, and B.",
"zh": "A stationary electric dipole **p** = p**ẑ** is situated at the origin. A negative point charge -q (mass m) moves in a uniform magnetic field **B** = B**ẑ**, superimposed with the dipole’s electric field. The charge executes a circular orbit of radius s above x-y plane, and its angular momentum has positive projection along $\\hat{a}$. Find: The speed of the charge, expressed in terms of p, q, m, s, and B."
} | v=\frac{qBs}{2m}(1+\sqrt{1+\frac{\sqrt{2}pm}{3\sqrt{3}\pi\varepsilon_0qB^2s^4}}) |
PHY-058 | Physics | null | {
"en": "A uniform disk of mass has a radius of $R_2$. There is a concentric mirror region with a radius of $R_1 < R_2$ on the disk's surface, while the remaining annular region is a perfect blackbody. The disk is fully illuminated by a uniform parallel laser beam perpendicular to the plane of the disk. The laser is circularly polarized light with a wavelength of $\\lambda$. Initially, the disk is suspended stationary in a gravity-free space laboratory. After being illuminated by the laser beam, it starts to move and rotate, and every part of the disk except the center moves along a helical orbit. Try to find the pitch $h$ of the helical orbit.",
"zh": "一个质量均匀圆盘半径为$R_2$，在圆盘表面上存在一同心半径为$R_1<R_2$的镜面区域，而其余的圆环区域均为理想黑体，该圆盘受到垂直于圆盘平面方向的均匀平行激光束的全面照射。激光是圆偏振光，波长为\\(\\lambda\\)。圆盘开始时在无重力的空间实验室中静止地悬浮，经激光束照射后开始平动和转动，圆盘除中心外的每一部位都沿螺旋轨道运动。试求螺旋轨道的螺距h。"
} | \[\frac{2\pi^2R_2^2}{\lambda}\frac{R_2^2+R_1^2}{R_2^2-R_1^2}\] |
PHY-059 | Physics | null | {
"en": "An object flies vertically upward at a constant speed \\(v_f\\) from a platform at a height \\(h\\) above the ground. A missile is launched from the ground at a horizontal distance \\(L\\) from the take-off point of the flying object. The missile and the flying object are launched simultaneously. The speed of the missile \\(v > v_f\\) is constant, and the missile always points to the instantaneous position of the flying object. Try to find: The total time required for the missile to hit the flying object.",
"zh": "飞行物自相距地面高h的平台以匀速\\(v_f\\)向上垂直飞行。在地面离飞行物起飞点水平相距\\(L\\)处处发射一枚导弹，导弹与飞行物同时发射。导弹的速率\\(v>v_f\\)为常值，导弹始终指向飞行物瞬时位置。试求：导弹击中飞行物所需总时间"
} | t=\frac{\alpha^2-1+\gamma(\alpha^2+1)}{2\alpha\gamma(1-\gamma^2)}\cdot\frac{L}{v}, \alpha=\frac{h}{L}+\sqrt{1+(\frac{h}{L})^2}, \gamma=\frac{v_f}{v} |
PHY-060 | Physics | null | {
"en": "A light rigid rod of length \\(l\\) has two small balls 1 and 2 fixed at its two ends. The mass of each small ball is \\(m\\), and their charges are \\(\\pm q\\) respectively, with the positively charged ball 1 at the upper end and the negatively charged ball 2 at the lower end. Let's call it a “charged barbell”. There is an infinite uniform magnetic field perpendicular to the plane of the paper, with magnetic induction intensity \\(B\\). The acceleration due to gravity \\(g\\) satisfies \\(g = \\frac{B^{2}q^{2}l}{2m^{2}}\\). Initially, the charged barbell is placed vertically on a smooth horizontal surface and remains at rest. At this time, we give it a small perturbation so that the charged barbell has a clockwise angular velocity (very small), and the charged barbell starts to fall. Note: The following calculation results should be expressed only in terms of \\(m\\), \\(l\\), \\(B\\), and \\(q\\).\n(1) When the charged barbell makes an angle \\(\\theta\\) with the vertical direction, find the magnitude of the velocity of ball 1, \\(v_{1}\\), and the magnitude of the velocity of ball 2 relative to ball 1, \\(v_{2}\\).\n(2) Following the first question, find the normal force \\(N\\) exerted by the ground on the charged barbell at this time, and analyze whether it is possible to leave the ground.",
"zh": "一根长为l的轻质刚性杆两端固连着两个小球1和2，小球的质量均为m,电量分别为±q，正电小球1在上端负电小球2在下端，不妨称它为“带电杠铃”。空间中分布着垂直于纸面向里的无限大均匀磁场，磁感应强度为B。重力加速度g满足g=$\\frac{B^2*q^2*l}{2*m^2}$.初始时刻，带电杠铃竖直的放置在一个光滑的水平面上，保持静止，此时我们给它一个微扰，使带电杠铃具有一个顺时针方向的角速度（非常小），带电杠铃开始倒下。注意：以下计算结果只用m,l,B,q表示\n（1）当带电杠铃与竖直方向成θ角时，求小球1的速度大小$v_{1}$以及小球2相对于小球1的速度大小$v_{2}$\n(2)接第一问，求此时地面对带电杠铃的支持力N，并分析是否有可能脱离地面"
} | (1)$v_{1}=\frac{qBl}{m}*\sqrt{\frac{(\sin\theta)^2}{2}-\frac{\cos\theta}{2}+\frac{1}{2 + 2(\sin\theta)^2}+\frac{(1 - \cos\theta)\cos\theta\sin\theta}{2\sqrt{1+(\sin\theta)^2}}}$
$v_{2}=\frac{qBl}{m}*\frac{\sin\theta}{\sqrt{1+(\sin\theta)^2}}$
(2)$N=\frac{q^{2}*B^{2}*l}{m}*(1-\frac{(\sin\theta)^2*\cos\theta*(2 + (\sin\theta)^2)}{(1+(\sin\theta)^2)^2}-\frac{\cos\theta\sin\theta}{\sqrt{1+(\sin\theta)^2}})$
I won't deviate. |
PHY-061 | Physics | null | {
"en": "There are many stones launched from the origin \\((0,0,0)\\) of a three-dimensional rectangular coordinate system with the same magnitude of initial velocity \\(v_{0}\\), different launch angles \\(\\theta\\) (\\(\\arctan(\\sqrt{x^{2}+y^{2}}/z)\\)) and different azimuth angles \\(\\varphi\\) (\\(\\arctan y/x\\)), and the acceleration due to gravity is \\(g\\). Find the curve equation jointly formed by the points farthest from the origin in the orbits of each stone. If the angular velocity of the Earth's rotation \\(\\Omega\\ll g/v_{0}\\) and its direction is along the positive \\(z\\)-axis, try to find the first-order offset of this curve.",
"zh": "有许多石块以相同的初速度大小$v_{0}$，不同的发射角θ(arctan(\\sqrt{x^2+y^2}/z)以及不同的方向角φ(arctany/x)从三维直角坐标系原点(0,0,0)出射，重力加速度为g。求各个石块的轨道离原点最远处共同构成的曲线方程,如果给定地球自转角速度Ω<<g/$v_{0}$且方向沿z轴正方向,试求该曲线的一阶偏移"
} | The equations are:
\(x^{2}+y^{2}=(-4z^{2}+\frac{2v_{0}^{2}z}{g})\)
and the union of \(z = 0\) and \(0\leq x^{2}+y^{2}\leq\frac{v_{0}^{2}}{g}\)
Offset:
For points \((x,y)\) such that \((x,y)\in x^{2}+y^{2}=(-4z^{2}+\frac{2v_{0}^{2}z}{g})\)
\(\delta(x,y,z)=\omega*\sqrt{\frac{4v_{0}^{2}}{g^{2}}-\frac{6z}{g}}(y\hat{i}-x\hat{j})\)
For points \((x,y)\) such that \((x,y)\in0\leq x^{2}+y^{2}\leq\frac{v_{0}^{2}}{g}\)
Each point has two ways of deviation (this is because particles with two exit angles at the same position can both hit the target)
\(\delta_{\pm}(x,y)=\frac{\omega v_{0}(1\pm\sqrt{1 - \frac{g^{2}*r^{2}}{v_{0}^{4}}})(y\hat{i}-x\hat{j})}{g}\) |
PHY-062 | Physics | null | {
"en": "Inside the inner wall of a hollow spherical shell with radius $R$, there is a small ball that can be regarded as a particle moving at a constant speed along a fixed horizontal circular path. The angle between the line connecting the small ball and the center of the hollow spherical shell and the vertical line is $\\theta$. If during the stable motion process, the small ball is subjected to a small perturbation perpendicular to the direction of motion, causing it to deviate from the original orbit and perform small-amplitude vibrations in the vertical plane along the inner wall, try to find the period of this vibration.",
"zh": "在半径为R的空心球壳内壁，有一可当作质点的小球沿固定的水平圆周作匀速率运动，小球与空心球壳球心的连线与铅垂线的夹角为$\\theta$，若小球在稳定运动过程中受到垂直于运动方向的微小扰动，导致其偏离原轨道并沿内壁在竖直平面内做小幅度振动，试求此振动的周期"
} | $2\pi\sqrt{\frac{R}{g}\frac{\cos\theta}{1 + 3\cos^{2}\theta}}$ |
PHY-063 | Physics | null | {
"en": "A particle 1 collides with a target particle 2 at rest, generating and scattering \\(n - 2\\) new particles \\(3,4,\\cdots,n\\). To achieve this reaction with the minimum kinetic energy of particle 1, find the minimum threshold kinetic energy \\(T\\) of particle 1 under relativistic conditions. Given the speed of light is \\(c\\), and the rest mass of particle \\(i\\) is \\(m_{i0}\\).",
"zh": "以粒子1去碰撞处于静止状态的靶粒子2，生成并散射出n-2个新粒子3，4,.....n.想要用最小的粒子动能发生这种反应，试在相对论条件下求粒子1的最小阈值动能T，已知光速为c,粒子i的静止质量为$m_{i0}$"
} | $c^2*\frac{-m_1^2-m_2^2-2m_1*m_2+(\sum_{i=3}^{n} m_i)^2}{2*m_2}$ |
PHY-064 | Physics | null | {
"en": "In a two-dimensional world, inside a rectangular container, there are a large number of identical small balls randomly moving in all directions. The small balls can be regarded as mass points, and elastic collisions occur between the small balls and between the small balls and the end faces of the container (perpendicular to its length direction). The speed of the small balls and whether they move forward or backward are both random. We can introduce the concept of “temperature” $T$, which is defined by the pressure-temperature relationship of a two-dimensional ideal gas $p = nkT$, where $p$ is defined as the force per unit length and $n$ is defined as the areal number density. Now, turn one of the short sides of the container into an elastic piston, and slowly push the piston inward along the length direction of the container. The moving small balls increase their kinetic energy due to elastic head-on collisions with the piston, thereby increasing the “temperature” of the system.\n\nDerive the relationship between the “temperature” $T$ and the area $A$ during the slow pushing process of the piston. Assume that gravity, gravitational force, all kinds of resistance and friction can be ignored.",
"zh": "在二维世界中一长方形容器内，有大量相同的小球在朝着各个方向随机运动。小球可视为质点，小球之间以及小球与容器端面（与其长度方向垂直）之间作弹性碰撞。小球的速率以及向前还是向后运动都是无规的。可引入 “温度”T的概念，它是由二维理想气体压强温度关系$p=nkT$定义的，其中p定义为单位长度受力，n定义为数量面密度。现在将容器的一短边变为弹性活塞，并使活塞缓慢地沿容器长度方向向内推进，运动的小球因与活塞作弹性正碰撞增加了动能，从而使系统的 “温度” 升高。\n\n试导出在活塞缓慢推进过程中，“温度”T与面积A的关系式。设重力，引力，各种阻力和摩擦力均可忽略。"
} | TA=const |
PHY-065 | Physics | null | {
