P2214_F08_PS7SolutionV3

P2214_F08_PS7SolutionV3 - PHYS 2214 Fall 2008 PS7 Solutions...

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PHYS 2214 - Fall 2008 PS7 Solutions PHYS 2214: Waves, Optics, and Particles — Fall 2008 Problem Set 7 – Solutions October 22, 2008 c Joe P. Chen 1. (a) Recall that for a sound wave travelling in the ± x direction, s ( x, t ) = f ( x vt ), the first partial derivatives of s ( x, t ) are related via ∂s ( x, t ) ∂t = v ∂s ( x, t ) ∂x We’ll refer to this as the “pulse equation.” So for a right-moving wave, we can rewrite the instantaneous power per unit cross-sectional area along + x as I inst ( x, t ) = - B ∂s ∂x ∂s ∂t = - 1 2 B ∂s ∂x ∂s ∂t - 1 2 B ∂s ∂x ∂s ∂t = - 1 2 B - 1 v ∂s ∂t ∂s ∂t - 1 2 B ∂s ∂x - v ∂s ∂x = 1 2 B v ∂s ∂t 2 + 1 2 Bv ∂s ∂x 2 = v 1 2 B v 2 ∂s ∂t 2 + 1 2 B ∂s ∂x 2 = v 1 2 ρ ∂s ∂t 2 + 1 2 B ∂s ∂x 2 = v [ u k ( x, t ) + u p ( x, t )] = vu ( x, t ) In the penultimate line we used the fact that the sound speed v = B/ρ . Similarly we can show that for a left-moving wave s ( x, t ) = s ( x + vt ), the instantaneous power per cross-sectional area along + x satisfies I inst ( x, t ) = - B ∂s ∂x ∂s ∂t = · · · = - vu ( x, t ) Figure 1: Setup of the continuity equation for sound. We can interpret this result by considering a small volume of the fluid which ranges from x to x + Δ x and spans a small cross-sectional area Δ A (Fig. 1). Without loss of generality, assume that sound is travelling in the + x direction. First, we repeat the derivation of the continuity equation done in Lecture 14 (10/16/08): At time t , the total energy E ( t ) contained in this small volume is E ( t ) = x x x x u ( x , t A Δ x At a slightly later time t + Δ t , the total energy contained in this volume is E ( t + Δ t ) = x x x x u ( x , t + Δ t A Δ x 1
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PHYS 2214 - Fall 2008 PS7 Solutions By local energy conservation, the change in energy contained in this volume E ( t + Δ t ) - E ( t ) must be equal to the energy flow into the volume between times t and t + Δ t . There are two contributions to the energy flow: 1) Net energy flow through Δ A at x into the volume, which is just the (instantaneous power per unit cross-sectional area along + x ) times Δ A times Δ t , or I inst ( x, t A Δ t ; 2) Net energy flow through Δ A at x + Δ x out of the volume, which is equal to - I inst ( x + Δ x, t A Δ t . Putting it altogether we have E ( t + Δ t ) - E ( t ) = I inst ( x, t A Δ t - I inst ( x + Δ x, t A Δ t or x x x x [ u ( x , t + Δ t ) - u ( x , t )] Δ A Δ x = [ I inst ( x, t ) - I inst ( x + Δ x, t )] Δ A Δ t If we divide both sides of the equation by Δ x Δ t Δ A : x x x x [ u ( x , t + Δ t ) - u ( x , t )] Δ t = [ I inst ( x, t ) - I inst ( x + Δ x, t )] Δ x and take the limit Δ x, Δ t 0, we obtain the continuity equation: ∂u ( x, t ) ∂t = -
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