HW 35 - Morby, Grant Homework 35 Due: May 1 2006, noon...

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Unformatted text preview: Morby, Grant Homework 35 Due: May 1 2006, noon Inst: Drummond 1 This print-out should have 11 questions. Multiple-choice 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 A standing wave is a superposition of two harmonic waves described by y 1 = 6 sin( k x + t ) , and y 2 = 6 sin( k x- t ) . where k = 17 . 5929 m- 1 and = 17 . 5929 s- 1 . Determine the smallest positive value of x corresponding to an antinode. Correct answer: 8 . 92857 cm. Explanation: On a standing wave, the distance of the nearest antinode and node is d = 4 , where is the wavelength. In this problem, x = 0 is a node, which can be seen by adding y 1 and y 2 , and the wavelength is 2 /k . So the smallest positive value of x corresponding to an antinode is x = / 2 k . 002 (part 1 of 2) 10 points A standing wave of frequency 5 hertz is set up on a string 2 meters long with nodes at both ends and in the center, as shown. 2 meters Find the speed at which waves propagate on the string. 1. 10 m / s correct 2. 5 m / s 3. 2 . 5 m / s 4. . 4 m / s 5. 20 m / s Explanation: Let : f = 5 Hz and = 2 m . The wavelength is = 2 m, so the wave speed is | ~v | = f = (5 Hz)(2 m) = 10 m/s . 003 (part 2 of 2) 10 points Find the fundamental frequency of vibration of the string. 1. 1 Hz 2. 7 . 5 Hz 3. 5 Hz 4. 10 Hz 5. 2 . 5 Hz correct Explanation: 2 meters The fundamental wave has only two nodes at the ends, so its wavelength is = 4 m and the fundamental frequency is f = v = 10 m / s 4 m = 2.5 Hz ....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.

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HW 35 - Morby, Grant Homework 35 Due: May 1 2006, noon...

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