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USAPhO 2023, Полуфинал — A2. Time is a Flat Circle

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2023 USA Physics Olympiad Exam · 25 т.

Внимание: Бележка към темата

The archive contains the complete question packet and the official interleaved solution packet. The separate official answer sheets referred to in the instructions are absent from this archive folder. Original printed errors are retained: A1(c)'s solution uses $z'_x$ although the question defines $y(x)$ ; A2(b.iii)'s solution writes radius $r+\delta r$ although the question defines $r=r_0+\delta r$ ; A3's key refers to “part (b)” in its part a.iii check, and B1's key calls the rod a “disc”. B2's printed energy expression uses $2.0\times10^7\ \mathrm{J}$ although the question gives energy density in J/kg. The “this paper” references in B2 and B3 have no recoverable URI in the supplied PDF.

AAPT UNITED STATES PHYSICS TEAM — AIP 2023; USA Physics Olympiad Exam

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Important Instructions for the Exam Supervisor

This examination has two parts. Each part has three questions and lasts for 90 minutes.
For each student, print out one copy of the exam and one copy of the answer sheets. Print everything single-sided, and do not staple anything. Divide the exam into the instructions (pages 2–3), Part A questions (pages 4–8), and Part B questions (pages 9–11).
Begin by giving students the instructions and all of the answer sheets. Let the students read the instructions and fill out their information on the answer sheets. They can keep the instructions for both parts of the exam. Also give students blank sheets of paper to use as scratch paper throughout the exam.
Students may bring calculators, but they may not use symbolic math, programming, or graphing features of these calculators. Calculators may not be shared, and their memory must be cleared of data and programs. Cell phones or other electronics may not be used during the exam or while the exam papers are present. Students may not use books or other references.
To start the exam, give students the Part A questions, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the Part A questions and answer sheets.
Then give students a 5 to 10 minute break. Then give them the Part B questions, and allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, collect everything, including the questions, the instructions, the answer sheets, and the scratch paper. Students may not keep the exam questions. Everything can be returned to the students after April 19th, 2023.
After the exam, sort each student’s answer sheets by page number. Scan every answer sheet, including blank ones.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvili, Kellan Colburn, Samuel Gebretsadkan, Abi Krishnan, Natalie LeBaron, Kye Shi, Brian Skinner, and Kevin Zhou.

Instructions for the Student

You should receive these instructions, the reference table on the next page, answer sheets, and blank paper for scratch work. Read this page carefully before the exam begins.
You may use a calculator, but its memory must be cleared of data and programs, and you may not use symbolic math, programming, or graphing features. Calculators may not be shared. Cell phones or other electronics may not be used during the exam or while the exam papers are present. You may not use books or other outside references.
When the exam begins, your proctor will give you the questions for Part A. You will have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. If you finish all of the questions, you may check your work, but you may not look at Part B during this time.
After 90 minutes, your proctor will collect the questions and answer sheets for Part A. You may then take a short break.
Then you will work on Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time. When the exam ends, you must return all papers to the proctor, including the exam questions.
Do not discuss the questions of this exam, or their solutions, until after April 19th, 2023. Violations of this rule may result in disqualification.

Below are instructions for writing your solutions.

All of your solutions must be written on the official answer sheets. Nothing outside these answer sheets will be graded. Before the exam begins, write your name, student AAPT number, and proctor AAPT number as directed on the answer sheets.
There are several answer sheets per problem. If you run out of space for a problem, you may use the extra answer sheets, which are at the end of the answer sheet packet. To ensure this work is graded, you must indicate, at the bottom of your last answer sheet for that problem, that you are using these extra answer sheets.
Only write within the frame of each answer sheet. To simplify grading, we recommend drawing a box around your final answer for each subpart. You should organize your work linearly and briefly explain your reasoning, which will help you earn partial credit. You may use either pencil or pen, but sure to write clearly so your work will be legible after scanning.

Reference table of possibly useful information

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm{C}$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$	$(1+x)^n\approx1+nx$ for $\vert x\vert \ll1$
$\sin\theta\approx\theta-\theta^3/6$ for $\vert \theta\vert \ll1$	$\cos\theta\approx1-\theta^2/2$ for $\vert \theta\vert \ll1$

Possibly useful integrals

\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}(x)+C

\int\frac{dx}{1-x^2}=\tanh^{-1}(x)+C

\int\frac{dx}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}(x)+C

\int\frac{dx}{1+x^2}=\tan^{-1}(x)+C

\int\frac{dx}{(1+x^2)^{3/2}}=\frac{x}{\sqrt{1+x^2}}+C

\int\sqrt{\frac{1+x}{1-x}}\,dx=-\sqrt{1-x^2}-2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{\sqrt{1+x}}{(1-x)^{3/2}}\,dx=2\sqrt{\frac{1+x}{1-x}}+2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{1/2}}=\sqrt{\frac{1+x}{1-x}}+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{1/2}}=\frac{(2-x)\sqrt{1+x}}{3(1-x)^{3/2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{3/2}}=\frac{1+2x-2x^2}{3(1-x)^{3/2}(1+x)^{1/2}}+C

You may use this sheet for both parts of the exam.

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Условие

A particle of mass $m$ and negative charge $-q$ is constrained to move in a horizontal plane. In the situations described below, this particle can either oscillate back and forth in a straight line or move in a circle. (These two modes of motion are interesting because they generate linearly and circularly polarized radiation, respectively, but in this problem you may ignore any energy lost to radiation.)

a.

