hw4solutions - Optics Waves and Particles Physics 214 HW#04...

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Optics, Waves, and Particles Physics 214, HW #04 Spring 2008 Problem 1. a) The initial length of the segment is L o = ( x + Δ x ) - x = Δ x The length after the ends are displaced is L = [( x + Δ x ) + s ( x + Δ x, t )] - [ x + s ( x, t )] Therefore, the change in length Δ L is Δ L = L - L o = s ( x + Δ x, t ) - s ( x, t ) b) F = A Y Δ L L o = A Y s ( x + Δ x, t ) - s ( x, t ) Δ x c) Recalling the definition of the derivative, taking the limit Δ x 0 gives F ( x, t ) = A Y ∂s ( x, t ) ∂x d) Apply Newton’s Second Law, where m = ρ o A Δ x summationdisplay F = F ( x, t ) + F ( x + Δ x, t ) = A Y parenleftbigg - ∂s ( x, t ) ∂x + ∂s ( x + Δ x, t ) ∂x parenrightbigg = ρ o A Δ x 2 s ( x cm , t ) ∂t 2 Dividing both sides by ρ o A Δ x gives, 2 s ( x cm , t ) ∂t 2 = Y ρ o parenleftBig ∂s ( x + Δ x, t ) ∂x - ∂s ( x, t ) ∂x parenrightBig Δ x Taking the limit Δ x 0 gives 2 s ( x, t ) ∂t 2 = Y ρ o 2 s ( x, t ) ∂x 2 1
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Therefore, the wave speed is Y / ρ o radicalbig . Problem 2. a) Recall that f n = v λ n = n v 2 L = n f 1 for n = 1 , 2 , 3 , We can find v with the information given for the case n = 1 . We know the fundamental frequency is f 1 = 150 Hz and the fundamental wavelength is λ 1 = 2 L = 2 ( 25 in ) = 2( 0.635 m ) = 1.27 m. Therefore, v = f 1 λ 1 = 150 ( 1.27 ) m / s = 190.5 m / s b) f 2 = 2 f 1 = 300 Hz f 3 = 3 f 1 = 450 Hz c) If we press firmly, the length is reduced to L = 4 L /5 , and therefore λ 1 = 2 L = 8 L /5 .
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