Grant Morby, Homework 33 Due: Apr 26 2006, noon Inst: Drummond This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A transverse wave in a wire has form y(x, t) = (1.6 cm) sin (5.08 m1 ) x (7110 s1 ) t The wire has linear density = 7.59 g/m. What is its tension? Correct answer: 14868 N. Explanation: A sinusoidal transverse wave of the general form y(x, t) = A sin(kx t) travels in the positive x direction at speed v= . k At the time t = 0 +1 A (meters)
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1 x (meters) 10 m 20 m 30 m
Which wave function corresponds best to the diagram? 2 4 xt 15 m 3 2 2 2. y = A sin xt 3m 3 2 5 3. y = A sin xt 9m 3 2 2 4. y = A sin xt 9m 3 2 1 5. y = A sin xt 3m 3 2 4 6. y = A sin xt 9m 3 2 4 7. y = A sin xt 3m 3 2 1 8. y = A sin x t 15 m 3 correct 1. y = A sin 9. y = A sin 10. y = A sin 2 15 m 2 15 m xt xt 2 3 5 3
For the wave in question, this speed is v= 7110 s1 = 1399.61 m/s. 5.08 m1
For a transverse wave in a wire, this speed follows from the wires tension and linear density according to v= hence F = v 2 = 14868 N. 002 (part 1 of 1) 10 points A harmonic wave y = A sin[k x t ] , where A = 1 meter, k has units of m1 , has units of s1 , and has units of radians, is plotted in the diagram below. F ,
Explanation: From the diagram of the wave function the wave-length = 15 m (6 horizontal scale divisions of 2.5 m each, see diagram below). Notice: Since one wave-length is 2 ra 2 = dians, each horizontal division is 6 3 radians.
Morby, Grant Homework 33 Due: Apr 26 2006, noon Inst: Drummond The given wave function (sine function with t = 0) y = A sin k x = A sin = A sin 2 x 2 x 15 m 1. y = log[b (x v t)] correct 2. y = exp(v x v t) 3. y = log(v x t) 4. y = exp(v x t) 5. y = exp(b x v t) 6. y = log(v x v t)
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(dark curve in diagram below) is shifted 1 divisions to the right (negative phase shift) of a no-phase-shift sine function y = sin 2 15 m x
(gray curve in diagram below), therefore =1 1 = radians . 3 3
Explanation: Any function of the form f (x v t) is the solution of 1 2y 2y =2 2 x2 v t Since y = log[b (x v t)] is of this form, its the solution of the wave equation. 004 (part 1 of 1) 10 points It is possible to hear an approaching train before you can see it by listening to the sound wave through the track. The speed of sound in air is 343 m/s. The elastic modulus is 2 1011 N/m2 and the density of steel is 3700 kg/m3 . Approximately how many times faster is the speed of sound in the track than in air? Correct answer: 21.4348 . Explanation: First , we can get the speed of sound in the track by viron = B 2 1011 N/m2 3700 kg/m3
Checking the wave function y at x = 0, we have agreement with the diagram below y = sin 0 1 3 = 0.866025 m .
0 +1 A (meters)
At the time t = 0 2
15 m x (meters) = 15 m Therefore, the wave function is y = A sin 2 15 m xt 1 3
003 (part 1 of 1) 10 points Which of the functions is a solution to 1 2y 2y =2 2 x2 v t
30 m .
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=
= 7352.15 m/s . Then we have n= viron vair 7352.15 m/s = 343 m/s = 21.4348
Morby, Grant Homework 33 Due: Apr 26 2006, noon Inst: Drummond 005 (part 1 of 1) 10 points If air pressure remains constant and temperature rises, the speed of sound 1. varies erratically. 2. falls. 3. Unable to determine. 4. stays the same. 5. rises. correct Explanation: At constant pressure, the speed of sound is proportional to the temperature of the air, so the speed increases as the temperature rises. 006 (part 1 of 1) 10 points Consider a harmonic sound wave in air of density = 1.2 kg/m3 . The displacement wave has form s(x, t) = smax cos(k x t) where k = 1.4 rad/m, = 481.255 rad/s and smax = 1.51 108 m. Calculate the sound pressure P (x, t) of this wave at x = 1.72 m and t = 0.00909 s. Correct answer: 0.00276588 Pa. Explanation: Basic Concept: The pressure and the displacement in a sound wave are related to each other according to P (x, t) = B s x hence vs = and (481.255 rad/s) = = 343.754 m/s k (1.4 rad/m)
