6bmid2sol[1]

6bmid2sol[1] - Solutions to Practice Midterm 2 Problem 1 We have the two waves y 1 x,t = A sin kx ωt y 2 x,t = A sin kx ωt = A sin kx ωt π(1

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Unformatted text preview: Solutions to Practice Midterm 2 Problem 1: We have the two waves y 1 ( x,t ) = A sin( kx- ωt ) y 2 ( x,t ) =- A sin( kx + ωt ) = A sin( kx + ωt + π ) (1) Wave 1 is traveling in the direction of positive x , due to the kx- ωt term. Wave 2 is traveling in the direction of negative x , due to the kx + ωt term. (2) The phase difference between the two waves is Δ φ = φ + π . We make the definitions θ 1 = kx- ωt θ 2 = kx + ωt + φ + π (3) The resultant wave is a standing wave, given by y ( x,t ) = y 1 ( x,t ) + y 2 ( x,t ) = A sin θ 1 + A sin θ 2 = 2 A sin θ 2 + θ 1 2 ¶ cos θ 2- θ 1 2 ¶ = 2 A sin kx + φ 2 + π 2 ¶ cos ωt + φ 2 + π 2 ¶ =- 2 A cos kx + φ 2 ¶ sin ωt + φ 2 ¶ (4) The resultant wave has amplitude 2 A . (5) The maximum transverse position of an element of the string is the maximum value of y ( x,t ) =- 2 A cos kx + φ 2 ¶ sin ωt + φ 2 ¶ This value is y max = 2 A . This value is attained at ( x n ,t m ) for which cos kx n + φ 2 ¶ = 1 , x n = 1 k 2 nπ- φ 2 ¶ , n = 0 , 1 , 2 ,..... sin ωt m + φ 2 ¶ =- 1 , t m = (4 m + 3) π- φ 2 ω , m = 0 , 1 , 2 ,..... or at ( x n ,t m ) for which cos kx n + φ 2 ¶ =- 1 , x n = 1 k • (2 n + 1) π- φ 2 ‚ , n = 0 , 1 , 2 ,..... sin ωt m + φ 2 ¶ = 1 , t m = (4 m + 1) π- φ 2 ω , m = 0 , 1 , 2 ,..... 1 Note that the x n are the locations of the antinodes of the resultant wave. (6) The resultant wave is a standing wave, and therefore does not travel in either direction. (7) The nodal positions are at x j , where- 2 A cos kx j + φ 2 ¶ sin ωt + φ 2 ¶ = 0 cos kx j + φ 2 ¶ = 0 kx j + φ 2 = (2 j + 1) π 2 , j = 0 , 1 , 2 ,.... x j = (2 j + 1) π- φ 2 k , j = 0 , 1 , 2 ,.... (8) The antinode positions are at x n , where cos kx n + φ 2 ¶ = ± 1 kx n + φ 2 = nπ, n = 0 , 1 , 2 ,..... x n = 2 nπ- φ 2 k , n = 0 , 1 , 2 ,..... Note that this is the same set of x n as obtained by combining the two sets from the answer to question (5). (9) The distance d between successive nodes, or between successive antinodes, is half a wavelength, which is d = λ 2 = π k We can verify this explicitly. The distance between successive nodes is d nodes = x j +1- x j = [2( j + 1) + 1] π- φ 2 k- (2 j + 1) π- φ 2 k = π k The distance between successive antinodes is d anti = x n +1- x n = 2( n + 1) π- φ 2 k- 2 nπ- φ 2 k = π k 2 The derivatives of y ( x,t ) are as follows: ∂y ∂x = 2 Ak sin kx + φ 2 ¶ sin ωt + φ 2 ¶ ∂ 2 y ∂x 2 = 2 Ak 2 cos kx + φ 2 ¶ sin ωt + φ 2 ¶ ∂y ∂t =- 2 Aω cos kx + φ 2 ¶ cos ωt + φ 2 ¶ ∂ 2 y ∂t 2 = 2 Aω 2 cos kx + φ 2 ¶ sin ωt + φ 2 ¶ Therefore, we have ∂ 2 y ∂x 2 = k 2 ω 2 ∂ 2 y ∂t 2 = 1 v 2 ∂ 2 y ∂t 2 where v = ω/k is the speed of each of the oppositely-directed traveling waves of which the resultant standing wave is composed....
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This note was uploaded on 01/24/2011 for the course PHYSICS 6b taught by Professor Gruner during the Winter '10 term at UCLA.

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6bmid2sol[1] - Solutions to Practice Midterm 2 Problem 1 We have the two waves y 1 x,t = A sin kx ωt y 2 x,t = A sin kx ωt = A sin kx ωt π(1

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