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Unformatted text preview: )] 1 If we plug these back into the original dierential equation, we nd the trial solution works if . . . B = b 2 m and = r 2( b 2 m ) 2 where = r k m So we get a solution that looks like. . . x ( t ) = A ebt 2 m cos( s 2( b 2 m ) 2 t + ) In plain words, the mass oscillates at a frequency slightly dierent than the natural frequency , with a timedependent amplitude that decays at a rate proportional to the size of the drag factor. As the strength of the drag force ( b ) goes to zero, the solution reduces to the solution for a simple harmonic oscillator. The amplitude and phase shift must either be given or found from initial conditions. 2...
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 Spring '10
 Corbin

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