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Условие Решение

USAPhO 1998, Полуфинал — B1

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Всички материали от USAPhO 1998

1998 Exam 2 · 40 т.

1998 Exam 2 — INSTRUCTIONS

United States Physics Team

Entia non multiplicanda sunt praeter necessitatem

DO NOT OPEN THIS TEST UNTIL YOU ARE GIVEN THE SIGNAL TO BEGIN.
Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all work, as partial credit may be earned.
Begin each problem on a fresh sheet of paper. In the upper right-hand corner of each page, write your name (last name first), the problem number, and the page number / total number of pages for that particular problem, for example:

Einstein, Albert

A2 - 3/4 for “Problem A2, page 3 of 4”

A hand-held calculator may be used. Its memory must be cleared of data and programs. You may not use any books, notes, tables, or collections of formulas.
Best wishes for an outstanding exam!

Possibly useful information

Quantity	Value
Gravitational field at Earth’s surface	$g=9.8$ N/kg
Newton’s gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb’s constant	$k=1/4\pi\epsilon_o=9\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_o/4\pi=10^{-7}$ N·s $^2$ /C $^2$
Speed of light in vacuum	$c=3\times10^8$ m/s
Boltzmann’s constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro’s number	$N_A=6.02\times10^{23}$ mol $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/mol-K
Elementary charge	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ J
Planck’s constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.1\times10^{-31}\text{ kg}=\frac12\text{ MeV}/c^2$
Binomial expansion	$(1+z)^n\simeq1+nz$ whenever $\lvert z\rvert\ll1$
Small-angle approximations	$\cos\theta\simeq1-\frac12\theta^2$ , $\sin\theta\simeq\theta$ whenever $\lvert\theta\rvert\ll1$

Условие

Two sirens located on the x-axis are separated by a certain distance $D$ . As heard by an observer at rest, $f_L$ denotes the frequency emitted by the left-hand siren, and $f_R$ denotes the frequency emitted by the right-hand siren. You are moving with constant velocity at speed $v_o$ along the x-axis, and record the following observations:

When you are on the right side of both sirens, you hear a beat frequency of 1.01 Hz.

When you are on the left of both sirens, you hear a beat frequency of 0.99 Hz.

When you are between the sirens, the beat frequency is zero.

Sirens fL and fR separated by distance D along the x-axis.

a. Determine in which direction you are moving along the x-axis. [5 т.]

b. Evaluate your speed $v_o$ as a fraction of the speed of sound. [10 т.]

c. Determine which frequency is greater, $f_L$ or $f_R$ . [5 т.]

d. Determine the numerical values of $f_L$ and $f_R$ . [10 т.]

e. Suppose you stop moving, and the siren that emits sound of frequency $f_L$ is taken aboard a helicopter. The helicopter hovers with zero velocity above you. The siren (assumed to be battery-powered) is dropped from the helicopter. In the case of negligible air resistance, sketch a graph of frequency as a function of time, that describes the frequency you hear from this falling siren. [If you did not obtain a value for $f_L$ in part (d), use 50 Hz.] Take the speed of sound to be 343 m/s. The siren falls for 30 seconds before striking the ground beside you. [10 т.]

Решение

a. (5 т.)

The beat frequency is “blue shifted” on the [text missing in original scan], and “red shifted” on the L — hence moving [text missing in original scan].

b. (10 т.)

The beats are a “source” of frequency $\lvert f_L-f_R\rvert\equiv\Delta$ [remainder missing in original scan].

When on L: $\Delta f'_L=\Delta f_o(1-v/c)$ , moving away; $c=$ speed of sound.

Using given data,

0.99\text{ Hz}=\Delta f_o(1-v/c).\qquad (1)

When on R: $\Delta f'_R=\Delta f_o(1+v/c)$ , moving towards.

1.01\text{ Hz}=\Delta f_o(1+v/c).\qquad (2)

(1) and (2) $\Rightarrow$

\beta\equiv\frac{1.01}{0.99}=\frac{1+v/c}{1-v/c}\qquad (3)

\Rightarrow\frac vc=0.01.\qquad (4)

c. (5 т.)

In between, $f_R$ is red-shifted:

f'_R=f_R-\epsilon_R\quad\text{for some }\epsilon_R>0.

In between, $f_L$ is blue-shifted:

f'_L=f_L+\epsilon_R\quad\text{for some }\epsilon_L>0.

In between, $f'_L=f'_R$ .

f_L+\epsilon_R=f_R-\epsilon_R

\Rightarrow f_R=f_L+\epsilon_R+\epsilon_R>f_L

so $f_R>f_L$ .

d. (10 т.)

In between,

f'_L=f_L\left(1+\frac vc\right)\quad\text{blue shifted},\qquad f'_R=f_R\left(1-\frac vc\right)\quad\text{red-shifted}.

And $f'_L=f'_R$ , so $f_L(1+v/c)=f_R(1-v/c)$ .

I.e.,

\frac{f_R}{f_L}=\frac{1+v/c}{1-v/c}=\beta\quad\text{[from (3)]}.\qquad (5)

Also, from either (1) or (2), and using (4) [ $v/c=0.01$ ], it follows that

f_R-f_L=1\text{ Hz}.\qquad (6)

Solving (5) and (6) gives $f_R=50.5\text{ Hz}$ , $f_L=49.5\text{ Hz}$ .

e. (10 т.)

Source moving: $v=-gt$ .

f'=f_o\left(1+\frac vc\right)^{-1}=f_o\left(1-\frac{gt}{c}\right)

$g=9.8\text{ m/s}^2$ , $f_o=50\text{ Hz}$ , $c=343\text{ m/s}$ .

f'=\frac{50\text{ Hz}}{1-\frac{t}{35\text{ s}}}.

Official key frequency versus time graph beginning at50Hz and rising through30s.

Оригинал в Архива: Semifinal 1998.pdf · официални решения: Semifinal 1998 Solutions.pdf