A large positive charge $Q \gg q$ is fixed in place a distance $R$ directly below the origin of the plane.

i. When the particle is a distance $r \ll R$ from the origin, find an approximate expression for its potential energy due to the charge $Q$ to second order in $r/R$ , up to an arbitrary constant. You may use this result for the rest of the problem.

ii. If the particle oscillates linearly with amplitude $a \ll R$ , what is its angular frequency $\omega_\ell$ ?

iii. If the particle performs circular motion with radius $r \ll R$ , what is its angular frequency $\omega_c$ ?

b.

Now an additional negative charge $-q$ is fixed in place at the origin of the plane.

i. What is the equilibrium distance $r_0$ of the particle from the origin?

ii. If the particle oscillates linearly with amplitude $a \ll r_0$ , what is its angular frequency $\Omega_\ell$ ? Is it higher or lower than $\omega_\ell$ ?

iii. Now suppose the particle performs circular motion with radius $r = r_0 + \delta r$ , where $\delta r \ll r_0$ . What is its angular frequency $\Omega_c$ , in terms of $\omega_c$ , $r$ , and $\delta r$ ? Is it higher or lower than $\omega_c$ ?

Решение

a.i. The potential energy of the particle, relative to infinity, is

V(r) = -\frac{qQ}{4\pi\epsilon_0\sqrt{R^2+r^2}} \approx -\frac{qQ}{4\pi\epsilon_0 R} + \frac{qQr^2}{8\pi\epsilon_0 R^3}

where we have used the binomial theorem on the square root, since $r \ll R$ . The constant doesn't matter, so we could equivalently write this as

V(r) \approx \frac{qQr^2}{8\pi\epsilon_0 R^3} + \text{const.}

a.ii. For simple harmonic motion, the potential energy $V-V_0=\frac{1}{2}kx^2$ , where the "spring constant" $k$ is related to the angular frequency of oscillation by $\omega=\sqrt{k/m}$ . Reading the value of $k$ from the expression above for $V(r)$ gives $k=qQ/(4\pi\epsilon_0 R^3)$ , and therefore

\omega_\ell = \sqrt{\frac{qQ}{4\pi\epsilon_0 m R^3}}.

As usual for simple harmonic motion, the amplitude does not alter the frequency.

a.iii. For a radially symmetric, parabolic potential, a circular orbit can be thought of as simultaneous harmonic oscillation in the $x$ direction and the $y$ direction. Both oscillations have the same angular frequency $\omega_\ell$ as derived in part 1. So the answer is the same:

\omega_c = \sqrt{\frac{qQ}{4\pi\epsilon_0 m R^3}} = \omega_\ell.

Note that the orbit radius does not matter, so long as it is small enough ( $r\ll R$ ) to still correspond to simple harmonic motion.

b.i. With the additional charge at the origin, the total potential energy of the particle as a function of $r$ is

V(r) = V_0 + \frac{qQr^2}{8\pi\epsilon_0 R^3} + \frac{q^2}{4\pi\epsilon_0 r}.

Notice that the potential energy as a function of position has a ring of minima with a certain radius $r_0$ . When the particle is not in motion, its rest position is somewhere along the ring. The value of $r_0$ can be found from $V'(r_0)=0$ , which gives $r_0 = R(q/Q)^{1/3}$ .

b.ii. A linear oscillation involves the radius $r$ oscillating around the value $r_0$ . Taylor expanding the expression for $V(r)$ in part b.i around $r=r_0$ gives

V \approx V(r_0) + \frac{1}{2}\frac{3qQ}{4\pi\epsilon_0 R^3}\delta_r^2,

where $\delta_r = r - r_0$ . One can read from this expression the value of the spring constant $k$ , which gives for the angular frequency $\omega=\sqrt{k/m}$ the result

\Omega_\ell = \sqrt{\frac{3qQ}{4\pi\epsilon_0 m R^3}} = \sqrt{3}\,\omega_\ell

so that $\Omega_\ell > \omega_\ell$ . Evidently, adding the charge at the origin slightly stiffens the oscillation in the radial direction.

b.iii. For circular motion with radius $r+\delta r$ and $\delta r \ll r_0$ , the orbit is above, but is very close to the bottom of the ring of minima, $r \approx r_0$ . For such small $\delta r$ the force linearly depends on $r-r_0$ , as in Hooke's law, with

k = V''(r_0) = \frac{3Qq}{4\pi\epsilon_0 R^3}

Now Newton's second law gives

-k\delta r = -m\Omega_c^2 r,

with $r\approx r_0$ . This equation has a solution for $\delta r>0$ (so that the centripetal force is inward):

\Omega_c \approx \omega_c\sqrt{\frac{3\,\delta r}{r_0}}

Notice that the orbit frequency now explicitly depends on the orbit radius through $\delta r$ . So, unlike in the case of the usual harmonic oscillator, in this case one can generate circularly polarized light with a wide range of frequencies by exciting circular motion with different radii.

Since $\delta r \ll r_0$ , the angular frequency $\Omega_c \ll \omega_c$ , so that adding the charge $-q$ at the origin has made the orbit much slower. Intuitively, this is because the set of potential minima is a "flat circle," so that the orbit frequency can get arbitrarily small for arbitrarily small angular momenta. This is connected to Goldstone's theorem in quantum field theory, which states that spontaneously broken symmetries give rise to low frequency modes.

Оригинал в Архива: 2023-USAPhO-Exam.pdf · официални решения: 2023-USAPhO-solutions.pdf