ph2a_quiz2_soln

# 2 this allows one to see that a 1 8 h π 2 2 1 a 2 4

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2 . This allows one to see that A 1 = 8 h π 2 ( 2 - 1 ) , A 2 = 4 h π 2 , A 3 = 8 h 9 π 2 ( 2 + 1 ) , A 4 = 0. Part 4 [2 points] : Are any of these first four coefficients zero? If so, can you think of a reason why? (Hint: Think about the symmetry of the characteristic modes.) The coefficient A 4 is zero, for the following reason. Because the initial displacement is zero from L /2 x L , only the region between 0 and L /2 will contribute to the Fourier coefficients. Because the initial displacement of the string is an even function about x = L /4 while the 4 th mode sin ( 4 π x / L ) is odd about x = L /4, their product is an odd function, and this coefficient must vanish. (As an aside, one can see that this holds true for all modes of the form A 4 n , n N either by inspecting the formula of part 3, or through an argument of the type given above). 2

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Part 5 [2 points] : Sketch the approximate for y ( x , t = 0 ) obtained by using only the first four terms in the series expansion. At time t = 0, the Fourier series is given by y ( x , t = 0 ) = n = 1 A n sin n π x L , for the coefficients A n found in part 3. Since the series is being truncated after the first four terms, one has that y ( x , t = 0 ) 8 h π 2 ( 2 - 1 ) sin π x L + 4 h π 2 sin 2 π x L + 8 h 9 π 2 ( 2 + 1 ) sin
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