Damped Harmonic Oscillator

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Question: A damped harmonic oscillar loses 5.0 percent of its mechanical energy per cycle. (a) By what percentage does its frequency differ from the natural frequency [itex]\omega_0 = \sqrt{k/m}[/itex]? (b) After how many periods will the amplitude have decreased to 1/e of its original value?

(a) Let E(t) represent the mechanical energy at time t. Therefore E(T) = 0.95 E(0) where T is the period. At t = 0 and t = T, the mechanical energy is just spring potential energy so 1/2kx(T)2 = 0.95(1/2kx(0)2) which simplifies to x(T)2 = 0.95 x(0)2. Since [itex]x(t) = Ae^{-\alpha t}\cos{\omega t}[/itex], then the equation becomes

[tex]e^{-2\alpha T} = 0.95[/tex]

Solving for T yields T = ln(0.95)/(-2α) and the frequency is just the inverse of this. Note that α = b/(2m). Now how do I calculate the percent difference? Is it just (fN - f)/fN, where fN is the natural frequency and f is the frequency calculated above?

(b) Wouldn't the answer be 2m/b?
 

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