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# handout01 - ∂f 1 ∂u ∂u ∂t =-c f f 1 ∂f 2 ∂t =...

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PHYSICS 218 SOLUTION TO HANDOUT 1 Created: September 12, 2004 0:45am Last updated: September 12, 2004 8:48am Elmore & Heald, Question 5.1.1 Show that the most general solution of the wave equation (5.1.8), shown below 2 p = 1 c 2 f 2 p ∂t 2 , consisting of plane waves traveling in the direction of the unit vector n is given by p = f 1 ( n · r - c f t ) + f 2 ( n · r + c f t ) , (1) where f 1 and f 2 are arbitrary functions. Solution: We are asked to show that the most general plane wave solution of hte three-dimensional acoustic wave equation traveling in the direction n has the form in (1). Recall that r = x ˆ x + y ˆ y + z ˆ z , and let’s define u = n x x + n y y + n z z - c f t , v = n x x + n y y + n z z + c f t . For (1) to be a solution of the wave equation, it must satisfy the wave equation. So we check this first by doing the following derivatives. ∂f 1 ∂x = ∂f 1 ∂u ∂u ∂x = n x f 1 , ∂f 2 ∂x = ∂f 2 ∂v ∂v ∂x = n x f 2 , ∂f 1 ∂y = ∂f 1 ∂u ∂u ∂y = n y f 1 , ∂f 2 ∂y = ∂f 2 ∂v ∂v ∂y = n y f 2 , ∂f 1 ∂z = ∂f 1 ∂u ∂u ∂z = n z f 1 , ∂f 2 ∂z = ∂f
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Unformatted text preview: ∂f 1 ∂u ∂u ∂t =-c f f 1 , ∂f 2 ∂t = ∂f 2 ∂v ∂v ∂t =-c f f 2 , ∂ 2 f 1 ∂x 2 = n 2 x f 00 1 , ∂ 2 f 2 ∂x 2 = n 2 x f 00 2 , ∂ 2 f 1 ∂y 2 = n 2 y f 00 1 , ∂ 2 f 2 ∂y 2 = n 2 y f 00 2 , ∂ 2 f 1 ∂z 2 = n 2 z f 00 1 , ∂ 2 f 2 ∂z 2 = n 2 z f 00 2 , ∂ 2 f 1 ∂t 2 = c 2 f f 00 1 , ∂ 2 f 2 ∂t 2 = c 2 f f 00 2 . Therefore, we ﬁnd ∇ 2 p = ∂ 2 p ∂x 2 + ∂ 2 p ∂y 2 + ∂ 2 p ∂z 2 = ( n 2 x + n 2 y + n 2 z ) ( f 00 1 + f 00 2 ) = f 00 1 + f 00 2 , where we have made used of the fact that n is a unit vector | n | = 1. On the other hand, we also ﬁnd 1 c 2 ∂ 2 p ∂t 2 = 1 c 2 f ( c 2 f f 00 1 + c 2 f f 00 2 ) = f 00 1 + f 00 2 . Thus, equation (1) is indeed a solution of the acoustic wave equation. 1...
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