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Unformatted text preview: 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 = AY Δ L L o = AY s ( x + Δ x,t ) s ( x,t ) Δ x c) Recalling the definition of the derivative, taking the limit Δ x → gives F ( x,t ) = AY ∂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 ) = AY 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 → gives ∂ 2 s ( x,t ) ∂t 2 = Y ρ o ∂ 2 s ( x,t ) ∂x 2 1 Therefore, the wave speed is Y / ρ o radicalbig . Problem 2. a) Recall that f n = v λ n = nv 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...
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This note was uploaded on 03/31/2009 for the course PHYS 214 taught by Professor Thorne during the Spring '08 term at Cornell.
 Spring '08
 THORNE
 Physics

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