phy293_l07.page1 - • It can be tempting to identify the...

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PHY293 Oscillations Lecture #7 September 25, 2009 1. Second problem set now assigned (http://www.physics.utoronto.ca/ phy293h1f/waves/phy293 ps2.pdf) Start of material 1. More on Power Absorbed by an Oscillator Looking back at what forces are acting on the mass: F = m ¨ x = - ˙ x - 2 0 x + ma 0 ω 2 0 cos( ωt ) = - ˙ x - sx + F 0 cos( ωt ) The first term is the damping force, second is the spring and third is the drive P Damp = F Damp · v = - ˙ x 2 = - mγA 2 ( ω ) ω 2 sin 2 ( ωt - δ ) Can average this over one cycle to get < P Damp > = - 1 2 mγA 2 ( ω ) ω 2 This is the same magnitude as the < P Drive > we saw last time, but opposite sign Can also look at the spring: P Spring = F Spring · v = - sx · ˙ x = sA 2 ( ω ) ω cos( ωt - δ ) sin( ωt - δ ) But averaging this over complete cycles gives 0 since R cos φ sin φdφ = 0 The spring can’t absorb (or generate!) any power, except at various times inside a cycle Just to beat a (hopefully) dead horse a little further The instantaneous power is given by: P = F 0 A ( ω ) ω [ - cos δ sin( ωt ) cos( ωt ) + sin δ cos 2 ( ωt )] The first term is the in-phase piece, while the second is out of phase
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Unformatted text preview: • It can be tempting to identify the sin( ωt ) part as giving a clue to the phase, but it just arises because the power is F · v and v has one time derivative (changing a cos to a sin ). • A better way to “remember” which term is which is to use cos δ (in-phase) and sin δ (out-of-phase) to tag the two pieces. • After the time differentiation to get ˙ x = v we see that only the out-of-phase piece can absorb power, on-average, because only this term has a non-vanishing average over a complete cycle. 2. Quality of a Resonance • We can re-cast many of the results we’ve derived in terms of: Q = ω /γ ω m = ω p 1-1 / 2 Q 2 Resonant Frequency A ( ω m ) = a Q 1 p 1-1 / 4 Q 2 Maximium Amplitude V ( ω ) = a ω Q Maximum Velocity < P ( ω ) > = 1 2 ma 2 ω 3 Q Maximum Power Transfer...
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