Homework 20 Solutions

Homework 20 Solutions - lim (kl9356) – Homework 20 –...

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Unformatted text preview: lim (kl9356) – Homework 20 – Weathers – (17104) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points It is found that a 6 . 1 m segment of a long string contains 3 complete wavelengths and has a mass of 190 g. At one point it is vibrat- ing sinusoidally with a frequency of 55 Hz and a peak to valley displacement of 0 . 8 m. What power is being transmitted along the string? Correct answer: 33279 W. Explanation: Let : l = 6 . 1 m , λ = 6 . 1 m 3 , m = 190 g , f = 55 Hz , and A = . 8 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 190 g 6 . 1 m parenrightbigg [2 π (55 Hz)] 2 × parenleftbigg . 8 m 2 parenrightbigg 2 parenleftbigg 6 . 1 m 3 parenrightbigg (55 Hz) = 33279 W . 002 (part 1 of 2) 10.0 points A point source emits sound waves with an average power output of 1 . 24 W. Find the intensity 3 . 15 m from the source. Correct answer: 0 . 00994468 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 = 3 . 15 m from the source is I = P av 4 π R 2 = 0 . 00994468 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: 198 . 201 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 . Using this value for I and the equation I = P av 4 π r 2 , we find for the distance r = radicalbigg P av 4 π I = radicalBigg 1 . 24 W 4 π (2 . 51189 × 10 − 6 W / m 2 ) = 198 . 201 m ....
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This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

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Homework 20 Solutions - lim (kl9356) – Homework 20 –...

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