ph2a_quiz2_soln

# Since the string has no initial velocity the phases

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• AgentStarOkapi9220
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1, 2, 3, 4 at the end (this ultimately will require less work). Since the string has no initial velocity, the phases δ n can be chosen to be zero and the coefficients are given by A n = 2 L L 0 y ( x , t = 0 ) sin n π x L dx . 1

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Because we are dealing with a piecewise-continuous function the integral breaks into the following parts A n = 2 L L /4 0 4 h L x sin n π x L dx + 2 L L /2 L /4 2 h sin n π x L dx - 2 L L /2 L /4 4 h L x sin n π x L dx . Let’s evaluate each integral separately, and add the results to get the final answer. Inte- grating by parts, one has that 2 L L /4 0 4 h L x sin n π x L dx = - 8 h n π L x cos n π x L x = L /4 x = 0 - L /4 0 cos n π x L . One can then evaluate this integral to find 2 L L /4 0 4 h L x sin n π x L dx = - 2 h n π cos n π 4 - 4 n π sin n π 4 . The second integral can be evaluated immediately to find 2 L L /2 L /4 2 h sin n π x L dx = - 4 h n π cos n π 2 - cos n π 4 . The third integral is similar to the first, and by integrating by parts, one can find that - 2 L L /2 L /4 4 h L x sin n π x L dx = 4 h n π cos n π 2 - 1 2 cos n π 4 - 8 h n 2 π 2 sin n π 2 - sin n π 4 . There is a large amount of cancellation when the three integrals are added, leaving just A n = 8 h n 2 π 2 2 sin n π 4 - sin n π
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