{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

6bmid2sol[1]

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

This preview shows pages 1–5. Sign up to view the full content.

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 - 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 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 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 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 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 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 , the electric field at ( x, 0) is E (0 , 0) = E x ( x, 0) ˆ i + E y ( x, 0) ˆ j where E x ( x, 0) = 2 kQ ( x + a ) 2 - kQ ( x - a ) 2 E y ( x, 0) = 0 (4) The electric field is zero at a position x = x 1 . Regardless of where x 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}