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Unformatted text preview: Solving differential equations with Fourier transforms Consider a damped simple harmonic oscillator with damping and natural frequency and driving force f ( t ) d 2 y dt 2 + 2 b dy dt + 2 y = f ( t ) At t = 0 the system is at equilibrium y = 0 and at rest so dy dt = 0 We subject the system to an force acting at t = t , f ( t ) = ( t t ), with t > We take y ( t ) = R  g ( ) e i t d and f ( t ) = R  f ( ) e i t d Example continued Substitute into the differential equation and we find 2 2 + 2 ib g ( ) = f ( ) We find also f ( ) = 1 2 R  ( t t ) e i t dt = 1 2 e i t We find a relationship between the g ( ) and f ( ), and then we can write for the response g ( ) g ( ) = 1 2 e i t 2 2 + 2 ib Then with y ( t ) = 0 for t < t , we get y ( t ) for t > t y ( t ) = 1 2 Z  e i ( t t ) 2 2 + 2 ib d Example continued The integral is hard to do (we might get to later), but the point is we have reduced the problem to doing an integral Assume b < , then we find for y ( t ) with t > t , y ( t ) = e...
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This note was uploaded on 07/30/2011 for the course PHZ 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida.
 Spring '03
 Staff

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