"en": "Consider a single electron outside the atomic nucleus, which is in the ground state. In this state, let the mean value of the square of the distance from the electron to the center of the atomic nucleus be denoted by $r_0^2$, and the mean value of the square of the electron's momentum (defined as the square of the uncertainty in the momentum component) be denoted by $p_0^2$. Find the inequality satisfied by $r_0^2$ and $p_0^2$, and the result can be expressed in terms of $\\hbar$.",
"zh": "考虑一个原子核外只有唯一一个电子，其处于基态，在此态中用\\(r_0^2\\)表示电子到原子核中心距离平方的平均值，用\\(p_0^2\\)表示电子动量平方的平均值（定义为动量分量不确定量平方），求$r_0^2$和$p_0^2$满足的不等式，结果可用$\\hbar$相关式子表达。"
} | r_0^2p_0^2\geq\frac{9}{4}\hbar^2 |
PHY-066 | Physics | null | {
"en": "Two separated thin lenses are combined into a lens group. Their focal lengths are \\(f_1\\) and \\(-f_2\\) respectively, where \\(0 < f_2 < f_1\\). Their average refractive indices within the used light wave band are \\(n_1\\) and \\(n_2\\) respectively, and their average dispersive powers are \\(D_1\\) and \\(D_2\\) respectively. The question is: What should the separation between the two lenses be so that the lens group has the minimum chromatic aberration?",
"zh": "两个分离的薄透镜组合成透镜组，它们的焦距分别为\\(f_1\\)和\\(-f_2\\)其中\\(0<f_2<f_1\\)，在所使用的光波波段内的平均折射率分别为\\(n_1\\)和\\(n_2\\)，平均色散率分别为\\(D_1\\)和\\(D_2\\)。试问：为使透镜组具有最小的色差，两透镜的间隔应为多大？"
} | \[d=\frac{\frac{f_1D_2}{n_2-1}-\frac{f_2D_1}{n_1-1}}{\frac{D_1}{n_1-1}+\frac{D_2}{n_2-1}}\] |
PHY-067 | Physics | null | {
"en": "An incident light irradiates a light-transmitting surface at an incident angle \\(i\\). The energy flux density of the incident light is \\(I\\). The complex reflectivity of the surface to light is \\(r\\) (including phase), and the complex transmittance to light is \\(t\\), with an exit angle \\(\\varphi\\). Try to find the light pressure and tangential force per unit area of this surface.",
"zh": "入射光以入射角 i 照射到一透光表面上，入射光的能流密度为 I，表面对光的复反射率为 r(包含相位），对光的复透射率为t，出射角φ，为试求该表面单位面积所受的光压和切向力。"
} | P = I((1+rr*)cosi-tt*cosφ)/c
F_切 = I((1-rr*)sini-tt*sinφ)/c |
PHY-068 | Physics | null | {
"en": "The volume of the gas storage tank is \\(V\\), and the gas pressure inside the tank is \\(p\\). The gas storage tank is connected to a vacuum chamber with volume \\(V_0\\) via a valve. The valve is opened to charge the vacuum chamber. After reaching equilibrium, the valve is closed. Then, the gas storage tank is used to continue charging a new vacuum chamber;... and so on for continuous charging until the gas pressure in the gas storage tank reaches \\(p_0 (p_0 < p)\\). Assume that the temperature remains constant during the charging process. How many vacuum chambers are needed in total?",
"zh": "贮气罐的体积为V，罐内气体压强为p。贮气罐经阀门与体积为\\(V_0\\)的真空室相连，打开阀门，为真空室充气，达到平衡后，关闭阀门；然后用该贮气罐继续为一个新的真空室充气；… 。如此不断充气，直到储气罐中气体的压强达到\\(p_0 (p_0 < p)\\)为止 。设充气过程中温度恒定不变 。试问共需多少个真空室 。"
} | \left\lceil \frac{\ln\left( \dfrac{p}{p_0} \right)}{\ln\left( 1 + \dfrac{V_0}{V} \right)} \right\rceil |
PHY-069 | Physics | null | {
"en": "On the Oxy plane, there are two spheres with centers at $(-2R,0,0)$ and $(2R,0,0)$ respectively, both with a radius of $R$. The two spheres are uniformly charged with volume charge densities $\\pm\\rho$ respectively, and they have the same angular velocity $\\omega(t)$ pointing in the positive $z$-axis direction, with $\\frac{d\\omega}{dt}=\\beta$ (a positive constant). Find the vector distribution of the induced electric field throughout space $E(x,y,z) = E_x\\hat{i}+E_y\\hat{j}+E_z\\hat{k}$.",
"zh": "Oxy平面上两个球心分别在 (−2R,0,0) 和 (2R,0,0)，半径同为 R 的球，\n。两球分别均匀带电电荷体密度±ρ，并且拥有相同的方向指向z轴正方向的角速度ω(t)，$\\frac{dω}{dt}$=β（正的常量）。求全空间感生电场矢量分布E(x,y,z) = E_x\\hat{i}+E_y\\hat{j}+E_z\\hat{k}"
} | If $(x + 2R)^2 + y^2+z^2>R^2$ and $(x - 2R)^2 + y^2+z^2>R^2$
then
$E=\frac{\mu_0*(\pi)^2\rho\beta R^6}{15}*(\frac{y}{((x + 2R)^2 + y^2+z^2)^{\frac{3}{2}}}-\frac{y}{((x - 2R)^2 + y^2+z^2)^{\frac{3}{2}}})\hat{i}+\frac{\mu_0*(\pi)^2\rho\beta R^6}{15}*(\frac{(x - 2R)}{((x - 2R)^2 + y^2+z^2)^{\frac{3}{2}}}-\frac{(x + 2R)}{((x + 2R)^2 + y^2+z^2)^{\frac{3}{2}}})\hat{j}$
If $(x + 2R)^2 + y^2+z^2\leq R^2$,
then
$E=\frac{\mu_0*(\pi)^2\rho\beta R^6}{15}*(y*(\frac{2}{R^3}-\frac{((x + 2R)^2 + y^2+z^2)^{\frac{3}{2}}}{R^6})-\frac{y}{((x - 2R)^2 + y^2+z^2)^{\frac{3}{2}}})\hat{i}+\frac{\mu_0*(\pi)^2\rho\beta R^6}{15}*(\frac{(x - 2R)}{((x - 2R)^2 + y^2+z^2)^{\frac{3}{2}}}-(x + 2R)*(\frac{2}{R^3}-\frac{((x + 2R)^2 + y^2+z^2)^{\frac{3}{2}}}{R^6}))\hat{j}$
If $(x - 2R)^2 + y^2+z^2\leq R^2$,
then
$E=\frac{\mu_0*(\pi)^2\rho\beta R^6}{15}*(-y*(\frac{2}{R^3}-\frac{((x - 2R)^2 + y^2+z^2)^{\frac{3}{2}}}{R^6})+\frac{y}{((x + 2R)^2 + y^2+z^2)^{\frac{3}{2}}})\hat{i}+\frac{\mu_0*(\pi)^2\rho\beta R^6}{15}*(-\frac{(x + 2R)}{((x + 2R)^2 + y^2+z^2)^{\frac{3}{2}}}+(x - 2R)*(\frac{2}{R^3}-\frac{((x - 2R)^2 + y^2+z^2)^{\frac{3}{2}}}{R^6}))\hat{j}$ |
PHY-070 | Physics | null | {
"en": "An interface is formed by medium 1 and medium 2, with the refractive indices of the two media being \\(n_1\\) and \\(n_2\\). Introduce \\(n = n_2/n_1\\), and the normal of the interface is aligned with the \\(x\\)-axis of the \\(S\\) frame. Now, the interface moves at a uniform velocity \\(\\beta c\\) along the normal relative to an observer in the \\(S\\) frame. In the \\(S\\) frame, an incident light ray from medium 1 strikes the interface at an incident angle \\(\\theta_i\\), and the reflection angle and refraction angle are denoted as \\(\\theta_r\\) and \\(\\theta_t\\) respectively. Derive the critical angle for total reflection \\(\\theta_f\\) as observed by the observer in the \\(S\\) frame.",
"zh": "由介质 1 和介质 2 构成一界面，两介质的折射率为\\(n_1\\)和\\(n_2\\)，引入$n=n_2/n_1$，界面的法线与S系的x轴一致。现界面相对S系中的观察者以速度\\beta c沿法线作匀速平行运动，在S系中入射光以入射角\\(\\theta_i\\)从介质 1 向界面入射，反射角和折射角分别用\\(\\theta_r\\)和\\(\\theta_t\\)表示。试导出S系中的观察者观测到的全反射临界角$\\theta_f$."
} | \[\theta_f=\arcsin\frac{n}{\sqrt{n^2\beta^2+\frac{1-\beta^2}{1-n^2\beta^2}}}-\arcsin\frac{n\beta}{\sqrt{n^2\beta^2+\frac{1-\beta^2}{1-n^2\beta^2}}}.\] |
PHY-071 | Physics | null | {
"en": "Take the rectangular coordinate system xyz. There is a conduction current in an infinite layer region where \\(-d\\leqslant x\\leqslant d\\). The direction of the current density is along the positive z-axis, and its magnitude varies linearly with x, that is \\( j(x) = j_0 \\frac{|x|}{d} \\). Try to find the distribution of the magnetic induction intensity \\(\\boldsymbol{B}\\) in vacuum.",
"zh": "取直角坐标系xyz，在\\(-d\\leqslant x\\leqslant d\\)的一层无穷区域内有传导电流，电流密度方向为z轴正方向，大小随x线性变化，即\\( j(x) = j_0 \\frac{|x|}{d} \\)。试求真空中磁感应强度\\(\\boldsymbol{B}\\)的分布。"
} | B=\frac{\mu_0j_0}{2d}\min\{x^2,d^2\} |
PHY-072 | Physics | null | {
"en": "It is known that the refractive index \\(n\\) of an optical fiber has a radial distribution given by\n\\(n = n_{0}(1-\\alpha^{2}r^{2})\\)\nwhere \\(n_0\\) is the refractive index at the center, and \\(\\alpha\\) is a positive number much smaller than 1. Try to find the trajectory of a light ray propagating in the fiber, assuming the initial incident slope \\(r_0'\\ll1\\).",
"zh": "已知光学纤维的折射率n沿径向的分布为\n\\(n=n_{0}(1 - \\alpha^{2}r^{2})\\)\n式中\\(n_0\\)为中心的折射率，\\(\\alpha\\)为比 1 小得多的正数 。试求光线在纤维中传播的轨迹 ，设光线的初始入射斜率 r_0'<<1 。"
} | \[r(z)=\frac{r_0'}{\sqrt{2}\alpha}\sin(\sqrt{2}\alpha)\] |
PHY-073 | Physics | null | {
"en": "The electric charge of a muon \\(\\mu\\) is \\(q = -e\\) (\\(e = 1.6\\times10^{-19}\\ C\\)), its rest mass is \\(m_0 = 100\\ MeV/c^2\\), and its lifetime at rest is \\(\\tau_0 = 10^{-6}\\ s\\). Suppose there is a muon at a height \\(h = 10^{4}\\ m\\) above the Earth's equator moving vertically downwards at a speed close to the speed of light in a vacuum. \n1. What is the minimum total energy this muon should have to reach the ground?\n2. If the Earth's magnetic field within the \\(10^{4}\\ m\\) height range above the equator is regarded as a uniform magnetic field with a magnetic induction intensity \\(B = 5\\times10^{-5}\\ T\\), and the direction of the magnetic field is parallel to the ground. Try to find the direction of deviation and the total deflection angle of the muon with the energy obtained in the first question when it reaches the ground.",
"zh": "\\(\\mu\\)子的电量\\(q = -e\\)（\\(e = 1.6×10^{-19}\\ C\\)），静止质量\\(m_0 = 100\\ MeV/c^2\\)，静止时的寿命\\(\\tau_0 = 10^{-6}\\ s\\)。设在地球赤道上空离地面高度为\\(h = 10^{4}\\ m\\)处有一\\(\\mu\\)子以接近于真空中光速的速度垂直向下运动。1.试问此\\(\\mu\\)子至少应有多大总能量才能到达地面？\n2.若把赤道上空\\(10^{4}\\ m\\)高度范围内的地球磁场看作匀强磁场，磁感应强度\\(B = 5\\times10^{-5}\\ T\\)，磁场方向与地面平行。试求具有第 1 问所得能量的\\(\\mu\\)子在到达地面时的偏离方向和总的偏转角。"
} | (1) $3.3\times10^3\text{MeV}$
(2) Deflect westward by $0.023\text{ rad}$ |
PHY-074 | Physics | null | {
"en": "For an electron gas obeying the Fermi distribution with a level degeneracy of 2, given the electron number density \\(n\\), electron mass \\(m\\), Planck constant \\(\\hbar\\), speed of light \\(c\\), and gas volume \\(V\\), at \\(T = 0\\ K\\):\n(1) Find the magnitude distribution \\(f(p)\\) of the electron gas momentum.\n(2) Find the internal energy \\(U\\) and pressure \\(P\\) of the electron gas (relativistic effects need to be considered).",
"zh": "对于服从费米分布的电子气体，其能级简并度为2，给定电子数密度n,电子质量m，普朗克常数$\\hbar$，光速c,气体体积V,在T=0K的情况下：\n(1)求电子气体的动量大小分布f(p)\n(2)求电子气体的内能U和压强P（需要考虑相对论效应）"
