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Условие Отговори Решение

USAPhO 2004, Полуфинал — Задача 3. A3

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Всички материали от USAPhO 2004

AAPT UNITED STATES PHYSICS TEAM AIP 2004 — 2004 Semi-Final Exam · 25 т.

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.
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 your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

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.
Do not discuss the contents of this exam with anyone until after the submission deadline.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

Sound sources S $_1$ and S $_2$ are located at $+x_0$ and $–x_0$ , respectively. The sources produce tones with the same phase and amplitude. The frequency can be varied from 175 Hz to 625 Hz, but both speakers have the same frequency. Use 340 m/s for the velocity of sound in air and $x_0 = 0.850$ m.

An x-axis with a y-axis arrow at the origin; source S2 marked by a dot at –x0 and source S1 at +x0.

a. For which frequency or frequencies in the above range is minimum sound intensity heard at all points along the $x$ -axis with $x > x_0$ . [8 т.]

b. Next consider the region on the $x$ -axis between the two sound sources S $_1$ and S $_2$ . Assuming that the amplitude $A$ does not decrease with distance from the source, write an equation that gives the resultant sound wavefunction in the form $\Psi(x,t) = 2A\psi(x)\phi(t)$ , where $\psi$ and $\phi$ are functions of $x$ and $t$ , respectively. The equation may also depend on the wavelength $\lambda$ and frequency $f$ of the sound waves and should be valid for all points on the $x$ -axis with $-x_0 < x < +x_0$ . [10 т.]

c. For the frequency or frequencies found in Part (a), at which points between $x = –\ 0.85$ m and $x = +0.85$ m does minimum sound intensity occur? [7 т.]

Отговори

Решение

A3. a. In order for minimum sound intensity to be heard in the region along the $x$ -axis with $x > x_0$ , the distance between sources must be an odd half-integer multiple of the wavelength $\lambda$ .

2x_0 = (2n - 1)\frac{\lambda}{2} = (2n - 1)\frac{v}{2f} \qquad n = 1, 2, 3, \ldots

where $v = 340\ \mathrm{m/s}$ , the velocity of sound in air, and $175\ \mathrm{Hz} \le f \le 625\ \mathrm{Hz}$ , the frequency of the sound that produces minimum intensity. Solving for the frequency

f_n = (2n - 1)\frac{v}{4x_0} = (2n - 1)\frac{(340\ \mathrm{m/s})}{4(0.85\ \mathrm{m})} = (2n - 1)(100\ \mathrm{Hz}).

The frequencies in the possible range are

$f_2 = 300\ \mathrm{Hz}$ and $f_3 = 500\ \mathrm{Hz}$

b. In the region between the sources, source S $_1$ emits a wave $\Psi_1$ that travels to the left and source S $_2$ emits a wave $\Psi_2$ that travels to the right.

\Psi_1 = A\sin(\omega t + k(x - x_0)) \qquad \Psi_2 = A\sin(\omega t - k(x + x_0))

where $k = \dfrac{2\pi}{\lambda}$ and $\omega = 2\pi f$ .

Adding the waves to determine the resultant wave

\Psi = \Psi_1 + \Psi_2 = A\sin(\omega t + k(x - x_0)) + A\sin(\omega t - k(x + x_0))

\Psi = A\sin(\omega t - kx_0)\cos(kx) + A\cos(\omega t - kx_0)\sin(kx)

+ A\sin(\omega t - kx_0)\cos(kx) - A\cos(\omega t - kx_0)\sin(kx)

\Psi = 2A\cos(kx)\sin(\omega t - kx_0)

Note: This equation has the correct $x$ -dependence. The waves travel the same distance to reach $x = 0$ . This point is an interference maximum. Any expression with

\Psi = 2A\cos(kx)\sin(\omega t - kx_0 + \delta)

where $\delta$ is a phase constant is valid.

c. Minimum sound intensity occurs when $\cos(kx) = 0$ , i.e., $kx = \pm(2n + 1)\pi/2$ . Solving for $x$ ,

x = \pm\frac{(2n + 1)\pi}{2k} = \pm\frac{(2n + 1)\pi}{2(2\pi/\lambda)} = \pm\frac{(2n + 1)\lambda}{4} = \pm\frac{(2n + 1)v}{4f}

For $f = 300$ Hz:

x = \pm\frac{(2n + 1)(340\ \mathrm{m/s})}{4(300\ \mathrm{Hz})} = \pm(2n + 1)(0.283\ \mathrm{m}) = \pm 0.283\ \mathrm{m} \qquad \text{with } n = 0.

For $f = 500$ Hz:

x = \pm\frac{(2n + 1)(340\ \mathrm{m/s})}{4(500\ \mathrm{Hz})} = \pm(2n + 1)(0.170\ \mathrm{m}) = \pm 0.170\ \mathrm{m},\ \pm 0.510\ \mathrm{m}

with $n = 0, 1$ .

Оригинал в Архива: Semifinal 2004.pdf · официални решения: Semifinal 2004 Solutions.pdf