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6bmid2sol

# 6bmid2sol - Solutions to Practice Midterm 2 Problem 1 We...

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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. We make the definitions θ 1 = kx - ωt θ 2 = kx + ωt + φ + π (2) The phase difference between the two waves is Δ φ = | θ 2 - θ 1 | = | kx + ωt + φ + π - kx + ωt | = 2 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 a position-dependent amplitude of A m ( x ) = 2 A cos kx + φ 2 This is the amplitude of the simple harmonic oscillations of the individual elements of the string. (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 1

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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 - 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 , ..... 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 - 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 2
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 - φ 2 k = π k 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 cos kx + φ 2 cos ωt + φ 2 2 y ∂t 2 = 2 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. Problem 2: (1) The electric field at (0 , 0) is 3

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E (0 , 0) = E x (0 , 0) ˆ i + E y (0 , 0) ˆ j where E x (0 , 0) = 2 kQ a 2 + kQ a 2 = 3 kQ a 2 E y (0 , 0) = 0 (2) The electric field at (0 , a ) is E (0 , a ) = E x (0 , a ) ˆ i + E y (0 , a ) ˆ j where E x (0 , a ) = 2 kQ a 2 + a 2 a a 2 + a 2 + k ( - Q ) a 2 + a 2 - a a 2 + a 2 = 2 kQ 2 a 2 2 + kQ 2 a 2 2 = 3 kQ 2 a 2 2 E y (0 , a ) = 2 kQ a 2 + a 2 a a 2 + a 2 + k ( - Q ) a 2 + a 2 a a 2 + a 2 = 2 kQ 2 a 2 2 - kQ 2 a 2 2 = kQ 2 a 2 2 (3) For x > a
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