This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 325, Fall 2006 Solutions to Homework #8 1) This is an exercise in finding the Fourier series of a periodic function. F ( t ) = F  sin f t  = 1 2 a + X n =1 a n cos( n f t ) + b n sin( n f t ) with a n = 2 Z / 2 / 2 F ( t ) cos( n f t ) dt b n = 2 Z / 2 / 2 F ( t ) sin( n f t ) dt The period = 2 / f so / 2 = / f and 2 / = f / . Actually, this is twice the period, but it is easier to apply the Fourier series formulas from class if we center the intervals on zero. Also, it is important to use this defintion of the period because the f = 2 / in the Fourier Series defintion is the same as the f in the driving function. All the b n s are zero by symmetry (they are the integrals of products of even and odd functions, which makes an odd function, from / 2 to / 2, so the integrals over the negative times cancel out the integrals over the positive times)....
View
Full
Document
This note was uploaded on 02/23/2009 for the course PHYS 325 taught by Professor Lamb during the Fall '09 term at University of Illinois at Urbana–Champaign.
 Fall '09
 Lamb
 mechanics, Work

Click to edit the document details