} | (1)f(p) = \frac{V}{\pi^{2}\hbar^{3}}*p^2*I_{0,(3\pi^{2})^{1/3}\left(\frac{N}{V}\right)^{1/3}\hbar}(p)
where \(I\) is the indicator function
(2)
\(U=\frac{cV}{8\pi^{2}\hbar^{3}} \left[ p_{F} (2p_{F}^{2} + m^{2}c^{2}) \sqrt{p_{F}^{2} + m^{2}c^{2}} - (mc)^{4} \mathrm{arsinh}\left(\frac{p_{F}}{mc}\right) \right]\)
\(P=\frac{c}{8\pi^{2}\hbar^{3}} \left\{ p_{F} \left( \frac{2}{3}p_{F}^{2} - m^{2}c^{2} \right) \sqrt{p_{F}^{2} + m^{2}c^{2}} + (mc)^{4} \mathrm{arsinh}\left(\frac{p_{F}}{mc}\right) \right\}\)
where, \(p_{F} = (3\pi^{2})^{1/3}\left(\frac{N}{V}\right)^{1/3}\hbar\) |
PHY-075 | Physics | null | {
"en": null,
"zh": null
} | None |
PHY-76 | Physics | images/PHY-76.png | {
"en": "Three homogeneous rigid rods are hinged together as shown in the figure. The angular coordinates are denoted as $\\theta_1$, $\\theta_2$, $\\theta_3$ as shown in the figure. The range of values for all three angular coordinates is $[0, 2\\pi]$. The initial hinge coordinate of the first rod is $(0,0)$, and the final hinge coordinate of the third rod is $(d,h)$. The lengths of the three rods are denoted as $l_1$, $l_2$, $l_3$ respectively, and their masses are denoted as $m_1$, $m_2$, $m_3$ respectively. There is a bug with mass $M$ at a distance $x$ from the midpoint of the second rod. If $x>0$, it means the bug is closer to rod 3, and if $x < 0$, it means the bug is closer to rod 1. The acceleration due to gravity in space is $g$. Solve the following problems:\n(1) If the system is in a state of force equilibrium at this moment, find the equations that the equilibrium position angular coordinates $\\theta_{10}$, $\\theta_{20}$, $\\theta_{30}$ need to satisfy.\n(2) Based on (1), assume the system is about a stable equilibrium and find the perturbation angular frequency $\\omega$.",
"zh": "三根匀质硬杆如图铰接在一起，角标如图所示分别记为θ_1,θ_2,θ_3,三个角标的取值范围均为[0,2$\\pi$]，第一根杆首端铰接坐标为(0,0),第三根杆末端铰接坐标为(d,h),三根杆的长度分别记为l_1,l_2,l_3,质量分别记为m_1,m_2,m_3,在与第二根杆中点相距x处有一质量为M的小虫，若x＞0代表小虫更靠近杆3，x<0代表小虫更靠近杆1.空间重力加速度为g,求解以下问题：\n(1)如果系统此时处于受力平衡状态，平衡位置的角标θ_10,θ_20,θ_30需要满足的方程。\n(2)在(1)的基础上，假设系统关于围绕稳定求微扰角频率ω。\n"
} | (1)
Three equations determine $\theta_{10}$, $\theta_{20}$, $\theta_{30}$:
$
l_1*\cos\theta_{10}+
l_2*\cos\theta_{20}+
l_3*\cos\theta_{30}=d
l_1*\sin\theta_{10}+
l_2*\sin\theta_{20}+
l_3*\sin\theta_{30}=h
(m_1 + m_2+M)*\cos\theta_{10}*(\sin(\theta_{30}-\theta_{20}))+(m_2 + M + 2M*x/l_2)*\sin(\theta_1-\theta_3)+m_3*\sin(\theta_1-\theta_2)=0$
(2)
$\omega = \sqrt{\frac{
\left[
-\left( \frac{1}{2}m_{1} + m_{2} + m_{3}+M \right) gl_1 \sin \theta_{1}
+ \left(
\frac{1}{2} \left( m_{2} +M*(1+2*x/l_2)+ 2m_{3} \right) gl \sin \theta_{2}
\cdot \frac{l_1^{2} \sin^{2} (\theta_{3} - \theta_{1})}{l_2 \sin^{2} (\theta_{2} - \theta_{3})}
+ \frac{1}{2} m_{3} gl_1^2 \sin \theta_{3}
\cdot \frac{l^{2} \sin^{2} (\theta_{3} - \theta_{2})}{l^{3} \sin^{2} (\theta_{2} - \theta_{3})}
\right)
\right]
}{\frac{1}{3}m_1l_1^2 + m_2l_1^2 + \frac{\cos(\theta_1-\theta_2)\sin(\theta_3-\theta_1)}{\sin(\theta_2-\theta_3)}m_2l_1^2 + \frac{\sin^2(\theta_3-\theta_1)}{3\sin^2(\theta_2-\theta_3)}m_2l_1^2 + \frac{\sin^2(\theta_1-\theta_2)}{3\sin^2(\theta_2-\theta_3)}m_3l_1^2
+M*l_1^2*(1+(1/2+x/l_1)^2*(\frac{\sin(\theta_1-\theta_3)}{\sin(\theta_2-\theta_3)})^2+2\cos(\theta_1-\theta_2)*(1/2+x/l_1)*(\frac{\sin(\theta_1-\theta_3)}{\sin(\theta_2-\theta_3)})}}$ |
PHY-77 | Physics | images/PHY-77.png | {
"en": "The crank \\(OA = r\\) rotates with a constant angular acceleration \\(\\alpha\\) about the fixed axis \\(O\\). The connecting rod \\(AB\\) is connected to the end point \\(A\\) of the crank by a hinge and can slide inside a hinged sleeve \\(N\\). When \\(\\varphi=\\pi\\), the angular velocity of the crank \\(OA\\) is \\(\\omega\\). Given that \\(AB = l>2r\\), find the magnitude of the velocity and acceleration of point \\(B\\) on the connecting rod, and the radius of curvature of the trajectory when \\(\\varphi = \\pi\\).",
"zh": "曲柄 \\(OA = r\\)，绕定轴 O 以\\(\\alpha\\)为角加速度匀加速转动，连杆 AB 用铰链与曲柄端点 A 连接，并可在具有铰链的滑套 N 内滑动。当 \\(\\varphi = \\pi\\) 时，曲柄OA的角速度为\\(\\omega\\)。已知 \\(AB = l>2r\\)，求当 \\(\\varphi = \\pi\\) 时，连杆上 B 点的速度、加速度的大小，和轨道的曲率半径。"
} | v=(l/2-r)\omega,
a=\sqrt{(r-l/4)^2\omega^4+(l/2-r)^2\alpha^2},
\rho=\frac{l/2-r}{l/4-r}. |
PHY-78 | Physics | images/PHY-78.png | {
"en": "Consider a symmetric ellipsoid that is in pure rolling on the XOZ plane with center $O'$. The angle between the major axis of the ellipsoid and the $y$ axis is $\\theta$. The equation of the ellipsoid is: $\\frac{x'^{2}+z'^{2}}{b^{2}}+\\frac{y'^{2}}{a^{2}} = 1$. The relationship between the coordinate axes $x',y',z'$ and $x,y,z$ is as shown in the figure.\n\nThe ellipsoid moves along the $OX$ direction with a spin angular acceleration of $\\beta$. There is no precession or nutation, and it makes contact with the $XZ$ plane at point $O$. \n\nFind the coordinates of point $O'$ in the $Xyz$ coordinate system.\n\nFind the velocity $v_O$ and acceleration $a_O$ of $O'$ in the ground system (the $XYZ$ system).",
"zh": "考虑位于 XOZ 平面上纯滚动的对称椭球，中心为 O′，椭球主轴与 y 轴夹角为 θ，椭球方程为：\n(x'^2+z'^2)/b^2+y'^2/a^2=1，坐标轴x',y',z'与x,y,z的关系如图所示\n椭球沿 OX 方向运动，自转角加速度 为β，无进动或章动，与 XZ 平面接触于 O 点。\n求点 O′ 在 Xyz 坐标系下的坐标。\n求 O′ 在地面系（XYZ 系）下的速度 v_O 与加速度 a_O。\n"
} | v = $ω(a^2 \cos^2 θ + b^2 \sin^2 θ)^1/2 \hat{x}-ω \frac{(a^2-b^2)cosθsinθ} {\sqrt{a^2 \cos^2 θ+b^2\sin^2 θ} }\hat{y}$
a = $β \sqrt{a^2 \cos^2{θ} + b^2 \sin ^2{θ}} - ω^2\frac{(a^2-b^2)cosθsinθ} {\sqrt{a^2 \cos^2 θ+b^2\sin^2 θ} }\hat{x}-( β \frac{(a^2-b^2)sinθcosθ}{\sqrt{(a^2 \cos^2 θ + b^2 \sin^2 θ) }}+ ω^2 \frac{(a^2-b^2)(a^2 cos^4 θ-b^2sin^4 θ}{(a^2 \cos^2 θ + b^2 \sin^2 θ)^3} )\hat{y} [(6)]
$ |
PHY-79 | Physics | images/PHY-79.png | {
"en": "As shown in the figure, there are two points A and B in the spherical coordinate system. Using the subscripts $\\theta$ and $\\varphi$, their coordinates are recorded as $(0,0)$ and $(\\frac{\\pi}{2},0)$ respectively, and the radius of the sphere is denoted as $R$. Now a missile is launched from point A, and upon hearing the news, the personnel at point B escape along the equator at a constant speed $v$, while the missile chases B at a constant speed $u$ in a direction always along the instantaneous geodesic from the missile to the personnel. Try to find the differential equations that the coordinates $(\\theta,\\varphi)$ of A need to satisfy.",
"zh": "如图所示，球坐标系下有两点A,B,用角标θ，φ记录二者坐标分别为(0,0)和(\\pi/2,0)，球体半径记为R。下面从A点发射一枚导弹，而B点的人员听到消息后以恒定的速率v沿着赤道逃跑，而导弹以大小恒定为u的，方向始终沿着导弹到人员的瞬时测地线的速度追击B，试求A的坐标(θ,φ)所需要满足的微分方程"
} | $
dθ/dt = v/(R \sqrt{1+\tan^2 (u t/R-φ)}
dφ/dt =\tan(u*t/R-φ)*v/(R*\tanθ* \sqrt{1+\tan^2 (u t/R-φ)}
$ |
PHY-80 | Physics | images/PHY-80.png | {
"en": "Consider a symmetric ellipsoid that is in pure rolling on the XOZ plane with its center at O′. The angle between the principal axis of the ellipsoid and the y-axis is θ, and the equation of the ellipsoid is: \\(\\frac{x'^{2}+z'^{2}}{b^{2}}+\\frac{y'^{2}}{a^{2}} = 1\\). The relationship between the coordinate axes \\(x',y',z'\\) and \\(x,y,z\\) is as shown in the first figure.\n\nThe ellipsoid moves along the OX direction, with an angular acceleration of \\(\\beta\\), and there is no precession or nutation. It makes contact with the XZ plane at point O.\n\nFind the coordinates of point O′ in the Xyz coordinate system.\n\nIn the \\(x'y'z'\\) coordinate system, the position of a certain point C on the sphere is as shown in the second figure:\n\nThe angle between the vector \\(\\overrightarrow{OC}\\) and the \\(y'\\) axis is \\(\\psi\\).\n\nThe angle between the projection of \\(\\overrightarrow{OC}\\) on the \\(x'z'\\) plane and the \\(x'\\) axis is \\(\\varphi\\).\n\nFind: The acceleration vector \\(\\mathbf{a}_{C}\\) of this point C in the XYZ ground coordinate system.",
"zh": "考虑位于 XOZ 平面上纯滚动的对称椭球，中心为 O′，椭球主轴与 y 轴夹角为 θ，椭球方程为：\n(x'^2+z'^2)/b^2+y'^2/a^2=1，坐标轴x',y',z'与x,y,z的关系如第一张图所示\n椭球沿 OX 方向运动，自转角加速度 为β，无进动或章动，与 XZ 平面接触于 O 点。\n求点 O′ 在 Xyz 坐标系下的坐标。\n在 x′y′z′ 坐标系下，球面上某一点 C 的位置如第二张图所示：\n向量 OC 与 y′ 轴的夹角为 ψ。\nOC 在 x′z′ 平面上的投影与 x′ 轴的夹角为 φ。\n求：该点 C 在 XYZ 地面坐标系下的加速度向量 a_C。"
} | a_C = $(\begin{aligned} =\beta\left[\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}+\frac{ab(\cos\psi \cos\theta-\sin\theta \sin\psi \sin\varphi)}{\sqrt{a^{2}\sin^{2}\psi+\cos^{2}\psi b^{2}}}\right] \\ & -w^{2}\left[\frac{\left(a^{2}-b^{2}\right)\cos\theta\sin\theta}{\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}}+\frac{ab(\cos\psi\sin\theta+\cos\theta\sin\psi\sin\psi)}{\sqrt{a^{2}\sin^{2}\psi+b^{2}\cos^{2}\psi}}\right] \end{aligned})\hat{i}+(\beta[-\frac{\left(a^2-b^2\right)\cos\theta\sin\theta}{\sqrt{a^2\cos^2\theta+b\sin^2\theta}}-\frac{ab\left(\sin\psi\sin\varphi\cos\theta+\sin\theta\cos\psi\right)}{\sqrt{a^2\sin^2\psi+b^2\cos^2\psi}}]\text{ 一}w^2[\frac{(a^2-b^2)(a^2\cos^4\theta-b^2\sin^4\theta}{\left(\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}\right)^{\frac{3}{2}}}+\frac{ab\left(\cos\psi\cos\theta-\sin\theta\sin\psi\sin\phi\right)}{\sqrt{a^2\sin^2\psi+b^2\cos^2\psi}}]\mathrm{(11)})\hat{j}$ |
PHY-81 | Physics | images/PHY-81.png | {
