# HW 15 - sabeel(was826 – HW15(Chap16 – Bouchalkha...

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Unformatted text preview: sabeel (was826) – HW15 (Chap16) – Bouchalkha – (191zx) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is HW15 for Chapter 16 (part2). It is due on Wednesday 19 May 2010 before 7 AM. 001 (part 1 of 2) 10.0 points Two pulses traveling on the same string are described by y 1 = 5 ( a x- b t ) 2 + 2 and y 2 =- 5 ( a x + b t- 6) 2 + 2 where a = 8 . 3 m − 1 and b = 9 s − 1 . At what time do the two cancel? Correct answer: 0 . 333333 s. Explanation: y 1 travels in the positive x direction, and y 2 travels in the negative x direction. To cancel each other, we require y 1 + y 2 = 0 which is 5 ( a x- b t ) 2 + 2 = 5 ( a x + b t- 6) 2 + 2 ( a x- b t ) 2 = ( a x + b t- 6) 2 a x- b t = ± ( a x + b t- 6) . (1) Using the positive square root in Eq. (1), t = 3 b = 3 9 s − 1 = . 333333 s . 002 (part 2 of 2) 10.0 points At what point do the two waves always can- cel? Correct answer: 0 . 361446 m. Explanation: Using the negative square root in Eq. (1), x = 3 a = 3 8 . 3 m − 1 = . 361446 m . keywords: 003 10.0 points A wave pulse on a string is described by the equation y 1 = A ( B x- C t ) 2 + D . A second wave pulse on the same string is described by y 2 =- A ( B x + C t- E ) 2 + D , where x is in meters and t in seconds, and A = 6 . 04 m, B = 5 . 82 m − 1 , C = 7 . 24 s − 1 , D = 3 . 03, and E = 8 . 66. At what time will the two waves exactly cancel everywhere? Correct answer: 0 . 598066 s. Explanation: Let : A = 6 . 04 m , B = 5 . 82 m − 1 , C = 7 . 24 s − 1 , D = 3 . 03 , and E = 8 . 66 . According to superposition principle y = y 1 ( x, t ) + y 2 ( x, t ) . Since y = y 1 + y 2 , y will be zero when y 1 =- y 2 A ( B x- C t ) 2 + D =- A ( B x + C t- E ) 2 + D B x- C t = ± ( B x + C t- E ) (5 . 82 m − 1 ) x- (7 . 24 s − 1 ) t = ± [(5 . 82 m − 1 ) x +(7 . 24 s − 1 ) t- 8 . 66] . (The ± arises from taking a square root.) Using the “+” sign, 2 C t = E t = E 2 C = 8 . 66 2 (7 . 24 s − 1 ) = . 598066 s . sabeel (was826) – HW15 (Chap16) – Bouchalkha – (191zx) 2 Because x cancels out, our solution tells us y is zero for all positions x at this particular time t . keywords: 004 10.0 points Two harmonic waves are described by y 1 = A sin( k x- ω t ) , y 2 = A sin( k x + ω t ) , where A = 2 . 9 mm, k = 1 . 3 rad / m and ω = 2200 rad / s. What is the amplitude y max ( x ) of the resul- tant wave at x = 1 . 98 m?...
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## This note was uploaded on 05/21/2010 for the course PHYS 191 taught by Professor Aboc during the Spring '10 term at The Petroleum Institute.

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HW 15 - sabeel(was826 – HW15(Chap16 – Bouchalkha...

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