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solutionshw4

# solutionshw4 - Solutions to Homework 4 Due October 21st...

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Solutions to Homework 4: Due October 21 st Chapter 16 Problem 40: Comparing 0.35sin 10 3 4 y t x π π π = - + with ( 29 ( 29 sin sin y A kx t A t kx ϖ φ ϖ φ π = - + = - - + we have 3 k m π = , 10 s ϖ π = , 0.35 m A = . Then 10 s 2 3.33 m s 2 3 m v f f k π λ ϖ λ π π π = = = = = . (a) The rate of energy transport is ( 29 ( 29 ( 29 2 2 2 2 3 1 1 75 10 kg m 10 s 0.35 m 3.33 m s 15.1 W 2 2 A v μϖ π - = = × = P . (b) The energy per cycle is ( 29 ( 29 ( 29 2 2 2 2 3 1 1 2 m 75 10 kg m 10 s 0.35 m 3.02 J 2 2 3 E T A λ π μϖ λ π π - = = = × = P . Problem 44: The linear wave equation is 2 2 2 2 2 1 y y x v t = If ( 29 b x vt y e - = then ( 29 b x vt y bve t - = - and ( 29 b x vt y be x - = ( 29 2 2 2 2 b x vt y b v e t - = and ( 29 2 2 2 b x vt y b e x - = Therefore, 2 2 2 2 2 y y v t x = , demonstrating that ( 29 b x vt e - is a solution

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Problem 48: Compare the given wave function ( 29 4.00sin 2.00 3.00 cm y x t = - to the general form ( 29 sin y A kx t ϖ = - to find (a) amplitude 4.00 cm 0.040 0 m A = = (b) 1 2 2.00 cm k π λ - = = and cm 0.031 4 m λ π = = (c) 1 2 3.00 s f ϖ π - = = and 0.477 Hz f = (d) 1 2.09 s T f = = (e)
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