hw05_solutions - Physics 214 PS#5 Solutions by Stephen...

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Physics 214 PS #5 Solutions by Stephen Poprocki 2008–02–26 Problem 1. a) A standing wave has the form y ( x, t )= A f ( kx ) cos ( ωt + ϕ ) , where f is either cosine or sine. Since the string is fixed at both ends, then y (0 , t )=0 and y ( L, t )= 0 , at all times t . Thus f (0) = f ( kL ) = 0 . This cannot be satisfied if f = cos, so we must have f = sin and hence sin ( kL ) = 0 kL = , for n = 1 , 2 , 3 , . Since the possible k ’s depend on n , we write k n = / L , so y ( x, t )= A sin parenleftBig L x parenrightBig cos ( ωt + ϕ ) . i. The nodes occur when sin parenleftBig L x parenrightBig =0 L x = mπ, m =0 , 1 , 2 , , n Hence the first node (besides at x =0 ( m =0 )) corresponds to m =1 and occurs at x = L n , n =1 , 2 , 3 , . ii. The antinodes occur when sin parenleftBig L x parenrightBig = ± 1 L x = 2 , m =1 , 3 , 5 , , 2 n 1 . The first antinode corresponds to m =1 and occurs at x = L 2 n , n =1 , 2 , 3 , . b) Take the ideal free end to be at x = L . The node at x = 0 requires f (0) = 0 so again f = sin, but now the antinode at x = L requires sin ( kL )= ± 1 k n = 2 L , n =1 , 3 , 5 , , so y ( x, t )= A sin parenleftBig 2 L x parenrightBig cos ( ωt + ϕ ) . i. The first node (excluding the wall at x =0 ) occurs when 2 L x = π x = 2 L n , n =1 , 3 , 5 , . 1
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ii. The first antinode occurs when 2 L x = π 2 x = L n , n =1 , 3 , 5 , . c) Let L be the length of the string with fixed ends at both sides. It suffices to make the first nodes occur at the same place. Let us reindex the free end solutions by n = 2 1 , = 1 , 2 , 3 , , so that the two enumeration methods match. Then L = 2 L 2 1 L = 2 2 1 L , =1 , 2 , 3 , so the standing wave on the string of length L with both ends fixed, y fixed ( x, t )= A sin parenleftbigg ℓπ L x parenrightbigg cos ( ωt + ϕ ) , is the same (between 0 and L ) as the string of length L with a free end at x = L : y free ( x, t )= A sin parenleftbigg (2 1) π 2 L x parenrightbigg cos ( ωt + ϕ ) , for a given =1 , 2 , 3 , . 0 L 6 5 L 4 3 L 2L Equal 3 Equal 2 Equal 1 Figure 1. Matching standing wave patterns for a fixed end (blue) and free end (red) for =1 , 2 , 3 .
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