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Unformatted text preview: PHY 5346 HW Set 5 Solutions Kimel 1. 2.13 The system is pictured in the following figure: a) Notice from the figure, , , ; thus from Eq. (2.71) in the text, , a 0 n1 a n n cos n a 0 V1 V2 2 /2 b, 3/2 2a 0 V 1 V 2 Using /2 cos m cos nd nm 3/2 Applying this to , only odd terms m contribute in the sum and m 1 2 V1 V2 am 1 2 mb m Thus 2 V1 V2 i m m e im , V 1 V 2 Im 2 mb m
m odd Using 2
m odd x m ln m 1x 1 x
1 and Im ln A iB tan we get V1 V2 V V2 , 1 tan 2 as desired. b) 0 1 B/A 2 b cos 1
2 b2 , b V1 V2 0 V1 V2 tan
2 1 2 b cos 1
2 2 b2 b 0 2 : 0 b cos b 4 b b 2b 4 4 2 b 2 cos 2 2 2 V1 V2 b cos ...
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This note was uploaded on 01/10/2012 for the course PHYSIS 506 taught by Professor Smith during the Spring '08 term at University of Arkansas Community College at Morrilton.
 Spring '08
 smith

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