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Unformatted text preview: Patel, Kinal – 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 2 m segment of a long string contains 3 complete wavelengths and has a mass of 180 g. At one point it is vibrat ing sinusoidally with a frequency of 48 Hz and a peak to valley displacement of 0 . 39 m. What power is being transmitted along the string? Correct answer: 4980 . 51 W. Explanation: Let : l = 2 m , λ = 2 m 3 , m = 180 g , f = 48 Hz , and A = . 39 m 2 . Since k ≡ 2 π λ , ω ≡ 2 π f , v ≡ ω k , P = 1 2 μω 2 A 2 v = 1 2 ‡ m l · (2 π f ) 2 A 2 λf = 1 2 µ 180 g 2 m ¶ [2 π (48 Hz)] 2 × µ . 39 m 2 ¶ 2 µ 2 m 3 ¶ (48 Hz) = 4980 . 51 W . keywords: 002 (part 1 of 2) 10 points A point source emits sound waves with an average power output of 1 . 26 W. Find the intensity 7 . 02 m from the source. Correct answer: 0 . 00203463 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 . 02 m from the source is I = P av 4 π R 2 = 0 . 00203463 W / m 2 . 003 (part 2 of 2) 10 points Find the distance at which the sound reduces to a level of 65 dB. Correct answer: 178 . 066 m. Explanation: We can find the intensity at the β = 65 dB level by using the equation β = 10 log µ I I ¶ with I = 1 × 10 12 W / m 2 : 65 dB = 10 log µ I 1 × 10 12 W / m 2 ¶ therefore I = 10 β/ 10 I = 10 6 . 5 (1 × 10 12 W / m 2 ) = 3 . 16228 × 10 6 W / m 2 ....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 1710 taught by Professor Weathers during the Spring '07 term at North Texas.
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