2 B = vs = 141800 Pa .
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pressure wave (2) involves the sine function of the running phase = kxt (4)
while the displacement wave involves the cosine function of the same running phase (4). For the wave in question, the running phase at x = 1.72 m and t = 0.00909 s is = (1.4 rad/m) (1.72 m) (481.255 rad/s) (0.00909 s) = 1.96661 rad , hence according to Eq. 2 the sound pressure at this point in space and time is simply P (x = 1.72 m, t = 0.00909 s) = (5) = Pmax sin(1.96661 rad) where the pressure amplitude Pmax is given in Eq. 3. All we need is the bulk modulus B of the air, and we can nd it from the speed of the sound wave: vs = k = B , (6)
Consequently, the pressure amplitude is Pmax = smax B k = (1.51 108 m) (141800 Pa) (1.4 rad/m) = 0.00299765 Pa and hence P (x = 1.72 m, t = 0.00909 s) = (0.00299765 Pa) sin(1.96661 rad) = 0.00276588 Pa . 007 (part 1 of 1) 10 points
where B is the bulk modulus of the uid in which the sound is propagating. For a harmonic sound wave s(x, t) = smax cos(k x t), P (x, t) = Pmax sin(k x t) where Pmax = smax B k. (3) Note the 90 phase dierence between the displacement and the pressure waves: The (1) (2)
Morby, Grant Homework 33 Due: Apr 26 2006, noon Inst: Drummond The yellow-green light emitted by street lights matches the yellow-green color to which the human eye is most sensitive. Consequently, a 100-watt street light emits light that is better seen at night. Similarly, the monitored sound intensities of television commercials are louder than the sound from regular programming, yet dont exceed the regulated intensities. At what frequencies do advertisers concentrate the commercials sound? 1. The sound of commercials is concentrated at frequencies to which the ear is most sensitive. correct 2. The sound of commercials is concentrated at the low-frequency region of audible sound frequencies. 3. The sound of commercials is concentrated at the frequency of 60 Hz. 4. The sound of commercials is concentrated at the high-frequency region of audible sound frequencies. Explanation: The sound of commercials is concentrated at frequencies to which the ear is most sensitive; whereas the overall sound meets regulations, our ears perceive the sound as distinctly louder. 008 (part 1 of 1) 10 points The sound of a man shouting at the top of his lungs from a rather large distance away from your ear has loudness of only 20 decibels. What would be the decibel level of four men shouting at the top of their (equally powerful) lungs from the same distance away from you ear? Assume that there is no interference from superposed waves and round o your answer to the nearest integer. 1. 6 db 2. 160 db 3. 80 db 4. 40 db 5. 26 db correct 6. 5 db 7. 10 db 8. 60 db 9. 14 db 10. 20 db
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Explanation: The decibel level is a logarithmic measure of sound intensity I, specically I = 10 log10 . I0 Since we have assumed that there is no interference from superposed waves, four men shouting together produce sound intensity four times greater than that of just one man, I4 = 4I1 , hence 4I1 4 = 10 log10 I0 I1 = 10 log10 + 10 log10 4 I0 = 1 + 6 db = 26 db. 009 (part 1 of 2) 10 points The area of a typical eardrum is about 4.5 105 m2 . Find the sound power (the energy per second) incident on an eardrum at the threshold of hearing I0 = 1 1012 W/m2 . Correct answer: 4.5 1017 W. Explanation: a = 4.5 105 m2 and I0 = 1 1012 W/m2 . At the threshold of hearing, the power incident on the eardrum is P =IA = 1 1012 W/m2 4.5 105 m2 Given : = 4.5 1017 W .
Morby, Grant Homework 33 Due: Apr 26 2006, noon Inst: Drummond 010 (part 2 of 2) 10 points Find the sound power incident on an eardrum at the threshold of pain I1 = 1 W/m2 . Correct answer: 4.5 105 W. Explanation: Given : I1 = 1 W/m2 . At the threshold of pain P = 1 W/m2 = 4.5 10
5
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Solution: The intensity of the sound waves is given by W I= . 4 r2 012 (part 2 of 2) 10 points The average sound level of a typical musical instrument in a musical group is dB. What is the sound level in dB if 10 instruments are simultaneously playing? 1. = 2 2. = 5
4.5 105 m2 W.
011 (part 1 of 2) 10 points A point source emits sound waves uniformly in all directions with a power output W .
rock concert r A
3. = + 20 4. = + 10 correct 5. = 20 6. = + 2 7. = 8. = 10 9. = + 1 10. = + 5
Assume: The receiver has a cross section A perpendicular to the direction of the incident sound. What is the intensity of the sound waves detected by a receiver a distance r from the source? W 1. I = 2 r 2. I = r W 3. I = W r2
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4. I = 2 r W W correct 5. I = 4 r2 W 6. I = 2rr 7. I = 4 r 2 W 8. I = r W Explanation: Basic Concept: = 10 log I I0
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Explanation: The intensity of the sound waves generated by a single instrument is I = I0 10/10 . The intensity generated by 10 such instruments is I = 10 I = I0 10[1+(/10)] , so = 10 log I I0 = 10 + .
013 (part 1 of 1) 10 points An underwater sound source emits waves of frequency 25.8 kHz in all directions. How does the intensity of the waves vary with distance r from the source? 1. 1 r3/2
Morby, Grant Homework 33 Due: Apr 26 2006, noon Inst: Drummond 1 r3 1 3. r 1 4. 2 correct r Explanation: As the wave travels spherically, the emerge E on the spherical wave front remains the same. But the area of the wave front is 4 r 2 . Then from 2. I S = I 4 r 2 = E, we know I 1 . r2
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