"en": "Three homogeneous rigid rods are hinged together as shown in the figure. The subscripts are denoted as $\\theta_1$, $\\theta_2$, $\\theta_3$ as shown in the figure, and the value range of all three subscript angles is $[0, 2\\pi]$. The lengths of the three rods are denoted as $l_1$, $l_2$, $l_3$ respectively. The hinge coordinate at the head end of the first rod is $(0,0)$, and the hinge coordinate at the end of the third rod is $(d,h)$. Given the values of $\\theta_1$, $\\theta_2$, $\\theta_3$ at time $t$, find the radius of curvature $\\rho(\\theta_1,\\theta_2,\\theta_3)$ of the trajectory of the midpoint of the second rod at time $t$.",
"zh": "三根匀质硬杆如图铰接在一起，角标如图所示分别记为θ_1,θ_2,θ_3,三个角标的取值范围均为[0,2$\\pi$]三根杆的长度分别记为l_1,l_2,l_3,第一根杆首端铰接坐标为(0,0),第三根杆铰接末端坐标为(d,h),\n已知t时刻θ_1,θ_2,θ_3的取值，求第二根杆中点t时刻的运动轨迹的曲率半径ρ(θ_1,θ_2,θ_3)"
} | ρ = $\frac{(sin^2 (θ_2-θ_3)+sin^2 (θ_3-θ_1)/4+sin(θ_2-θ_3)sin (θ_3-θ_1)cos(θ_1-θ_2))^{3/2}}{sin^3 (θ_2-θ_3)/(l_1)+cos(θ_1-θ_2)sin(θ_2-θ_3)sin (θ_3-θ_1)(sin(θ_2-θ_3)/(l_1)+sin (θ_3-θ_1)/(l_2))+sin^3 (θ_3-θ_1)/(4*l_2)+sin(θ_1-θ_2)*(sin^2 (θ_1-θ_2)/l_3+sin^2 (θ_3-θ_1)cos(θ_2-θ_3)/l_2+sin^2 (θ_2-θ_3)cos(θ_3-θ_1)/l_1)/2}$ |
PHY-82 | Physics | images/PHY-82.png | {
"en": "As shown in the figure is a binary star system. Two stars with masses $M_1$ and $M_2$ are located at points $\\alpha_1$ and $\\alpha_2$ in the figure respectively, and the distance between them is $2\\sigma$. The gravitational constant is $G$. Now an object with mass $m$ moves in the above field.\n(1) Now assume that the two stars are stationary. Use the Hamilton-Jacobi equation to construct conserved quantities in the motion, which can use the parameters in the figure and known constants.\n(2) If the binary stars can be regarded as rotating about their center of mass as a perturbation, find the angular velocity of rotation $\\omega$.\n(3) Under the assumption of (2), find the equations satisfied by the equilibrium positions of the object, expressed in terms of the parameters $r_1$ and $r_2$ in the figure.",
"zh": "如图是一个双恒星系统，两个质量分别为M_1,M_2的恒星分别位于图中α_1,α_2两个点，相距为2σ，万有引力常数为G，现有一物体质量为m在如上场中运动\n(1)现假设两个恒星静止不动，利用哈密顿雅可比方程构造运动中的守恒量，可以用图中的参数和已知常数。\n(2)如果双恒星可以看作微扰着二者的质心进行转动，求转动角速度ω\n(3)在(2)的假设下，求物体平衡位置所满足的方程，用图中的r_1,r_2参数表示。\n"
} | (1)$\begin{aligned} & \beta=\sigma^{2}\left(p_{\rho}^{2}+\frac{p_{\varphi}^{2}}{\rho^{2}}\right)-M^{2}-2m\sigma(G M_{1}\mathrm{cos}\theta_{1}+G M_{2}\mathrm{cos}\theta_{2}), \\ & \text{where} \\ & M^{2}=(r\times p)^{2}=p_{\rho}^{2}z^{2}+p_{z}^{2}\rho^{2}+\frac{r^{2}p_{\varphi}^{2}}{\rho^{2}}-2z\rho p_{z}p_{\rho}, \end{aligned}$
$p_z$, $p_{\rho}$, $p_{\varphi}$ are momenta defined in cylindrical coordinates.
(2)$\omega = \sqrt{\frac{G(M_1 + M_2)}{8\sigma^3}}$
(3) There are three equilibrium positions, which are respectively:
(3.1) $r_1 + r_2 =2\sigma$
$M_1\left(\frac{1}{(r_1)^2}-\frac{r_1}{8\sigma^3}\right)=M_2\left(\frac{1}{(r_2)^2}-\frac{r_2}{8\sigma^3}\right)$
(3.2) $r_1 - r_2 =2\sigma$
$M_1\left(\frac{1}{(r_1)^2}-\frac{r_1}{8\sigma^3}\right)+M_2\left(\frac{1}{(r_2)^2}-\frac{r_2}{8\sigma^3}\right)=0$
(3.3) $r_2 - r_1 =2\sigma$
$M_1\left(\frac{1}{(r_1)^2}-\frac{r_1}{8\sigma^3}\right)+M_2\left(\frac{1}{(r_2)^2}-\frac{r_2}{8\sigma^3}\right)=0$ |
PHY-83 | Physics | images/PHY-83.png | {
"en": "As shown in the figure, an orbital station with mass \\(M\\) and a spaceship with mass \\(m\\) docked to it are moving together in a circular orbit around the Earth. The orbital radius is \\(n\\) times the Earth's radius \\(R\\), where \\(n = 1.5\\). At a certain instant, the orbital station launches the spaceship in the direction of motion. After the launch, the spaceship and the orbital station will move along their respective new elliptical orbits. It is known that the apogee of the spaceship's elliptical orbit is at a distance of \\(6nR\\) from the center of the Earth. Assume that the mass of the orbital station remains unchanged after launching the spaceship (i.e., the mass of the ejected gas is ignored).\n\nHere's the question: For what value of the mass ratio \\(\\frac{m}{M}\\) will the spaceship just meet the orbital station after one full revolution around the Earth?",
"zh": "如图，质量为 M 的宇航站和与其对接上的质量为 m 的飞船一起绕地球沿圆轨道运动，轨道半径是地球半径 R 的 n 倍，n = 1.5。某一瞬时，宇航站将飞船沿运动方向发射出去。发射后，飞船和宇航站将分别沿各自新的椭圆轨道运行。已知飞船椭圆轨道的远地点离地球中心的距离为 6nR。设发射飞船后宇航站的质量不变（即忽略喷气的质量）。\n\n试问：当质量比\\(\\frac{m}{M}\\)为何值时，飞船绕地球一周后刚好与宇航站相遇。"
} | \frac{m}{M}=0.074或0.240 |
PHY-84 | Physics | images/PHY-84.png | {
"en": "As shown in the figure, for a horizontally placed parallel-plate capacitor, one plate is below the dielectric liquid surface and the other is above the liquid surface. The relative permittivity of the liquid is $\\epsilon_r$, and its density is $\\rho$. After delivering surface charge densities $\\sigma$ and $-\\sigma$ to the upper and lower plates of the capacitor respectively, the liquid level inside the capacitor rises by $h$ (unknown) and then remains in an equilibrium state.\n\nTry to find:\n(1) Calculate the change in the system's electric potential energy and the change in gravitational potential energy before and after the liquid level rises, and thereby obtain the height $h_1$ of the liquid level rise.\n(2) Directly calculate the force exerted on the liquid between the plates, and thereby obtain the height $h_2$ of the liquid level rise.\n(3) Calculate the energy-momentum tensor of the electric field between the plates, and calculate the force on the liquid surface, thereby obtaining the height $h_3$ of the liquid level rise.",
"zh": "如图所示，水平放置的平行板电容器，一块板在电介质液面下方，另一块在液面上方，液体的相对介电常数为ε_r,密度为ρ，传送给电容器上下极板电荷面密度分别为σ,-σ后，电容器内液面升高h(未知)后保持平衡状态。\n试求\n(1)通过升高前后系统电势能的的改变量和重力势能的改变量，由此得到液面升高的高度h_1\n(2)直接计算极板之间液体的受力，由此得到液面升高的高度h_2\n(3)计算极板之间电场的能量动量张量，并计算液面受力，由此得到液面升高的高度h_3"
} | (2) Solution: $h = (\varepsilon _{\mathrm{r} }^{2}- 1) \sigma ^{2}/ 2\varepsilon _{\mathrm{r} }^{2}\varepsilon _{0}\rho g$,
(1)(3) Solution: $h=(\varepsilon_r-1)\sigma^2/2\varepsilon_r\varepsilon_0\rho g.$ |
PHY-85 | Physics | images/PHY-85.png | {
"en": "An infinitely long coaxial and co - apical double cone made of thin sheets are insulated from each other at the vertex \\(O\\). The angles between their generatrices and the cone axis are \\(\\theta_1\\) and \\(\\theta_2\\) respectively, as shown in the figure.\n\nIf both thin sheets are conductors, with the initial potential of the inner sheet being \\(\\varphi_1\\) and that of the outer one being \\(\\varphi_2\\). Between the two sheets, there is a dielectric with electrical conductivity \\(\\sigma\\) and permittivity \\(\\varepsilon\\). Quasistatically find the potential distribution \\(V(r,\\theta,t)\\) between the plates.",
"zh": "用薄片制成的无限长同轴共顶双锥，在顶点O 是彼此绝缘的，它们的母线与锥轴的夹角 分别为θ_1 和θ_2 ，如图所示.\n若两个薄片均为导体，初始时内部薄片的电势为φ_1，外部为φ_2,两个薄片之间装有电导率为σ，介电常数为ε的电介质，试准静态地求极板间的电势分布V(r,θ,t)"
} | $V(r,\theta,t)=\frac{\exp{-\sigma t/\varepsilon}(\varphi_1 - \varphi_2)\ln\left(\tan\frac{\theta}{2}\right)-\varphi_2\ln\left(\tan\frac{\theta_2}{2}\right)+\varphi_2\ln\left(\tan\frac{\theta_1}{2}\right)}{\ln\left(\tan\frac{\theta_1}{2}\right)-\ln\left(\tan\frac{\theta_2}{2}\right)}$ |
PHY-86 | Physics | images/PHY-86.png | {
"en": "As shown in the figure, at a distance \\(h\\) from an infinitely large uncharged conducting plane, there is an uncharged conducting sphere with a radius of \\(R\\). A uniform electric field \\(E\\) exists throughout space, directed from top to bottom. Using the infinite method of images, assume that the new image charge generated on the conducting spherical surface by the image charge produced by the conducting plane at the \\(n\\)th time (which should obviously be located inside the sphere) is at a distance \\(x_n\\) from the center of the sphere, with an equivalent charge amount of \\(q_n\\) and an equivalent electric dipole moment of \\(p_n\\).\n(1) Determine the initial values \\(x_0\\), \\(p_0\\), \\(q_0\\) of the above three quantities when \\(n = 0\\), as well as the recurrence relations.\n(2) Determine \\(x_n\\), \\(p_n\\), \\(q_n\\).",
"zh": "如图，距离无限大不带电导体平面h处有一半径为R的不带电导体球，全空间存在从上向下的均匀电场E,采取无穷电像法，即假设第n次导体平面产生的像电荷对于导体球面产生的新像电荷(显然应该位于球体内部)距离球心x_n,等效电荷量为q_n,等效电偶极子为p_n\n(1)确定n=0时，上述三者的初值x_0,p_0,q_0以及递推式\n(2)确定x_n,p_n,q_n\n"
} | (1)$p_0 = 4 \pi ε_0 E R^3$
$x_0 = 0$
$q_0 = 0$
$x_{n+1} = \frac{R^2}{2h-x_n}$
$p_{n+1} = (\frac{x_{n+1}}{R})^3*p_n$
$q_{n+1} =q_n*\frac{x_{n+1}}{R}+p_n*(\frac{x_{n+1}}{R})^2$
(2)
$x_n = R^2*\frac{(h+\sqrt{h^2-R^2})^n-(h-\sqrt{h^2-R^2})^n}{(h+\sqrt{h^2-R^2}）^{n+1}-(h-\sqrt{h^2-R^2})^{n+1}}$
$p_n = \frac{8 \pi ε_0 E R^{n+3}*\sqrt{h^2-R^2}}{(h+\sqrt{h^2-R^2})^{n+1}-(h-\sqrt{h^2-R^2})^{n+1}}$
$q_n = \frac{8 \pi ε_0 E R^{n+3}*\sqrt{h^2-R^2}}{(h+\sqrt{h^2-R^2)}^n-(h-\sqrt{h^2-R^2})^n}*\sum_{k=1}^{n} (\frac{(h+\sqrt{h^2-R^2})^k-(h-\sqrt{h^2-R^2})^k}{(h+\sqrt{h^2-R^2})^{k+1}-(h-\sqrt{h^2-R^2})^{k+1}})^2$ |
PHY-87 | Physics | images/PHY-87.png | {
"en": "Two parallel infinite conducting planes are separated by a distance $d$. At a point $P$ which is at a distance $a$ from the surface of one of the conductors, there is a point charge with charge $q$. Given that the electric potential of both conductors is zero, find the induced charges on the two conducting planes respectively.",
"zh": "两个平行的无穷大导体平面，相距为d ，在离其中一导体表面为a 处的 P 点，有一电荷量为q 的点电荷.已知两导体的电势都是零，试分别求两导体平面 上的感应电荷量."
