# hw15 - karna (pk4534) HW 15 coker (58245) This print-out...

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karna (pk4534) – HW 15 – coker – (58245) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A harmonic wave is traveling along a rope. The oscillator that generates the wave com- pletes 46 . 0 vibrations in 22 . 1 s. A given crest oF the wave travels 412 cm along the rope in a time period oF 11 . 0 s. What is the wavelength? Correct answer: 0 . 179945 m. Explanation: Let : n = 46 . 0 vibrations , t = 22 . 1 s , Δ x = 412 cm = 4 . 12 m , and Δ t = 11 . 0 s . λ = v f = Δ x Δ t n t = t Δ x n Δ t = (22 . 1 s)(4 . 12 m) 46 (11 s) = 0 . 179945 m . 002 (part 1 oF 2) 10.0 points Transverse waves with a speed oF 45 . 3 m / s are to be produced on a stretched string. A 6 . 29 m length oF string with a total mass oF 0 . 0659 kg is used. What is the required tension in the string? Correct answer: 21 . 4996 N. Explanation: Let : m = 0 . 0659 kg , = 6 . 29 m , and v = 45 . 3 m / s . The linear density is μ = m , so v = r F 1 μ F 1 = μ v 2 = m v 2 = 0 . 0659 kg 6 . 29 m · (45 . 3 m / s) 2 = 21 . 4996 N . 003 (part 2 oF 2) 10.0 points ±ind the wave speed in the string iF the tension is 24 N. Correct answer: 47 . 8617 m / s. Explanation: Let : F 2 = 24 N . v = r F 2 μ = R F 2 · m = r (24 N) (6 . 29 m) 0 . 0659 kg = 47 . 8617 m / s . keywords: 004 (part 1 oF 5) 10.0 points The wave Function For a linearly polarized wave on a taut string is y ( x, t ) = A sin( ω t - k x + φ ) , where A = 0 . 52 m, ω = 5 . 9 s - 1 , k = 6 . 1 m - 1 , φ = 0 . 31, t is in seconds and x and y are in meters. What is the speed oF the wave? Correct answer: 0 . 967213 m / s. Explanation: ±or the propagation speed oF a wave, v = ω k = 5 . 9 s - 1 6 . 1 m - 1 = 0 . 967213 m / s . 005 (part 2 oF 5) 10.0 points What is the vertical displacement oF the string at x 0 = 0 . 14 m, t 0 = 3 s? Correct answer: - 0 . 516087 m. Explanation: Let : x 0 = 0 . 14 m and t 0 = 3 s .

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karna (pk4534) – HW 15 – coker – (58245) 2 y ( x 0 , t 0 ) = A sin[ ω t 0 - k x 0 + φ ] = (0 . 52 m) sin b (5 . 9 s - 1 ) (3 s) - (6 . 1 m - 1 ) (0 . 14 m) + 0 . 31 B = - 0 . 516087 m . 006
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## This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw15 - karna (pk4534) HW 15 coker (58245) This print-out...

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