Answer to Question #89130 in Physics for Shivam Nishad

Question

For a weakly damped harmonic oscillator, the
instantaneous displacement is given by
x(t) = ao exp (-bt) cos (wd t + phy).
Calculate its average energy.

Answer 1

$E=K+U=0.5m(\frac{dx}{dt})^2+0.5kx^2$

$\frac{dx}{dt}=-a_o \exp{(-bt)} [b \cos{(w_d t + \phi)}+w_d \sin{(w_d t + \phi)}]$

Thus

$K=0.5ma_0^2e^{-2bt} [b^2 \cos^2{(w_d t + \phi)}+w_d^2 \sin^2{(w_d t + \phi)}+bw_d \sin{2(w_d t + \phi)}]$

We have:

$k=mw_d^2$

$U=0.5ma_0^2w_d^2 e^{-2bt} \cos^2{(w_d t + \phi)}$ $\langle E \rangle=0.5ma_0^2 e^{-2bt} \langle A(t) \rangle$

$A(t)=(b^2+w_d^2) \cos^2{(w_d t + \phi)}+w_d^2 \sin^2{(w_d t + \phi)}+bw_d \sin{2(w_d t + \phi)}$

$\langle A(t) \rangle=0.5((b^2+w_d^2)+w_d^2)$

As for weakly damped harmonic oscillator

$b\ll w_d$

$\langle A(t) \rangle=w_d^2$

$\langle E \rangle=0.5ma_0^2w_d^2 e^{-2bt} =E_0e^{-2bt}$