ph2a_quiz2_soln

ph2a_quiz2_soln - Quiz 2 Solutions Kevin Setter November 6...

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Quiz 2 Solutions Kevin Setter November 6, 2009 Note: The correct solutions to parts 3 and 4 involve a subtlety that a Ph2a student is not expected to be able to get. Therefore, full credit was awarded to the “naive” solution to 3 and 4 described below. The correct solution is then described in 3’ and 4’. 1 At t = L/ 2 v , the string looks like Figure 1: Sketch of y ( x, 0). 2 At t = 0 the displacement is y ( x, 0) = ( 8 A L x, if 0 < x < L/ 8 0 , otherwise 3 At t = 0, the vertical string velocity is given by ∂y ∂t ( x, 0) = ( - 8 Av L , if 0 < x < L/ 8 0 , otherwise (See 3’ below for the correct answer.) 1
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Ph2a Solutions Quiz 2 November 6, 2009 4 Using the trigonometric identity written in the statement of problem 7, we rewrite the Fourier expansion in the more tractable form y ( x, t ) = X n =1 A n cos ( ω n t + δ n ) sin L x = X n =1 A 0 n sin L x cos ( ω n t ) + B 0 n sin L x sin ( ω n t ) where the phases δ n and the amplitudes A n are obtained from the coefficients A 0 n , B 0 n by ( A n = p ( A 0 n ) 2 + ( B 0 n ) 2 δ n = - tan - 1 B 0 n A 0 n If f ( x ) = y ( x,
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