} | Q_2 = -aq/d
Q_1 = -(d-a)q/d |
PHY-88 | Physics | images/PHY-88.png | {
"en": "As shown in the figure, there is an electric dipole $p_1$ at the origin, and another non-fixed electric dipole $p_2$ at a distance $r$ from this point. The relative angle of the position is denoted as $\\theta$, and the deflection angles of the electric dipole vectors $p_1$ and $p_2$ are denoted as $\\varphi_1$ and $\\varphi_2$ respectively. Assume that the masses of the electric dipoles are $m_1$ and $m_2$ respectively, and the moments of inertia about their own centers of mass are $I_1$ and $I_2$ respectively. The two electric dipoles are connected by a spring with an original length of 0 and a spring constant of $k$.\n\nFind:\n(1) The equilibrium position of the system.\n(2) Initially, the two are in equilibrium and the electric dipole vectors are in the same direction as the line connecting them. Find the perturbation modes of the system.",
"zh": "如图所示，在原点有一个电偶极子p_1,在距离该点r有另一非固定电偶极子p_2,位置的相对角标记作θ，再记p_1,p_2电偶极子矢量偏转角度为φ_1,φ_2,假定电偶极子的质量分别为m_1,m_2,相对于自身质心的转动惯量为I_1,I_2,两个电偶极子之间用一根原长为0的劲度系数为k的弹簧连接，\n求\n(1)系统的平衡位置。\n(2)初态二者平衡且电偶极子矢量与二者之间的连线均同向，求系统的微扰模式。"
} | (1) r = $(\frac{3p_1 p_2}{2\pi ε_0 k})^{(1/5)}$
2θ - φ_1 - φ_2 = 0 or $\pi$
φ_1 - φ_2 = 0 or $\pi$
(2)
The angular frequency of r's vibration is $\sqrt{\frac{5 k m_1 m_2}{m_1 + m_2}}$
The three angular frequencies of θ, φ_1, φ_2 are
ω_1 = 0
ω_23^2 is the solution of the following quadratic equation
$abc x^2+(2ab + 2ac+6bc)x + 3(a + b + c)=0$
where,
$a=\frac{m_1 m_2}{k(m_1 + m_2)}$
$b = 3I_1/(k r^2)$
$c = 3I_2/(k r^2)$
$r = (\frac{3p_1 p_2}{2\pi ε_0 k})^{(1/5)}$
|
PHY-89 | Physics | images/PHY-89.png | {
"en": "As shown in the figure is the generation process of a Koch curve, and the figure shows the boundary of the Koch curve. Now, connect the bases of the sub-triangles generated at each step to obtain a fractal resistor network. Assume that the resistance per unit length is $\\lambda$, and the side length of the triangle in Figure 1 is $a$. Find the fractal dimension $d$ of this figure and the resistance $R$ between the two endpoints in Figure 1 of this fractal resistor network.",
"zh": "如图是一个科赫曲线的生成过程，图中展示了科赫曲线的边界，现在把每一次生成的子三角形底边相连，得到一个分形电阻网络。假设单位长度的电阻为λ，图一中的三角形边长为a，求该图形的分形维度d并求该分形电阻网络在图一的两个端点之间的阻值R"
} | d=$\frac{ln(4)}{ln(3)}$
R= 11λa/36 |
PHY-90 | Physics | images/PHY-90.png | {
"en": "As shown in the figure, it is a resistance network of a regular N-sided prism. There are \\(n\\) resistors with resistance \\(R\\) on each edge, and there are also \\(n\\) regular N-sided cross-sections located between every two resistors, all with a resistance of \\(R\\). (The figure shows the case where \\(N = 5\\) and \\(n=3\\), and each edge is a resistor with resistance \\(R\\)). Now, a cylindrical coordinate system \\((\\theta,z)\\) is established to mark the position of each node, with the first plane being \\(z = 0\\). Find the resistance value \\(r\\) between the two points \\((0,0)\\) and \\((\\frac{2k\\pi}{N},z)\\).",
"zh": "如图为一正N边形柱体电阻网络，每条棱上有n个阻值为R的电阻，同时也有n个正N边形截面位于两两电阻之间，阻值也均为R。(图示为N=5，n=3的情况，每条边都是阻值为R的电阻)现在建立柱坐标系(θ,z)来标定每个节点的位置，则第一个平面为z=0,求(0,0)与(2k\\pi/N,z)两点间的电阻阻值r"
} | r = $\sum_{p=0}^{n} \sum_{q=0}^{N-1} \frac{2 R \cos(p\pi/(n+1))(1-\cos(2 k q\pi/N))\cos((p \pi*(2z+1)/(n+1))}{N(n+1)(2-\cos(2p\pi/(n+1))-\cos(2q\pi/N))}$ |
PHY-91 | Physics | images/PHY-91.png | {
"en": "As shown in the figure, it is one iteration of a resistor network, that is, the original resistor network composed of nine small triangles has its internal six small triangles refined into smaller nine triangles. What is the resistance value between points B and C after n such processes? Given that the side length of the outermost largest triangle is a, and the resistance per unit length is λ.",
"zh": "如图为电阻网络的一次迭代，即将原来由九个小三角形组成的电阻网络的内部六个小三角形细化成更小的九个三角形，请问n次这样的过程后，BC两点的电阻值是多少？已知最外层最大的三角形边长为a，单位长度电阻为λ"
} | r = $\frac{2*λa*5^(n+1)}{3*7^(n+1)}$ |
PHY-92 | Physics | images/PHY-92.png | {
"en": "As shown in the figure is a source-containing circuit. The relationship between each power source and time is a periodic function $\\epsilon(t)=at^{2}\\ (-T/2 < t < T/2)$, $\\epsilon(t + T)=\\epsilon(t)$. Try to find the steady-state current $I$ in each branch.",
"zh": "如图为一含源电路，各个电源与时间的关系为周期函数ε(t) =$at^(2)$ (-T/2<t<T/2)，ε(t+T) = ε(t)\n试求每条边上稳态电流I"
} | ["$\\begin{cases}i_1=\\sum_{n=1}^{∞}\\frac{4\\left(3Z_L+2Z_C\\right)}{\\left(3Z_L+Z_C\\right)\\left(4Z_L+3Z_C\\right)}\\widetilde{u}, \\\\ \\\\ i_2=\\sum_{n=1}^{∞}\\frac{4Z_L}{\\left(3Z_L+Z_C\\right)\\left(4Z_L+3Z_C\\right)}\\widetilde{u}. \u0026 \\end{cases}$\nwhere $Z_L = i*L*2n\\pi/T$\n$Z_C = 1/(i*C*2n\\pi/T)$\n\n$\\widetilde{u} = 4*(-1)^n*T^2/(n\\pi)^2\\exp(2ni\\pi t/T)$\nwhere $i$ is the imaginary unit with $i^2=-1$\nThe actual current is the real part of the complex current"] |
PHY-93 | Physics | images/PHY-93.png | {
"en": "As shown in the figure is a finite-length solenoid coil with a current magnitude of \\(I\\) and the number of turns per unit length of \\(n\\). Use the parameters given in the figure to calculate the magnitude of the magnetic field at this location.",
"zh": "如图是一个有限长螺线圈，电流大小为I，单位长度的匝数为n，利用图中所给参数计算该处磁场大小"
} | B= μ_0*n*I(cosθ_1+cosθ_2)/2 |
PHY-94 | Physics | images/PHY-94.png | {
"en": "As shown in the figure, in the $Oxyz$ coordinate space of a certain inertial frame, the distribution of the steady current is\n\\(\\boldsymbol{j}=\n\\begin{cases}\nj_0\\cdot(x/a)^2n\\boldsymbol{e}_j, & \\vert x\\vert\\leqslant a, \\\\\n0, & \\vert x\\vert>a, & & \n\\end{cases}\\) \nwhere \\(j_0\\) is a constant, and \\(\\boldsymbol{e}_j\\) is the direction vector of the \\(y\\)-axis. A particle with mass \\(m\\) and charge \\(q > 0\\) is at the position \\((-a,0,0)\\) at \\(t = 0\\), and has a velocity \\(v\\) pointing in the positive direction of the \\(y\\)-axis, satisfying \\(\\frac{q\\cdot\\mu_0\\cdot j_0\\cdot a^2}{m\\cdot(n + 1)\\cdot(n+2)}=v\\). Find the time \\(t\\) it takes for the particle to reach \\(x = a\\).",
"zh": "如图所示,在某惯性系的 Ox yz 坐标空间中,稳恒电流的分布为\n\\boldsymbol{j}=\n\\begin{cases}\nj_0*(x/a)^2n\\boldsymbol{e}_j, & \\mid x\\mid\\leqslant a, \\\\\n0, & \\mid x\\mid>a, & & \n\\end{cases} & j_0为常量\n其中 e_j 为y 轴的方向矢量. 质量 m,电量 q>0 的质点,在t=0时刻处于 （-a,0,0) 位置,具有指向y轴正方向的速度v，满足q*μ_0*j_0*a^2/(m*(n+1)*(n+2))=v\n求到达x=a处粒子的耗时t"
} | t = $\frac{a*\Gamma(1/2)*\Gamma(1/(4n+4))}{v*(2n+2)*\Gamma(1/2+1/(4n+4))}$ |
PHY-95 | Physics | images/PHY-95.png | {
"en": "The refractive index of a semi-elliptical cylindrical glass is \\(n = \\sqrt{2}\\), and the eccentricity \\(e = 3/4\\). It is placed in air. In the plane perpendicular to the axis of the semi-elliptical cylinder, light is incident at an angle of \\(45^{\\circ}\\) on the planar surface of the major axis of the cross-section of the semi-elliptical cylinder. Determine within what range of angles relative to the left focus does the light emerge from the semi-elliptical cylinder (expressed as a range of polar angles).",
"zh": "半椭圆柱形玻璃的折射率\\(n = \\sqrt{2}\\)，离心率\\(e=3/4\\)，放置在空气中。在垂直于半椭圆柱体轴的平面内，光线以\\(45^{\\circ}\\)角入射在半椭圆柱体的截面长轴平表面上。试问光线从半椭圆柱体的什么范围内透出（以相对于左侧焦点的极角范围表示）。"
