hw2solns

# hw2solns - / by a factor of 2 . (f) The power resonance...

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Phys 273 Homework 2 solutions

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4-8

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4-14. (a) the full-width-half-max (FWHM) of a resonance curve is ω 0 /Q , and given that FWHM = 0.04 ω 0 from the figure, then Q = 1/0.04 = 25 (b) the damping constant γ is also the FWHM of the power spectrum, so γ = 0.04 ω 0 (c) the energy decreases in time as exp ( −γ t ), so in 1 cycle ( t = 2π/ω 0 ), the fraction of energy remaining is exp ( 0.04*2 π ) = 0.78, so 22% of the energy is lost per cycle (d) doubled spring constant increases the resonant frequency (k/m) 1/2 by a factor of 2 . (e) doubled spring constant increases Q=

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Unformatted text preview: / by a factor of 2 . (f) The power resonance curve is 2 2 1 2 ) ( Q Q m F P + = , so the maximum (on resonance) power that can be put into the system (and dissipated) is 2 2 ) ( m Q F P = = . Since both and Q increase by 2 , this is unchanged. (g) since the average power is the same, the energy in the system is also unchanged....
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## This note was uploaded on 12/28/2011 for the course PHYSICS 273 taught by Professor Monroec during the Fall '11 term at Maryland.

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hw2solns - / by a factor of 2 . (f) The power resonance...

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