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Unformatted text preview: Physics 325, Fall 2010 Solutions to Homework #6 Please note that, in these solutions, we use for the natural frequency of the oscillator, not , as we have switched to using in class. 1) ( 10 points total ) The amplitude of a damped, driven oscillator is D ( f ) = A q ( 2 2 f ) 2 + 4 2 f 2 If f and (that is, the driving frequency is close to the resonance frequency, and the resonance is narrow), then D ( f ) = A q (  f ) 2 ( + f ) 2 + 4 2 f 2 D ( f ) A p (  f ) 2 4 2 + 4 2 2 = A 2 p (  f ) 2 + 2 Since D R = A/ (2 ), we can write D ( f ) D R p (  f ) 2 + 2 With f = 3, then (  f ) 2 = 3 2 . So D ( f ) = D R p 3 2 + 2 = D R 2 = tan 1 2 f 2 2 f ! = tan 1 2 f ( + f )(  f ) But + f 2 2 f . So tan 1  f = tan 1 3 = tan 1 r 1 3 ! 2) ( 15 points total ) Our base equation is x + 2 x + 2 x = F ext ( t ) m where F ext ( t ) = R e ( F e ( + i f ) t ) The most general solution to this equation is given by a sum of a particular and a com...
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 Fall '08
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 Physics, mechanics, Work

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