} | 121.4^{\circ}\leq\phi\lea 176.2^{\circ} |
PHY-96 | Physics | images/PHY-96.png | {
"en": "As shown in the figure, a plano-convex thin lens made of the same glass is attached to the outer wall of a thin-walled glass liquid tank filled with kerosene. The focal length of the lens in air is \\(f_0\\). Given that the refractive indices of kerosene, glass, and air are \\(n_1=\\frac{7}{5}\\), \\(n = \\frac{8}{5}\\), \\(n_2 = 1\\) respectively. There is an object point \\(Q\\) on the optical axis of the lens at a distance \\(s=\\frac{f_0}{2}\\) from the inner wall. Try to find: 1. The position of the image and the lateral magnification. 2. What is the result if the lens is attached to the inner wall of the water tank?",
"zh": "如图所示，在装满煤油的薄壁玻璃液体槽的外壁贴着由同种玻璃制成的平凸薄透镜，透镜在空气中的焦距为\\(f_0\\)。已知煤油，玻璃，空气的折射率分别为\\(n_1 = \\frac{7}{5}\\)，\\(n = \\frac{8}{5}\\)，\\(n_2 = 1\\)。在透镜光轴上距内壁\\(s = f_0/2\\)处有一物点 Q。试求：1. 像的位置和横向放大率。2. 若将透镜贴在水槽内壁，结果如何？"
} | 1. s'=-\frac{5}{9}f_0, M=14/9
2. s'=-\frac{15}{37}f_0, M=42/37 |
PHY-97 | Physics | images/PHY-97.png | {
"en": "As shown in the figure, a converging lens is placed behind the double holes \\(S_1\\) and \\(S_2\\) with a spacing of \\(d\\), and a screen is placed on the back focal plane of the lens. This interference device faces a distant extended light source. The extended light source is rectangular, with a total length of \\(b\\) along the direction of the double-hole connection line. Taking the center as the origin, the \\(x\\)-axis is established along the vertical direction of the double holes. The luminous intensity of this rectangle is \\(I(x) = i_0\\cos(\\frac{\\pi x}{b})\\). Find the fringe contrast on the screen.",
"zh": "如图所示，间距为 d 的双孔\\(S_1\\)和\\(S_2\\)后放置一会聚透镜，透镜后焦面上放一屏幕。该干涉装置正对远处的扩展光源。扩展光源为矩形，其顺着双孔连线的方向上总长度为b，以中心为原点，顺着双孔竖直方向建立x轴，该矩形的发光强度为\\(I(x)=i_0\\cos(\\frac{\\pi x}{b})\\)，求屏幕上的条纹反衬度。"
} | \(\gamma=\left|\frac{\cos(\frac{\pi db}{\lambda R})}{1-(\frac{2db}{\lambda R})^2}\right|\) |
PHY-98 | Physics | images/PHY-98.png | {
"en": "As shown in the figure, there is a magnetic dipole inside a spherical magnetic medium with a magnetic moment magnitude of \\(m\\). The permeability inside the sphere is \\(\\mu_1\\), and outside it is \\(\\mu_2\\). The radius of the sphere is \\(R\\). Find the magnetic field distribution \\(B(r,\\theta)\\) in the entire space in spherical coordinates.",
"zh": "如图所示，球体磁介质内部有一磁偶极子，磁矩大小为m,球体内部的磁导率为μ_1,外部为μ_2,球体半径为R,\n求球坐标系下全空间磁场分布B(r,θ)"
} | If \(r > R\)
\(B(r,\theta)=\frac{3m\mu_1\mu_2\cos\theta}{2\pi r^3(\mu_1 + 2\mu_2)}\hat{r}+\frac{3m\mu_1\mu_2\sin\theta}{4\pi r^3(\mu_1 + 2\mu_2)}\hat{\theta}\)
If \(r < R\)
\((\frac{m\mu_1\cos\theta}{2\pi r^3}-\frac{3m\mu_1(\mu_1 - \mu_2)\cos\theta}{2\pi R^3(\mu_1 + 2\mu_2)})\hat{r}+(\frac{m\mu_1\sin\theta}{4\pi r^3}+\frac{3m\mu_1(\mu_1 - \mu_2)\sin\theta}{2\pi R^3(\mu_1 + 2\mu_2)})\hat{\theta}\) |
PHY-99 | Physics | images/PHY-99.png | {
"en": "As shown in the figure, it is a spiral whose polar coordinate equation can be written as \\(r = a\\exp{\\theta}\\) (\\(0\\leq\\theta<\\infty\\)).\nTry to calculate the magnetic vector potential at a point directly above the origin and at a distance \\(z < a\\). Express the answer in the form of a series. Use the gauge with divergence equal to 0.",
"zh": "如图为一螺旋线，极坐标方程可以记作r = a $\\exp{θ}$ （0<=θ<∞)\n试计算在原点正上方，距离z<a处的点处的磁矢势矢量，答案用级数表示.用散度为0的规范。\n"
} | A = $\sum_{n=0}^{∞} \frac{(2n-1)(-1)^n \Gamma{1/2+n}z^{2n}}{a^{2n} n!*\sqrt{\pi}(4n^2+1)}\hat{i}+\sum_{n=0}^{∞} \frac{(2n+1)(-1)^n \Gamma{1/2+n}z^{2n}}{a^{2n} n!*\sqrt{\pi}(4n^2+1)}\hat{j}$ |
PHY-100 | Physics | images/PHY-100.png | {
"en": "As shown in the figure, it is a five-pointed star frame composed of five metal rods with length \\(l\\) and resistance \\(r\\) connected end-to-end. The metal rods intersect but do not make contact. Now, this frame rotates around one of its left-right axes of symmetry with an angular velocity \\(\\omega\\). There is a background magnetic field \\(B = B_0\\cos\\omega t\\), and the direction of the magnetic field is perpendicular to the axis of rotation of the frame.\n(1) Initially, the magnetic field is perpendicular to the plane of the wire frame. Find the charge \\(q\\) flowing through the loop after one full rotation.\n(2) At time \\(t\\), find the external torque \\(M\\) required to maintain the uniform rotation of the frame.",
"zh": "如图所示为一五角星框架，框架由5根长为l,电阻为r的金属棒首尾相接构成，金属棒相交但不接触。现在该框架以自转角速度ω绕自己的一个左右对称轴进行旋转。旋转背景有磁场B=B_0 cosωt,磁场方向与框架自转轴垂直。\n(1)初态磁场与线框平面垂直，求旋转一圈后流过回路的电量q。\n(2)t时刻维持框架匀速转动所需要施加的外力矩M。"
} | (1)
q = $\frac{B_0 l^2 sin(\pi/10) cos(\pi/10)}{r(1+sin(\pi/10))}$
(2)
M = $\frac{5B_0^2 ω l^4 sin^2(\pi/10) cos^2(\pi/10) sin^2(2ωt)}{16r(1+sin(\pi/10))^2}$ |
PHY-101 | Physics | images/PHY-101.png | {
"en": "As it is an electromagnetic system, there is a charge of \\(q/4\\) at the origin, and there are two charges of \\(-q\\) with mass \\(m\\) above and below the point at a distance \\(a\\) from the origin. They are connected to the charge at the origin by springs with a spring constant of \\(k\\). Initially, the system is at rest. Try to find:\n\nAfter releasing the system, the radiation power per unit area \\(P\\) received at point \\(P\\) at a position \\(r\\gg a\\) in the figure.",
"zh": "如同为一电磁系统，原点处有一电量为q/4的电荷，距离原点a处上下各有一个带电量为-q质量为m的电荷，且与原点处电荷以劲度系数为k的弹簧相连接,初态系统静止。试求：\n释放系统后，在图中P点的位置r>>a处接受到的单位面积辐射功率P。"
} | P = $\frac{q^2 a^4 k^3 sin^2(2θ)}{2(\pi)^2 ε_0 c^5 r^2 m^3}$ |
PHY-102 | Physics | images/PHY-102.png | {
"en": "As shown in the figure, the coordinates of a point in the original spherical coordinate system are denoted as $(r', \\theta', \\varphi')$, and the metric in this coordinate system is the normal metric $(ds)^2 = c^2(dt)^2-(dx)^2-(dy)^2-(dz)^2$. Now, relative to the original reference frame, a rotating reference frame is selected with $t' = t$, $r = r'$, $\\theta = \\theta'$, $\\varphi=\\varphi'+\\omega t$. Considering the range of $\\omega r\\sin\\theta < c$, try to find:\n\n(1) The metric of the new reference frame and the shape of a sphere with a radius of $R$ obtained by measuring distances in the original reference frame.\n\n(2) Points $A$ and $B$ are both located on a sphere with a radius of $R$. If an observer stationary at point $A$ in the new reference frame observes that an object at point $A$ rotates one full circle around the equatorial plane, what is the difference $\\Delta t$ between the elapsed time measured by the stationary object at point $A$ and the proper time of object $A$?\n\n(3) Try to find the force on a particle with an instantaneous velocity vector $\\mathbf{v}$ at the position $(r,\\theta,\\varphi)$ in the new rotating reference frame.",
"zh": "如图所示，原球坐标系下点坐标记为(r',θ',φ'),在该坐标系下的度规为正常的度规(ds)^2=c^2(dt)^2-(dx)^2-(dy)^2-(dz)^2.现在相对于原参考系选取转动参考系t'=t,r=r',θ=θ',φ=φ'+ωt\n在ωrsinθ<c的范围内考量，试求：\n(1)新参考系的度规和原参考系下通过测量距离得到的半径为R的球体的形状\n(2)A，B两点均位于半径为R的球体上，如果在新参考系下A点处静止的观察者观测到A点 的物体绕着赤道平面转动了一圈，试问A点静止的物体测得的耗时与物体A的本征时的差值Δt\n(3)试求在新的转动参考系下，位于(r,θ,φ）处拥有瞬时速度矢量v的粒子的受力\n"
} | (1) The new metric is $g_{00}=c^2 - \omega^2 r^2\sin^2(\theta), g_{11}=-1, g_{22}=-r^2, g_{33}=-r^2\sin^2(\theta), g_{03}=-\omega r^2\sin^2(\theta)$, and the rest are 0.
