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Unformatted text preview: roofner (bar784) – Homework 20 – Weathers – (17104) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points It is found that a 4 . 5 m segment of a long string contains 4 complete wavelengths and has a mass of 180 g. At one point it is vibrat ing sinusoidally with a frequency of 43 Hz and a peak to valley displacement of 0 . 53 m. What power is being transmitted along the string? Correct answer: 4959 . 52 W. Explanation: Let : l = 4 . 5 m , λ = 4 . 5 m 4 , m = 180 g , f = 43 Hz , and A = . 53 m 2 . Since k ≡ 2 π λ , ω ≡ 2 π f , v ≡ ω k , P = 1 2 μ ω 2 A 2 v = 1 2 parenleftBig m l parenrightBig (2 π f ) 2 A 2 λ f = 1 2 parenleftbigg 180 g 4 . 5 m parenrightbigg [2 π (43 Hz)] 2 × parenleftbigg . 53 m 2 parenrightbigg 2 parenleftbigg 4 . 5 m 4 parenrightbigg (43 Hz) = 4959 . 52 W . 002 (part 1 of 2) 10.0 points A point source emits sound waves with an average power output of 0 . 969 W. Find the intensity 7 . 73 m from the source. Correct answer: 0 . 00129049 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 πR 2 . Therefore, the intensity at a distance R = 7 . 73 m from the source is I = P av 4 π R 2 = 0 . 00129049 W / m 2 . 003 (part 2 of 2) 10.0 points Find the distance at which the sound reduces to a level of 64 dB. Correct answer: 175 . 209 m. Explanation: We can find the intensity at the β = 64 dB level by using the equation β = 10 log parenleftbigg I I parenrightbigg with I = 1 × 10 − 12 W / m 2 : 64 dB = 10 log parenleftbigg I 1 × 10 − 12 W / m 2 parenrightbigg therefore I = 10 β/ 10 I = 10 6 . 4 (1 × 10 − 12 W / m 2 ) = 2 . 51189 × 10 − 6 W / m 2 ....
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This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.
 Fall '08
 Weathers
 mechanics, Work

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