Съдържание

Условие a.–c.d.–e.Решение

USAPhO 2001, Полуфинал — A3

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2001

2001 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. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the 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	$k=1/4\pi\epsilon_0=9.0\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.0\times10^8$ 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)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
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	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

A plane wave traveling in the positive $z$ -direction can be described by the following wavefunction: $\psi(z,t)=A\sin(kz-\omega t+\phi)$ , where $A$ (the amplitude) and $\phi$ are constants, $k=\dfrac{2\pi}{\lambda}$ , and $\omega=2\pi f$ . The factor $(kz-\omega t+\phi)$ is called the phase of the wave.

a.–c.

a. Two sources, $S_1$ and $S_2$ , produce identical plane waves traveling in the same direction. The sources operate in phase and the waves have the same amplitude, frequency, and wavelength. Source $S_1$ is located a distance $z_1$ from point P and source $S_2$ is a distance $z_2$ from point P. What is the wavefunction of the resultant wave at point P? The following relation may be useful:

\sin\alpha+\sin\beta=2\cos\left[\frac12(\alpha-\beta)\right]\sin\left[\frac12(\alpha+\beta)\right].

[5 т.]

b. For what values of $\Delta z=|z_1-z_2|$ is the amplitude of the resultant traveling wave zero? [4 т.]

c. For what values of $\Delta z=|z_1-z_2|$ is the amplitude of the resultant traveling wave maximum? [4 т.]

d.–e.

A plane sound wave is normally incident on a coated surface as shown to the right. At surface 1, the wave is both transmitted and reflected. The reflected and transmitted waves have equal amplitudes. All of the wave reaching surface 2 is reflected. Assume the wave does not shift phase on reflection at either surface. In air, the wave has frequency $f$ , wavelength $\lambda$ , and speed $v_{so}=340$ m/s. In the material of thickness $t=0.200$ m, the wave has speed $v_{sm}=1020$ m/s.

Incident and reflected sound rays through two surfaces, coating thickness t and speeds 340m/s and1020m/s.

d. In terms of $f$ and $\lambda$ (the frequency and wavelength in air), what are the frequency and wavelength of the sound in the material? [4 т.]

e. List all frequencies smaller than 10,000 Hz for which the reflected sound intensity is a minimum? [8 т.]

Решение

a. (5 т.)

The resultant wave is the sum of the individual waves

\psi(z,t)=A\sin(kz_1-\omega t+\phi)+A\sin(kz_2-\omega t+\phi).

Using the trigonometric identity provided

\psi(z,t)=A2\cos\left[\frac12((kz_1-\omega t+\phi)-(kz_2-\omega t+\phi))\right]\sin\left[\frac12((kz_1-\omega t+\phi)+(kz_2-\omega t+\phi))\right]

\psi(z,t)=2A\cos\left[\frac k2(z_1-z_2)\right]\sin\left[\frac k2(z_1+z_2)-\omega t+\phi\right].

b. (4 т.)

The amplitude of the resultant traveling wave is zero if $\cos\left[\dfrac k2(z_1-z_2)\right]=0$ .

This occurs when the argument of the cosine is an odd half-integer multiple of $\pi$ .

\frac k2\Delta z=\frac{2n+1}{2}\pi\quad\text{for }n=0,1,2,\ldots

Since $k=\dfrac{2\pi}{\lambda}$

\frac12\frac{2\pi}{\lambda}\Delta z=\frac{2n+1}{2}\pi

\Delta z=\left(\frac{2n+1}{2}\right)\lambda\quad\text{for }n=0,1,2,\ldots

c. (4 т.)

The amplitude of the resultant traveling wave is maximum if $\cos\left[\dfrac k2(z_1-z_2)\right]=\pm1$ .

This occurs when the argument of the cosine is an integer multiple of $\pi$ .

\frac k2\Delta z=\frac12\frac{2\pi}{\lambda}\Delta z=n\pi\quad\text{for }n=0,1,2,\ldots

\Delta z=n\lambda\quad\text{for }n=0,1,2,\ldots

d. (4 т.)

The frequency does not change as the wave enters the new material. Letting $\lambda_m$ be the wavelength in the material

\frac{v_{so}}\lambda=\frac{v_{sm}}{\lambda_m}.

Solving for $\lambda_m$

\lambda_m=\frac{v_{sm}}{v_{so}}\lambda=\frac{1020\text{ m/s}}{340\text{ m/s}}\lambda=3\lambda

e. (8 т.)

The reflected sound intensity will be a minimum when the condition specified in part (b) is met. $\Delta z$ is twice the thickness (up and down) of the coating material and $\lambda$ is the wavelength in the material.

2t=\left(\frac{2n+1}{2}\right)\lambda_m\quad\text{for }n=0,1,2,\ldots

2t=\left(\frac{2n+1}{2}\right)\frac{v_{sm}}f

f=\left(\frac{2n+1}{2}\right)\frac{v_{sm}}{2t}=\left(\frac{2n+1}{2}\right)\frac{1020\text{ m/s}}{2(0.200\text{ m})}=(2n-1)1275\text{ Hz}

Inserting the integers into the above equation, we have

f=1275\text{ Hz},3825\text{ Hz},6375\text{ Hz},8925\text{ Hz}.

Оригинал в Архива: Semifinal 2001.pdf · официални решения: Semifinal 2001 Solutions.pdf