Physics 325 HW 6 Solutions

# Physics 325 HW 6 Solutions - Physics 325 Fall 2010...

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Unformatted text preview: Physics 325, Fall 2010 Solutions to Homework #6 Please note that, in these solutions, we use ω for the natural frequency of the oscillator, not ω , as we have switched to using in class. 1) ( 10 points total ) 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 p ( ω- ω f ) 2 4 ω 2 + 4 ω 2 β 2 = A 2 ω p ( ω- ω f ) 2 + β 2 Since D R = A/ (2 βω ), we can write D ( ω f ) ≈ βD R p ( ω- ω f ) 2 + β 2 With ω f = ω ± β √ 3, then ( ω- ω f ) 2 = 3 β 2 . So D ( ω f ) = βD R p 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 ± β √ 3 = ± tan- 1 r 1 3 ! 2) ( 15 points total ) Our base equation is ¨ x + 2 β ˙ x + ω 2 x = F ext ( t ) m where F ext ( t ) = R e ( F e (- α + iω f ) t ) The most general solution to this equation is given by a sum of a particular and a com-...
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Physics 325 HW 6 Solutions - Physics 325 Fall 2010...

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