How Does Damping Affect the Frequency and Amplitude of a Harmonic Oscillator?

[e(ho0n3]

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 \omega_0 = \sqrt{k/m}? (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 x(t) = Ae^{-\alpha t}\cos{\omega t}, then the equation becomes

e^{-2\alpha T} = 0.95

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 (f_N - f)/f_N, where f_N is the natural frequency and f is the frequency calculated above?

(b) Wouldn't the answer be 2m/b?

 

[wais]

After one period, the amplitude will decrease to 1/e of its original value. This is because the amplitude is given by A = A(0)e^{-\alpha T} and when T = 1 period, e^{-\alpha T} = e^{-\alpha}. So the amplitude decreases by a factor of e^{-\alpha} or 1/e of its original value. Therefore, after how many periods will the amplitude have decreased to 1/e of its original value is just 1 period.

To answer the first question, yes, the percentage difference can be calculated by (fN-f)/fN, where fN is the natural frequency and f is the frequency calculated above. This gives a percentage difference of 0.05 or 5%.

For the second question, the answer is indeed just 1 period. This is because the damping factor, e^{-\alpha T}, decreases the amplitude by a constant factor with each period. So after one period, the amplitude will decrease by a factor of e^{-\alpha} or 1/e of its original value. This is true for every subsequent period as well. So after one period, the amplitude will have decreased to 1/e of its original value.