handout01 - f 1 u u t =-c f f 1 , f 2 t = f 2 v v t =-c f f...

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PHYSICS 218 SOLUTION TO HANDOUT 1 Created: September 12, 2004 0:45am Last updated: September 12, 2004 8:48am 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 0 1 , ∂f 2 ∂x = ∂f 2 ∂v ∂v ∂x = n x f 0 2 , ∂f 1 ∂y = ∂f 1 ∂u ∂u ∂y = n y f 0 1 , ∂f 2 ∂y = ∂f 2 ∂v ∂v ∂y = n y f 0 2 , ∂f 1 ∂z = ∂f 1 ∂u ∂u ∂z = n z f 0 1 , ∂f 2 ∂z = ∂f 2 ∂v ∂v ∂z = n z f 0 2 , ∂f 1 ∂t =
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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 nd 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 nd 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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This note was uploaded on 09/28/2008 for the course PHYS 218 taught by Professor Pollack during the Fall '04 term at Cornell University (Engineering School).

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