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Unformatted text preview: y III = ( A 3 e ik 1 x + A 3 e ik 1 x ) e iωt Notice that A 3 = 0 because there are no reflected waves from infinity. The boundary condi tions are the continuity of the wave and its x –derivative at the two points of discontinuity. Specifically, A 1 + A 1 = A 2 + A 2 k 1 ( A 1 A 1 ) = k 2 ( A 2 A 2 ) A 2 e ik 2 L + A 2 e ik 2 L = A 3 e ik 1 L k 2 A 2 e ik 2 L k 2 A 2 e ik 2 L = k 1 A 3 e ik 1 L whose solution is A 3 = e ik 1 L cos k 2 L i k 2 1 + k 2 2 2 k 1 k 2 sin k 2 L A 1 so that 1 = A 3 A 1 2 = 4 k 2 1 k 2 2 4 k 2 1 k 2 2 cos 2 k 2 L + ( k 2 1 + k 2 2 ) sin 2 k 2 L implies k 2 L = nπ , or L = n 2 λ 2 which is the sought condition. Standing Waves If we have a sinusoidal wave (not a pulse as before) with a point held fixed, the wave will be a superposition of the wave and its inverted mirror image traveling in the opposite direction. Say the end is kept fixed at x=0. [Superposition of traveling waves Physlet Illustration 17.4 (1,2,3)] Say the incident wave to the right of the fixed point is given by y 1 = A cos( kx + ωt )? We then need y 2 = A cos( kx ωt ) (we choose a cosine function here so that it is easier to see the opposite sign). y 1 + y 2  x =0 = A cos( ωt ) A cos( ωt ) = A cos( ωt ) A cos( ωt ) = 0 as required. The total function ∀ x: y = y 1 + y 2 = A cos( kx + ωt ) A...
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This note was uploaded on 07/19/2011 for the course PH 113 taught by Professor Liorburko during the Spring '09 term at University of Alabama  Huntsville.
 Spring '09
 LIORBURKO
 Physics

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