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Unformatted text preview: 1. The following expression is proposed as a solution of the wave equation: y(x, t) = A sin(kx + ) cos(t), where A, k, , and are constants. a) Show that it is a solution of the wave equation, or more precisely, find the condition (involving some of the four constants) that ensures it is a solution. b) Suppose that this expression represents the motion of a string that is tied down at two ends (a standing wave). One end is at x = 0 and the coordinate y is zero at that point for all times. What does this condition determine about any of the four constants? c) The other end of the string at x = L is also tied down so that y is zero for all times. What does this condition determine about any of the four constants? Solution: a) The wave equations is on the front cover: 2y 1 2y = 2 2 x2 v t 2 y y = kA cos(kx+) cos(t); = -k 2 A sin(kx+) cos(t) = -k 2 y(x, t) 2 x x y 2y = -A sin(kx+) sin(t); = - 2 A sin(kx+) cos(t) = - 2 y(x, t) 2 t t Now put these results in the wave equation: -k 2 y(x, t) = - which is satisfied if v 2 = (/k)2 . b) y(0, t) = A sin cos(t) = 0 sin = 0 = m, m = 0, 1, 2, ... Though A = 0 would satisfy the condition, it is not a standing wave solution. Also, we cannot have cos(t) = 0 since for a nonzero this would only be true at particular times, and we must have y(0, t) = 0 for all times. c) y(L, t) = A sin(kL+) cos(t) = 0 sin(kL+) = 0 kL+ = n, n = 0, 1, 2, ... Given,from b), that = m, we could write: kL = (n - m) = q, q = 0, 1, 2, ... since q = n - m will take on the same integer values as either m or n. 1 2 y(x, t) v2 ...
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This note was uploaded on 04/26/2008 for the course PHY 206 taught by Professor Cohn during the Spring '08 term at University of Miami.
- Spring '08