Answer to Question #103276 in Mechanics | Relativity for sakshi

Question

Show that the average energy of a weakly damped oscillator is given by:
< E > = E0 exp (-2bt)

Answer 1

The energy of the system is

$$E(t)=KE(t)+U(t)=\frac{1}{2}m\bigg(\frac{dx}{dt}\bigg)^2+\frac{1}{2}kx^2.$$

The instantaneous displacement of a weakly damped harmonic oscillator is

$$x (t) = a_0 \text{exp} (- bt)\cdot \text{cos} (\omega_dt + \phi),\\ \space\\ \frac{dx(t)}{dt}= - a_0 \text{exp} (- bt) [b \text{cos} (\omega_dt + \phi) + \omega_d\text{sin} (\omega_dt + \phi)],\\ \space\\ KE(t)=\frac{1}{2}ma_0^2 \text{exp}(-2bt) [b\text{cos} (\omega_dt + \phi) + \omega_d\text{sin} (\omega_dt + \phi)]^2$$

(open the squared parentheses).

The potential energy is

$$U=\frac{1}{2}kx^2=\\ \space\\ =\frac{1}{2}ka_0^2 \text{exp} (-2bt)\cdot \text{cos}^2 (\omega_dt + \phi).$$

Remember that

$$k=m\omega_0^2.$$

$$U=\frac{1}{2}kx^2=\\ \space\\ =\frac{1}{2}m\omega_0^2a_0^2 \text{exp} (-2bt)\cdot \text{cos}^2 (\omega_dt + \phi).$$

During one oscillation, for small amplitudes, the average values of sine and cosine squared are equal

$$\text{sin}^2 (\omega_dt + \phi)=\text{cos}^2 (\omega_dt + \phi)=\frac{1}{2}.$$

Remember that the average values of sine per one oscillation is 0.

Therefore, after adding the KE with opened parentheses and U, we get

$$<E>=\frac{1}{2}ma_0^2 \text{exp}(-2bt)\bigg[\frac{b^2+\omega_0^2}{2}+\frac{\omega_d^2}{2}\bigg]=\\ \space\\ =\frac{1}{2}ma_0^2\omega^2_0\text{exp}(-2bt).$$

Since

$$\frac{1}{2}ma_0^2\omega^2_0=E_0,$$

we can finally write

$$<E>=E_0\text{exp}(-2bt).$$