damped_oscillators - )] 1 If we plug these back into the...

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DAMPER k M -kx -bv V Damped Oscillators Consider the system pictured above. It looks like a normal mass on a spring configuration, except for the motion damper (of drag constant b ) attached to the right side of the apparatus. Σ F x = ma x - kx - bv x = m d 2 x dt 2 0 = d 2 x dt 2 + b m dx dt + k m x This new differential equation has the two terms that show up in the free oscillator equation and it has a new term proportional to the drag coefficient. Trial Solution: It’s reasonable to expect the mass to oscillate back and forth if the drag coefficient isn’t too large. The drag force takes energy away from the system, however, so we expect the amplitude to decay with time. Accordingly, I propose we try a solution that looks like x ( t ) = A 0 e - Bt cos( ωt + δ ). x ( t ) = A 0 e - Bt cos( ωt + δ ) dx dt = A 0 e - Bt [ - B cos( ωt + δ ) - ω sin( ωt + δ )] d 2 x dt 2 = A 0 e - Bt [( B 2 - ω 2 ) cos( ωt + δ ) - 2 sin( ωt + δ
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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 e-bt 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 time-dependent 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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damped_oscillators - )] 1 If we plug these back into the...

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