(2) $\Delta t=\frac{2\pi\omega R^2}{c^2 - \omega^2 R^2}$
(3) $f=\frac{m\omega^2 r\sin\theta}{\sqrt{(1 - v^2/c^2)*(1-\omega^2 r^2\sin^2(\theta)/c^2)}}*(\sin\theta\hat{r}+\cos\theta\hat{\theta})+\frac{mv\sqrt{1-\omega^2 r^2\sin^2(\theta)/c^2}}{\sqrt{1 - v^2/c^2}}\times\nabla\times(\frac{\omega^2 r^2\sin^2(\theta)}{c^2-\omega^2 r^2\sin^2(\theta)}\hat{\varphi})$ where $\times$ represents the cross product. |
PHY-103 | Physics | images/PHY-103.png | {
"en": "As shown in the figure, a high-energy particle with an impact parameter \\(b\\) and velocity \\(v_0\\) is incident from infinity towards a target particle. The high-energy particle has a mass of \\(m\\) and a charge of \\(Z_1e\\), while the target particle has a mass of \\(M\\) and a charge of \\(Z_2e\\). Denote \\(k\\) as the electrostatic constant, \\(L = \\frac{m v_0 b}{\\sqrt{1 - v_0^2/c^2}}\\), \\(E=\\frac{mc^2}{\\sqrt{1 - v_0^2/c^2}}\\). Considering only the electrostatic interaction force between the particles, with the target particle fixed, find the deflection angle \\(\\theta\\) in the figure and the conditions for observing the outgoing particle.",
"zh": "如图所示，高能粒子以截距b速度v_0从无穷远处向靶粒子射入，高能粒子质量为m带电量为Z_1e,靶粒子质量为M带电量为Z_2e,记k为静电常数,L=$\\frac{m v_0 b}{\\sqrt(1-v_0^2/c^2)}$,\nE = mc^2/(\\sqrt{1-v_0^2/c^2})\n仅考虑粒子之间的静电相互作用力，固定靶粒子，求图中的偏转角θ以及能观测到出射粒子的条件。"
} |
\(\theta=\frac{2Lc\arccos\left(\frac{kZ_1Z_2e^{2}E}{\sqrt{L^{2}c^{2}E^{2}-m^{2}L^{2}c^{6}+k^{2}Z_1^{2}Z_2^{2}e^{4}m^{2}c^{4}}}\right)}{\sqrt{L^{2}c^{2}-k^{2}e^{4}Z_1^{2}Z_2^{2}}}\)
|
PHY-104 | Physics | images/PHY-104.png | {
"en": "As shown in the figure is the unit cell structure diagram of the NaCl crystal. Assume that the lattice constants $a = b = c$, $\\alpha=\\beta=\\gamma=\\frac{\\pi}{2}$, then at equilibrium, ions are arranged in turn at the vertices of the cubic lattice, that is: $R_{l_{1}l_{2}l_{3}}=(l_{1}a, l_{2}a, l_{3}a, l\\alpha),\\quad l_{1},l_{2},l_{3}=\\cdots, - 2,-1,0,1,2,\\cdots$. In the NaCl crystal, the dominant interaction potential between two particles is the electrostatic potential energy and the repulsive energy $V(r)=Ae^{-r/\\rho}$ caused by Pauli repulsion, where $A$ and $\\rho$ are regarded as known constants. Assume that the electron charge is $e$, the mass of the Na ion is $m$, the mass of the Cl ion is $M$, and the vacuum electrostatic constant is $\\epsilon_0$.\n(1) Try to derive the equation satisfied by the lattice constant $a$. It can be left in the form of an infinite series.\nLet $a$ be a known quantity below.\n(2) For a perturbation of one of the Na ions, regard all other ions as stationary, and find the perturbation period $T_1$; change the Na ion to a Cl ion, and find the perturbation period $T_2$. It can be left in the form of an infinite series.",
"zh": "如图为NaCl晶体的晶胞结构示意图。假定晶格常数a=b=c,α=β=γ=\\pi/2，那么平衡时离子依次排列在立方点阵的顶点上，即：$R_{l_{1}l_{2}l_{3}} = (l_{1}a, l_{2}a, l_{3}a, l\\alpha),\\quad l_{1},l_{2},l_{3} = \\cdots,-2,-1,0,1,2,\\cdots$.在NaCl晶体中，两两粒子之间占主导的相互作用势能为静电势能和由泡利排斥所导致的排斥能V(r)=Ae^{-r/ρ},其中A,ρ均视为已知常数。假定电子电荷量为e，Na离子的质量为m,Cl离子的质量为M,真空静电常数为ε_0\n(1)试导出晶格常数a满足的方程。可保留成无穷级数的形式。\n下设a为已知量\n(2)对于其中一个Na离子微扰，其他所有离子视为静止，求微扰周期T_1;将Na离子改为Cl离子，求微扰周期T_2.可保留成无穷级数的形式\n"
} | (1)\sum_{l_1,l_2,l_3≠0,0,0}^{∞} \frac{e^2 (-1)^{l_1+l_2+l_3+1}}{4 \pi ε_0 a^2 \sqrt{l_1^2+l_2^2+l_3^2}}-\frac{A \sqrt{l_1^2+l_2^2+l_3^2}e^{-\sqrt{l_1^2+l_2^2+l_3^2}a/ρ}}{ρ}=0
(2)T_1=$2\pi*\sqrt{\frac{m}{\sum_{l_1,l_2,l_3≠0,0,0}^{∞} {2e^{-\sqrt{l_1^2+l_2^2+l_3^2}a/ρ}/(3ρ^2)-5e^{-\sqrt{l_1^2+l_2^2+l_3^2}a/ρ}/(3ρ a \sqrt{l_1^2+l_2^2+l_3^2})}}}$ |
PHY-105 | Physics | images/PHY-105.png | {
"en": "As shown in the figure is a right-angle potential barrier.\nIf an incident particle with energy \\(E > U_0\\) is incident from left to right, let \\(k_1=\\frac{\\sqrt{2mE}}{\\hbar}\\), \\(k_2 = \\frac{\\sqrt{2m(E - U_0)}}{\\hbar}\\) Try to find the expression of the system wave function.",
"zh": "如图为一直角势垒，\n如果入射粒子以能量E>U_0从左向右入射，令k_1=$\\sqrt{2mE}/(\\hbar)$,k_2=$\\sqrt{2m(E-U_0)}/(\\hbar)$试求系统波函数表达式."
} | If \(x < 0\),
\(\psi(x)=e^{i k_1 x}+\frac{(k_1 - k_2)(1 - e^{i k_2 a})e^{-i k_1 x}}{k_1 + k_2+(k_1 - k_2)e^{i k_2 a}}\)
If \(0 < x < a\),
\(\psi(x)=\frac{2 k_1 (k_1 + k_2)e^{i k_2 x}}{(k_1 + k_2)^2-(k_1 - k_2)^2 e^{2 i k_2 a}}+\frac{2 k_1 (-k_1 + k_2)e^{-i k_2 x}}{(k_1 + k_2)^2-(k_1 - k_2)^2 e^{2 i k_2 a}}\)
If \(x > a\),
\(\psi(x)=\frac{4 k_1 k_2 e^{i k_2 a + i k_1 (x - a)}}{(k_1 + k_2)^2-(k_1 - k_2)^2 e^{2 i k_2 a}}\) |
PHY-106 | Physics | images/PHY-106.png | {
"en": "As shown in the figure, between two infinitely long concentric grounded conducting cylinders with radii \\( R_1 \\) and \\( R_2 \\) respectively, there is an infinitely long charged concentric elliptical cylinder with the equation \\( \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 \\). The surface charge density of this elliptical cylinder is related to the coordinates by: \\( \\sigma(x,y) = \\frac{\\lambda}{(x^2 + y^2) \\cdot \\sqrt{\\frac{x^2}{a^4} + \\frac{y^2}{b^4}}} \\). Try to find the total charges \\( \\lambda_1 \\) and \\( \\lambda_2 \\) per unit axial length on the two conducting cylinders respectively.",
"zh": "如图两个半径分别为R_1,R_2的可视为无限长同心接地导体圆柱体之间有一个方程为x^2/a^2+y^2/b^2=1的无限长带电同心椭球柱体，其带电量的面密度和坐标的关系为：$σ(x,y)=λ/((x^2+y^2)*\\sqrt{x^2/a^4+y^2/b^4})$,试分别求单位轴向长度上两导体圆柱体的总带电量λ_1和_2。"
} | λ_1 = -λ*$\frac{2\pi ln(\frac{2ab}{(b+a)R_1})}{ln(R_2/R_1)}$
λ_2 = λ*$\frac{2\pi ln(\frac{2ab}{(b+a)R_2})}{ln(R_2/R_1)}$ |
PHY-107 | Physics | images/PHY-107.png | {
"en": "\n\nAs shown in the figure, an infinitely long straight wire carries a current \\( I \\). Beside it is a uniformly charged ring with radius \\( r \\) and charge linear density \\( \\lambda \\), whose center is at a distance \\( d > r \\) from the long straight wire. In the current state, the ring is stationary and naturally in force equilibrium. The center of mass of the ring can move freely in translation, and the ring can also rotate freely. An impulsive perturbation is initially applied to the ring, giving the center of mass of the ring a small initial velocity within the plane of the ring without angular velocity. Solve the subsequent motion mode of the ring and the normal frequency.",
"zh": "如图一条无限长直导线通过电流为I，其旁边有一半径为r，电荷线密度为λ的均匀带电圆环，圆心距离长直导线为d>r.当前状态下圆环静止自然也受力平衡，圆环的质心可以自由平动，圆环也可以自由转动，初态给圆环一个冲量微扰，使得圆环质心获得一个小的位于圆环平面内的初速度而没有角速度，求解圆环之后运动的方式和简正频率。"
} |
The ring still maintains periodic translation in the plane and periodic rotation about an axis perpendicular to itself, with the normal frequency being
\[
\omega = \frac{\mu_0 I \lambda \left( \frac{d^2 + r^2}{d^2 - r^2} - \frac{2d}{\sqrt{d^2 - r^2}} + 1 \right)}{m r^2}
\] |
PHY-108 | Physics | images/PHY-108.png | {
"en": "\n\nIn a constant-temperature environment at temperature \\( T \\), there is a vertical infinite long straight cylinder containing an ideal gas separated by \\( K \\) smooth pistons with masses \\( M_1 = M \\), \\( M_2 = 2M \\), ..., \\( M_K = KM \\). The thickness of the pistons is negligible, and there is no air leakage between the pistons and the cylinder wall. From bottom to top, the total number of gas molecules in each part is \\( N_1 = N \\), \\( N_2 = 2N \\), ..., \\( N_K = KN \\), respectively, and the space above the top piston is a vacuum. Taking the bottom of the cylinder as the zero point of gravitational potential energy, find the total gravitational potential energy of the entire gas-piston system.",
"zh": "在温度为$T$的恒温环境中有一竖直无限长直筒，桶中有一种理想气体，被$K$个质量分别为$M_1=M, M_2=2M, \\ldots, M_K=KM$的光滑活塞所分隔开，活塞厚度忽略不计，活塞与桶壁之间不漏气。自下而上每一部分的气体分子总数分别为$N_1=N, N_2=2N, \\ldots，N_K=KN$，最上方活塞以上为真空，求整个气体-活塞系统总的重力势能，以桶底处为重力势能零点。"
} | \frac{K(K+1)}{2}NkT |
PHY-109 | Physics | images/PHY-109.png | {
"en": "\n\nAs shown in the figure, there is a vertical three-chamber cylinder with two pistons of masses \\( M_1, M_2 \\) and cross-sectional areas \\( S_1, S_2 \\), connected by a spring with a spring constant \\( k \\). Initially, all three chambers are in equilibrium at a temperature \\( T_0 \\). Chamber \\( C_1 \\) contains saturated water vapor with a small amount of liquefaction at a pressure \\( p_1 \\) (the volume and heat capacity of the liquid water are negligible, and the vibrational degrees of freedom of water vapor molecules are unexcited), with a known phase change latent heat \\( L \\) (considered independent of temperature). Chamber \\( C_2 \\) contains a monatomic gas, and chamber \\( C_3 \\) contains a diatomic gas with an initial pressure \\( p_3 \\). Except for a heat source at the bottom of chamber \\( C_3 \\) that maintains its gas temperature at \\( T_0 \\), the cylinder walls and pistons are completely non-conductive. The acceleration due to gravity is \\( g \\). The initial volumes of the three chambers are denoted as \\( V_1, V_2, V_3 \\). \n\n(1) Solve for the initial gas pressure \\( p_2 \\) in \\( C_2 \\) and the elongation \\( \\Delta x \\) of the spring. \n(2) Using the Clapeyron equation, derive the relationship between the volume \\( V \\) and water vapor pressure \\( P \\) in \\( C_1 \\) under adiabatic conditions. \n(3) Now, apply a small perturbation to this equilibrium system and determine the equation satisfied by the normal frequency \\( \\omega \\).",
"zh": "如图为一直立的三室气缸，内有质量分别为M_1,M_2，截面积分别为S_1,S_2,的两个活塞，之间以劲度系数为k的弹簧相连接，初态时三个气缸均位于温度为T_0的平衡状态。气室C_1中含有压强为p_1的已有小部分液化的饱和水蒸气(由于液化的数量远远小于气体数量,可忽略已经液态水的体积和热容，水蒸气分子振动自由度未被激发)，已知水蒸气的相变潜热为L(认为与温度无关)；气室C_2中含有一种单原子气体；气室C_3中含有一种双原子气体，初态压强为p_3，除了气室C_3底部存在热源可以维持气室C_3的气体温度恒为T_0外，气缸壁与活塞均视为完全不导热；重力加速度为g。初态三个气室体积分别记为V_1,V_2,V_3.\n(1)求解初态C_2内的气体压强p_2和弹簧的伸长量Δx.\n(2)利用克拉珀龙方程导出绝热条件下C_1气室体积V与水蒸气压强P的关系.\n(3)现给这一平衡系统一个微扰，试确定简正频率ω所满足的方程。\n"
