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Unformatted text preview: Nguyen, Thanh Homework 20 Due: Nov 16 2007, 7:00 pm Inst: D Weathers 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points It is found that a 5 . 2 m segment of a long string contains 3 complete wavelengths and has a mass of 170 g. At one point it is vibrat ing sinusoidally with a frequency of 51 Hz and a peak to valley displacement of 0 . 95 m. What power is being transmitted along the string? Correct answer: 33477 . 7 W. Explanation: Let : l = 5 . 2 m , = 5 . 2 m 3 , m = 170 g , f = 51 Hz , and A = . 95 m 2 . Since k 2 , 2 f , v k , P = 1 2 2 A 2 v = 1 2 m l (2 f ) 2 A 2 f = 1 2 170 g 5 . 2 m [2 (51 Hz)] 2 . 95 m 2 2 5 . 2 m 3 (51 Hz) = 33477 . 7 W . keywords: 002 (part 1 of 2) 10 points A point source emits sound waves with an average power output of 0 . 907 W. Find the intensity 2 . 25 m from the source. Correct answer: 0 . 0142571 W / m 2 . Explanation: A point source emits energy in the form of a spherical waves. At a distance R from the source, the power is distributed over the surface area of a sphere, 4 R 2 . Therefore, the intensity at a distance R = 2 . 25 m from the source is I = P av 4 R 2 = 0 . 0142571 W / m 2 . 003 (part 2 of 2) 10 points Find the distance at which the sound reduces to a level of 69 dB. Correct answer: 95 . 3232 m. Explanation: We can find the intensity at the = 69 dB level by using the equation = 10 log I I with I = 1 10 12 W / m 2 : 69 dB = 10 log I 1 10 12 W / m 2 therefore I = 10 / 10 I = 10 6 . 9 (1 10 12 W / m 2 ) = 7 . 94328 10 6 W...
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This note was uploaded on 10/24/2011 for the course PHYS 1710 taught by Professor Weathers during the Winter '08 term at North Texas.
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