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Unformatted text preview: )] 1 If we plug these back into the original diﬀerential equation, we ﬁnd the trial solution works if . . . B = b 2 m and ω = r ω 2( b 2 m ) 2 where ω = r k m So we get a solution that looks like. . . x ( t ) = A ebt 2 m cos( s ω 2( b 2 m ) 2 t + δ ) In plain words, the mass oscillates at a frequency slightly diﬀerent than the “natural frequency” ω , 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
 Simple Harmonic Motion, Drag coefficient, Normal mode, dt d2, free oscillator equation

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