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Unformatted text preview: bhappu (nkb284) Homework 18 florin (58140) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Sound waves travel through a liquid of density 4190 kg / m 3 at a speed of 447 m / s. What is the bulk modulus of this liquid? Correct answer: 8 . 372 10 8 Pa. Explanation: Let : = 4190 kg / m 3 and v sound = 447 m / s . In fluids, sound waves are pressure waves, and their speed depends on the fluids density and bulk modulus: v sound = B B = v 2 sound = (4190 kg / m 3 ) (447 m / s) 2 = 8 . 372 10 8 Pa . 002 (part 1 of 3) 10.0 points An outside loudspeaker (considered a small source) emits sound waves with a power out put of 93 W. Find the intensity 29 . 7 m from the source. Correct answer: 0 . 00838997 W / m 2 . Explanation: Let : P = 93 W and r = 29 . 7 m . The wave intensity is I = P 4 r 2 = 93 W 4 (29 . 7 m) 2 = . 00838997 W / m 2 . 003 (part 2 of 3) 10.0 points Find the intensity level in decibels at this distance. Correct answer: 99 . 2376 dB. Explanation: Let : I = 1 10 12 W / m 2 . The intensity level is = 10 log I I = 10 log . 00838997 W / m 2 1 10 12 W / m 2 = 99 . 2376 dB . 004 (part 3 of 3) 10.0 points At what distance would you experience the sound at the threshold of pain, 120 dB ? Correct answer: 2 . 72042. Explanation: Let : = 120 dB . = 10 log I I I = I 10 / 10 = ( 1 10 12 W / m 2 ) 10 12 = 1 W / m 2 , so the wave intensity is I = P 4 r 2 r = P 4 I = 93 W 4 (1 W / m 2 ) = 2 . 72042 m . 005 10.0 points The sound level produced by one singer is 69 . 4 dB. What would be the sound level produced by a chorus of 17 such singers (all singing at bhappu (nkb284) Homework 18 florin (58140) 2 the same intensity at approximately the same distance as the original singer)? Correct answer: 81 . 7045 dB. Explanation: Let : = 69 . 4 dB and n = 17 . The total sound intensity is the sum of the sound intensities produced by each individual singer: I n = n I 1 , so n = 10 log I n I = 10 log n I 1 I = 10 log I 1 I + 10 log n = 1 + 10 log n = 69 . 4 dB + 10 log 17 = 81 . 7045 dB . 006 10.0 points A copper rod is given a sharp compressional blow at one end. The sound of the blow, traveling through air at 3 . 6 C, reaches the opposite end of the rod 6 . 38 ms later than the sound transmitted through the rod. What is the length of the rod? The speed of sound in copper is 3680 m / s and the speed of sound in air at 3 . 6 C is 331 m / s. Correct answer: 2 . 3205 m....
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas.
 Spring '10
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