hw4solutions - Optics, Waves, and Particles Physics 214, HW...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 + x )- x = x The length after the ends are displaced is L = [( x + x ) + s ( x + x,t )]- [ x + s ( x,t )] Therefore, the change in length L is L = L- L o = s ( x + x,t )- s ( x,t ) b) F = AY L L o = AY s ( x + x,t )- s ( x,t ) x c) Recalling the definition of the derivative, taking the limit x gives F ( x,t ) = AY s ( x,t ) x d) Apply Newtons Second Law, where m = o A x summationdisplay F = F ( x,t ) + F ( x + x,t ) = AY parenleftbigg- s ( x,t ) x + s ( x + x,t ) x parenrightbigg = o A x 2 s ( x cm ,t ) t 2 Dividing both sides by o A x gives, 2 s ( x cm ,t ) t 2 = Y o parenleftBig s ( x + x,t ) x- s ( x,t ) x parenrightBig x Taking the limit x gives 2 s ( x,t ) t 2 = Y o 2 s ( x,t ) x 2 1 Therefore, the wave speed is Y / o radicalbig . Problem 2. a) Recall that f n = v 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 1 = 2 L = 2 ( 25 in ) = 2( 0.635 m ) = 1.27 m. Therefore, v = f 1 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...
View Full Document

Page1 / 4

hw4solutions - Optics, Waves, and Particles Physics 214, HW...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online