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

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Unformatted text preview: 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 ) = , 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 = nπ , for n = 1 , 2 , 3 , . Since the possible k ’s depend on n , we write k n = nπ / L , so y ( x,t ) = A sin parenleftBig nπ L x parenrightBig cos ( ωt + ϕ ) . i. The nodes occur when sin parenleftBig nπ L x parenrightBig = 0 nπ 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 nπ L x parenrightBig = ± 1 nπ L x = mπ 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 = nπ 2 L , n = 1 , 3 , 5 , , so y ( x,t ) = A sin parenleftBig nπ 2 L x parenrightBig cos ( ωt + ϕ ) . i. The first node (excluding the wall at x = 0 ) occurs when nπ 2 L x = π x = 2 L n , n = 1 , 3 , 5 , . 1 ii. The first antinode occurs when nπ 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 ℓ − 1 , ℓ = 1 , 2 , 3 , , so that the two enumeration methods match. Then L ′ ℓ = 2 L 2 ℓ − 1 L ′ = 2 ℓ 2 ℓ − 1 L , 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 and L ) as the string of length L with a free end at x = L : y free ( x,t ) = A sin parenleftbigg (2 ℓ − 1) π 2 L x parenrightbigg cos ( ωt + ϕ ) , for a given ℓ = 1 , 2 , 3 , ....
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This note was uploaded on 03/31/2009 for the course PHYS 214 taught by Professor Thorne during the Spring '08 term at Cornell University (Engineering School).

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

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