} | (1).$p_2=\frac{M_1 g +M_2 g +p_1 S_1 - p_3 S_2}{S_1-S_2}$
$Δx = \frac{(p_3-p_1) S_2 S_1 -M_2 g S_1 -M_1 g S_2}{k (S_1-S_2)}$
(2).$P V (1+\frac{L}{R T_0}-ln \frac{P}{P_1})^5 (\frac{L}{R T_0}-ln \frac{P}{P_1})^{-3}=P_1 V_1 (1+\frac{L}{R T_0})^5 (\frac{L}{R T_0})^{-3}$
(3)$M_1 M_2 ω^4 - (A M_2+B M_1) ω^2 + A B -C^2 = 0
其中：
$A = k + \frac{p_1 S_1^2 (L-R T_0 +3 R^2 T_0^2/L)}{V_1 (L+R T_0)}+\frac{5 S_1^2 (M_1 g +M_2 g +p_1 S_1 - p_3 S_2)}{3 V_2 (S_1-S_2)}$
$B= k + \frac{p_3 S_2^2 }{V_3}+\frac{5 S_2^2 (M_1 g +M_2 g +p_1 S_1 - p_3 S_2)}{3 V_2 (S_1-S_2)}$
$C = k + \frac{5 S_2 S_1 (M_1 g +M_2 g +p_1 S_1 - p_3 S_2)}{3 V_2 (S_1-S_2)}$ |
PHY-110 | Physics | images/PHY-110.png | {
"en": "\n\nA device deployed by a spacecraft falls vertically at a constant speed, approaching and reaching the surface of a certain planet. The curve of pressure \\( p \\) versus time \\( t \\) recorded by the device in conventional units is shown in the figure. When landing on the planet's surface, the device also measures the ambient temperature as \\( T = 850 \\, \\text{K} \\) and the free-fall acceleration on the planet's surface as \\( g = 10 \\, \\text{m/s}^2 \\). It is known that the planet's atmosphere consists of carbon dioxide. Determine the falling speed \\( v \\) of the device.",
"zh": "宇宙飞船投放的一个仪器以恒定速度垂直下落，接近并到达某行星的表面。该仪器用约定单位记录的压强p随时间t变化的曲线如图所示。落到行星表面时，该仪器还测出环境温度为\\(T = 850K\\)，行星表面自由落体加速度为\\(g = 10m/s^{2}\\). 已知该行星的大气由二氧化碳构成。试求仪器下落的速度v。"
} | Approximately 17.7m/s |
PHY-111 | Physics | images/PHY-111.png | {
"en": "\n\n\"Since the piston in a heat engine may undergo small vibrations, i.e., \\( P \\) and \\( V \\) fluctuate relative to their equilibrium positions, studying the thermodynamic cycle of this process has practical significance. Shown in the figure is the \\( P\\text{-}V \\) diagram of an off-axis elliptical thermodynamic cycle, where \\( a \\) and \\( b \\) are the semi-major and semi-minor axes, respectively. Due to the small perturbation, \\( a, b \\ll 1 \\), and \\( \\theta \\) is the angle between the semi-major axis of the ellipse and the \\( y \\)-axis. The thermodynamic cycle proceeds in a clockwise direction. Assume the gas is water vapor with unexcited vibrational degrees of freedom, and no phase change occurs during the thermodynamic cycle. Find: \n\n(1) The thermal efficiency \\( \\eta(\\theta, a, b) \\) of this thermodynamic cycle. \n\n(2) Suppose that due to additional factors (such as thermal expansion and nonlinear damping), the angular frequency of the vibration in the \\( P \\)-direction differs slightly from that in the \\( V \\)-direction in this heat engine. Correspondingly, in the thermodynamic diagram, the deflection angle \\( \\theta \\) of the ellipse has a small 'precession angular velocity,' i.e., after each elliptical cycle, the principal axis of the ellipse precesses by an angle of \\( 2\\pi/N \\), where \\( N \\gg 1 \\) and is a positive integer. \n\nDetermine the thermal efficiency \\( \\eta(a, b) \\) of this precessing elliptical thermodynamic cycle, where the perimeter of an ellipse with semi-major axis \\( a \\) and semi-minor axis \\( b \\) is a known function \\( C(a, b) \\).\"",
"zh": "由于热机中的活塞可能处于小振动，即P，V相对于平衡位置进行涨落，因此研究这一过程的热力循环有一定实际意义。如图为一偏轴椭圆热力循环过程的P-V图，其中a,b分别为半长轴和半短轴,由于是微扰a,b<<1，θ为椭圆半长轴偏离y轴的角度，热力循环方向为顺时针方向，假定气体为振动自由度未被激活的水蒸汽，热力循环过程中不会发生任何相变，求：\n(1)该热力循环的热机效率η(θ,a,b)\n(2)假定由于额外的因素(如热膨胀，非线性阻尼)，该热机的P方向振动角频率与V方向的振动角频率有一较小的差值，对应于该热力图中即椭圆的偏转角θ会有一个小的“进动角速度”，即每经过一次椭圆循环，椭圆主轴都会偏过\\2pi/N的角度，其中N>>1且为正整数。\n试求该进动椭圆的热力循环效率η(a,b)，半长轴半短轴分别为a,b的椭圆的周长为C(a,b)为一已知函数。"
} | (1)η=$\frac{\pi a b }{2 \sqrt{(3cosθ+4sinθ)^2 b^2 + (3sinθ-4cosθ)^2 a^2}}$
(2)η= $\frac{\pi^2 a b}{5 C(a,b)}$ |
PHY-112 | Physics | images/PHY-112.png | {
"en": "\n\n$\\nu$ moles of an ideal gas undergo a thermodynamic process as shown in the $S$-$T$ diagram. In the first half, $S$ is a quadratic function of $T$, with a horizontal tangent at $T_1 = T_0$. At $T' = 2T_0$, the process transitions to a linear segment along the tangent line and stops at $T_2 = 3T_0$. Derive the equation of state for this process.",
"zh": "$\\nu$摩尔理想气体经历的热力学过程如图$S-T$所示，前半段$S$为$T$的二次函数且$T_1=T_0$处切线水平，在$T'=2T_0$处沿着切线转变为直线过程直到$T_2=3T_0$停止，导出该过程的气体状态方程。"
} |
When \( T_1 < T < T' \),
\[
\ln TV^{\gamma-1} - \frac{S_2 - S_1}{3\nu C_V T_0^2}(T - T_0)^2 = \text{const.}
\]
When \( T' < T < T_2 \),
\[
\ln TV^{\gamma-1} - \frac{2(S_2 - S_1)}{3\nu C_V T_0}T = \text{const.}
\] |
PHY-113 | Physics | images/PHY-113.png | {
"en": "As shown in the figure, near the center of a uniform gaseous spherical planet O with mass volume density ρ, there is a celestial explorer of mass m captured by the planet's gravity. The explorer is at a distance d from the center, and the planet has a radius r. At a distance R from the planet's center, there is a space station M orbiting the planet. Now, the explorer wants to return to the space station. Assume that initially, the explorer gains an initial energy E through a chemical reaction (mass loss is negligible), and except for a mechanical energy loss δE when breaking through the gaseous planet's boundary, no other mechanical energy is lost. The explorer can change its velocity direction at any time by ejecting high-speed gas (mass and energy losses due to this are negligible). To ensure the explorer reaches the space station M as quickly as possible, with the gravitational constant being G and the space station's mass negligible. Given that in this scenario, the optimal emission angle of the explorer relative to the angle of Om is ψ, find the first-order differential equation satisfied by the explorer's trajectory equation (expressed in polar coordinates).",
"zh": "如图所示，质量体密度为ρ的均匀气体球形星球O中心附近有一被该星球引力捕获的质量为m的天体探测仪，探测仪距离中心距离为d,星球的半径为r，在距离星球中心R处有一绕该星球进行作业的空间站M，现在探测仪想要返回空间站，假定初态探测仪通过化学反应(质量损失可忽略不计)获得了初态能量E，除了在突破气体星球边界时会损失机器能δE外不会损失任何机械能。如果探测仪可以通过喷出高速气体的方式(因此而导致的质量和能量损失忽略不计）随时改变自身的速度方向，如果希望探测仪可以最快到达空间站M，万有引力常数为G，空间站质量忽略不计。如果已知这种情况下探测仪的最优出射角相对于Om的夹角为ψ\n求探测仪的轨迹方程所满足的一阶微分方程(用极坐标系表示）。"
} |
Taking \( x \) as the distance from the explorer to \( O \) and \( \varphi \) as the angle between the explorer's position vector to \( O \) and the initial position vector, a polar coordinate system is established, yielding:
If \( x < r \), then
\[
\frac{d\varphi}{dx} = \frac{1}{x \sqrt{\frac{x^2 \left( E - \frac{2G\pi\rho d^2}{3} \right)}{d^2 \left( E - \frac{2G\pi\rho x^2}{3} \right)}} - 1}
\]
If \( x > r \), then
\[
\frac{d\varphi}{dx} = \frac{1}{x \sqrt{\frac{x^2 \left( E - \frac{2G\pi\rho d^2}{3} \right)}{d^2 \left( E(1 - \delta) - \frac{2G\pi\rho r^3}{3x} \right)}} - 1}
\] |
PHY-114 | Physics | images/PHY-114.png | {
"en": "As shown in the figure, it is a schematic diagram of the first two iterations of a fractal pattern generated from an equilateral triangle with side length a (i.e., each iteration decomposes 6 small triangles into 9 smaller triangles, and fills the topmost small triangle among the remaining 3 small triangles with a transmissive film of transmittance t, which does not cause phase changes).After infinite iterations, if a laser with wavelength λ is incident on the center of the triangle, and the diffraction pattern is observed on a screen at a distance d, derive the diffraction intensity distribution I(x, y), assuming the intensity at (0, 0) is I₀ and d ≫ a²/λ.",
"zh": "如图所示为由边长为a的等边三角形生成的分形图样的前两次迭代的示意图(即每次迭代都会将其中6个小三角形分解成更小的9个小三角形，将剩下3个小三角形中位于最上方的小三角形填满透射率为t的透射膜，该膜不会导致相位变化),进行无限次迭代后，如果有一波长为λ的激光射入该三角形的中心，在距离为d的显示屏上观测衍射图样，试求衍射的光强分布I(x,y)，假定(0,0)处光强为$I_0$，假定有d>> a^2/λ"
} | I(x,y)= $I_0 \frac{t^2+4 cos \frac{\pi a x}{3 λ d}^2+4 t cos \frac{\pi a x}{3 λ d} cos \frac{\sqrt{3} \pi a y}{9 λ d}}{ (t+2)^2 (9+4 cos \frac{\pi a x}{3 λ d}^2+4(cos \frac{\pi a x}{3 λ d}+cos \frac{\sqrt{3} \pi a y}{3 λ d})^2-4 cos \frac{\pi a x}{3 λ d} cos \frac{2 \sqrt{3} \pi a y}{9 λ d}-4 (cos \frac{\pi a x}{3 λ d}+cos \frac{\sqrt{3} \pi a y}{3 λ d}) cos \frac{ \sqrt{3} \pi a y}{9 λ d}+4 cos \frac{\pi a x}{3 λ d}(cos \frac{\pi a x}{3 λ d}+cos \frac{\sqrt{3} \pi a y}{3 λ d}) cos \frac{ \sqrt{3} \pi a y}{3 λ d})}*\frac{\pi^4 a^4}{λ^4 d^4}(4/(3 y^2-x^2)^2+1/(x^2 (\sqrt{3}y+x)^2)+1/(x^2 (x-\sqrt{3} y)^2)+\frac{2 cos\frac{2 \pi a x}{d λ}}{x^2 (x^2-3 y^2)}+\frac{4 cos\frac{\pi a (\sqrt{3}y+x)}{λ a}}{(x^2-3 y^2) x (\sqrt{3} y + x)}+\frac{4 cos\frac{\pi a (\sqrt{3}y-x)}{λ a}}{(x^2-3 y^2) x (-\sqrt{3} y + x)})$ |
PHY-115 | Physics | images/PHY-115.png | {
"en": "As shown in the figure, it is a negative uniaxial crystal with a sphere of radius R removed. The optical axis of the crystal makes an angle θ with the normal to the crystal's bottom surface. The refractive indices for o-light and e-light are nₒ and nₑ, respectively. Light with wavelength λ is incident downward parallel to the normal of the crystal's bottom surface, and the interference pattern on the bottom surface of the crystal is observed from below.\n(1) Find the equation for the k-th order interference bright fringe.\n(2) If θ = π/2, adjust the radius R so that the interference fringes can still be observed under a microscope, and find the equation for the k-th order interference bright fringe.",
"zh": "如图为一被挖去半径为R的球体的负单轴晶体，晶体光轴方向与晶体底面的法相夹角为θ。o,e光对应的折射率分别为n_o和n_e。波长为λ的光平行于晶体底面法向量向下射入晶体，在下方观测晶体底面的干涉图样，\n(1)求k级干涉亮纹的方程。\n(2)如果θ=\\pi/2,调整半径R使得干涉条纹依然可以被显微镜观测，求k级干涉亮纹方程。\n"
} | (1)$x^2 + y^2 = \frac{2 k λ R}{n_0-\frac{1}{\sqrt{cos^2 θ/n_e^2 + sin^2 θ/n_o^2}}}$
(2)$(x^2+y^2)^2 = \frac{4 k λ R^3}{(n_o^2/n_e^2-1) (1-1/n_o)^2}$ |
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