hw7_solutions - Physics 325 Fall 2006 Solutions to...

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Physics 325, Fall 2006 Solutions to Homework #7 1) The amplitude of a damped, driven oscillator is D ( ω f ) = A q ( ω 2 - ω 2 f ) 2 + 4 ω 2 f β 2 If ω f ω and β ± ω (that is, the driving frequency is close to the resonance frequency, and the resonance is narrow), then D ( ω f ) = A q ( ω - ω f ) 2 ( ω + ω f ) 2 + 4 ω 2 f β 2 D ( ω f ) A q ( ω - ω f ) 2 4 ω 2 + 4 ω 2 β 2 = A 2 ω q ( ω - ω f ) 2 + β 2 Since D R = A/ (2 βω ), we can write D ( ω f ) βD R q ( ω - ω f ) 2 + β 2 With ω f = ω ² β 3, then ( ω - ω f ) 2 = 3 β 2 . So D ( ω f ) = βD R 3 β 2 + β 2 = D R 2 δ = tan - 1 ± 2 βω f ω 2 - ω 2 f ! = tan - 1 ± 2 βω f ( ω + ω f )( ω - ω f ) ! But ω + ω f 2 ω 2 ω f . So δ tan - 1 ± β ω - ω f ! = tan - 1 ± β ² β 3 ! = ² tan - 1 s 1 3 2) F ext ( t ) = R e ( F 0 e ( - α + f ) t ) Let’s solve the problem without the “Re” in there, adding in a fictitious imaginary compo- nent F ext ( t ). Since the resulting solution to the damped oscillator’s differential equation
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will be the sum of a real and an imaginary part, we’ll just remember to take the real part at the end of the problem. So let’s take F ext ( t ) = F 0 e ( - α + f ) t and solve ¨ x + 2 β ˙ x + ω 2 x = F ext ( t ) m Let’s take the hint and try a solution of the form x
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hw7_solutions - Physics 325 Fall 2006 Solutions to...

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