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Условие Отговори Решение

USAPhO 2008, Полуфинал — Задача 4. Question A4

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Всички материали от USAPhO 2008

AAPT AIP UNITED STATES PHYSICS TEAM 2008 — Semifinal Exam · 25 т.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

This examination consists of three parts.
Part A has four questions and is allowed 90 minutes.
Part B has two questions and is allowed 90 minutes.
Part C has one question and is allowed 20 minutes. The answer for Part C will not be used for team selection, but will be used for special recognition from the Optical Society of America.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for all three parts of the exam.
The three parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B or Part C. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students look at Part C or go back to Part A. Collect the answers to part B before allowing the examinee to begin Part C. Examinees are allowed a 10 to 15 minutes break between Parts B and C.
Allow 20 minutes to complete Part C. This part is optional; scores on Part C will not be used to select the US Team. Examinees may not go back to Part A or B.
Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-7), Part B (pages 8-10), and Part C (page 11). Examinees should be provided the parts individually, although they may keep the cover sheet.
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2008.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA's or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.
Please provide the examinees with graph paper for Part A.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Parts B or C during this time.
After you have completed Part A you may take a break.
Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Parts A or C during this time.
Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets.
Start each question on a new sheet of paper. Put your school ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

School ID #
Doe, Jamie
A1 - 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA's or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Questions with the same point value are not necessarily of the same difficulty.
Part C is an optional part of the test. You will be given 20 additional minutes to complete Part C. Your score on Part C will not affect the selection for the US Team, but can be used for special prizes and recognition to be awarded by the Optical Society of America.
In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after March 31, 2008.

Possibly Useful Information

Possibly Useful Information. You may use this sheet for all three parts of the exam.

$g = 9.8\ \mathrm{N/kg}$ $\qquad$ $G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$

$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$ $\qquad$ $k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$

$c = 3.00 \times 10^{8}\ \mathrm{m/s}$ $\qquad$ $k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$

$N_\mathrm{A} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}$ $\qquad$ $R = N_\mathrm{A} k_\mathrm{B} = 8.31\ \mathrm{J/(mol \cdot K)}$

$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$ $\qquad$ $e = 1.602 \times 10^{-19}\ \mathrm{C}$

$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$ $\qquad$ $h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$

$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$ $\qquad$ $(1+x)^n \approx 1 + nx$ for $|x| \ll 1$

$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $|\theta| \ll 1$ $\qquad$ $\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $|\theta| \ll 1$

Условие

A tape recorder playing a single tone of frequency $f_0$ is dropped from rest at a height $h$ . You stand directly underneath the tape recorder and measure the frequency observed as a function of time. Here $t = 0$ s is the time at which the tape recorder was dropped.

$t$ (s)	$f$ (Hz)
2.0	581
4.0	619
6.0	665
8.0	723
10.0	801

The acceleration due to gravity is $g = 9.80\ \mathrm{m/s^2}$ and the speed of sound in air is $v_{snd} = 340\ \mathrm{m/s}$ . Ignore air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air,

f = f_0\frac{v_{snd} \pm v_{obs}}{v_{snd} \pm v_{src}}

where $f_0$ is the emitted frequency as determined by the source, $f$ is the frequency as detected by the observer, and $v_{snd}$ , $v_{src}$ , and $v_{obs}$ are the speed of sound in air, the speed of the source, and the speed of the observer. The positive and negative signs are dependent upon the relative directions of the motions of the source and the observer.

a. Determine the frequency measured on the ground at time $t$ , in terms of $f_0$ , $g$ , $h$ , and $v_{snd}$ . Consider only the case where the falling tape recorder doesn't exceed the speed of sound $v_{snd}$ .

b. Verify graphically that your result is consistent with the provided data.

c. What (numerically) is the frequency played by the tape recorder?

d. From what height $h$ was the tape recorder dropped?

Отговори

Решение

a. The position of the tape recorder above the ground at a time $t$ is given by

y = h - \frac{1}{2}gt^2

and the speed of the tape recorder is given by

v_{src} = -gt

The observer "hears" the sound emitted from the tape recorder a time $\delta t$ earlier, since it takes time for the sound to travel to the listener. In this case,

y = v_{snd}\delta t

So at time $t$ the listener is hearing the tape recorder when it had emitted at time $t' = t - \delta t$ , or

t' = t - \frac{h}{v_{snd}} + \frac{g}{2v_{snd}}(t')^2

Solve this for $t'$ , first by rearranging,

\frac{g}{2}(t')^2 - v_{snd}t' + (v_{snd}t - h) = 0

and the by applying the quadratic formula

t' = \frac{v_{snd} \pm \sqrt{{v_{snd}}^2 + 2gh - 2gv_{snd}t}}{g}.

In the limit of small $h$ and large $v_{snd}$ , this reduces to the expected $t' = t$ if one keeps the negative result. Consequently,

v_{src} = \sqrt{{v_{snd}}^2 + 2gh - 2gv_{snd}t} - v_{snd}

gives the velocity of that source had when it emitted the sound heard at time $t$ . This result is negative, indicating motion down, and toward the observer, so one must use the positive sign in the denominator of the Doppler shift formula. Applying the Doppler shift formula,

f = f_0\frac{v_{snd}}{\sqrt{{v_{snd}}^2 + 2gh - 2gv_{snd}t}}.

b. First, note that in the limit of large $v_{snd}$ and small $h$ , $f$ reduces to

f = f_0\left(1 + \frac{g}{v_{snd}}t\right).

However, this is too inaccurate an approximation to get a reasonable answer. Keeping to the exact expression, we can rearrange it as

\frac{1}{f^2} = \frac{1}{{f_0}^2}\left(1 + \frac{2gh}{{v_{snd}}^2} - \frac{2g}{v_{snd}}t\right)

which would graph as a straight line by plotting $t$ horizontally and $1/f^2$ vertically. The slope and intercept are

-\frac{2g}{v_{snd}{f_0}^2},\quad \frac{1}{{f_0}^2}\left(1 + \frac{2gh}{{v_{snd}}^2}\right)

respectively. Now we may plot the following data:

$t$ (s)	$f$ (Hz)	$1/f^2$ ( $\times 10^{-6}$ s $^2$ )
2.0	581	2.96
4.0	619	2.61
6.0	665	2.26
8.0	723	1.91
10.0	801	1.56

When graphed, this indeed produces a line.

c. The slope is $-1.75 \times 10^{-7}$ s. Then

f_0 = \sqrt{\frac{2(9.8)}{(1.75 \times 10^{-7})(340)}}\,\mathrm{Hz} = 574\,\mathrm{Hz}.

d. The intercept is $3.31 \times 10^{-6}$ s $^2$ . This yields a height

h = (340)\frac{(3.31)}{(0.175)} - \frac{(340)^2}{2(9.8)} = 533\,\mathrm{m}.

Clearly, an impressive building, and a more impressive tape recorder, that it could be heard from such a distance!

Оригинал в Архива: olympiad_2008_semi-final.pdf · официални решения: USAPhO-2008-Solutions.pdf