# Lightly Damped Harmonic Oscillator

1. Jun 19, 2005

### e(ho0n3

Question:
(a) Show that the total mechanical energy of a lightly damped harmonic oscillator is

$$E = E_0 e^{-bt/m}$$

where $E_0$ is the total mechanical energy at t = 0.

(b) Show that the fractional energy lost per period is

$$\frac{\Delta E}{E} = \frac{2 \pi b}{m \omega_0} = \frac{2 \pi}{Q}$$

where $\omega_0 = \sqrt{k/m}$ and $Q = m \omega_0 / b$ is called the quality factor or Q value of the system. A larger Q value means the system can undergo oscillations for a longer time.

(a) When the velocity of the oscillator is 0, the total mechanical energy is purely potential energy, $U = 1/2kx^2$. Since I know that $x = Ae^{-bt/(2m)}\cos{\omega't}$ where t is some multiple of $2\pi/\omega'$, then

$$E = \frac{1}{2}kA^2e^{-bt/m}$$

and $E_0 = 1/2kA^2$. Of course, this is only valid when the velocity of the oscillator is 0, but since it is lightly damped the total mechanical energy should be approximately the same when the velocity is > 0. Right?

(b) Using some calculus, I can timidly state that

$$\frac{\Delta E}{\Delta t} \approx \frac{dE}{dt} = -\frac{E_0b}{m}e^{bt/m}$$

and since $\Delta t = 2\pi / \omega'$ then

$$\frac{\Delta E}{E} = -\frac{b\Delta t}{m} = -\frac{2\pi b}{m\omega'}$$

Since the oscillator is lightly damped, $\omega' \approx \omega_0$. However the result I get is negative. Should it be negative?

2. Jun 19, 2005

### OlderDan

Yes, it should be negative because you calculated the rate of change of energy, which is decreasing with time. The question asked for the fractional energy loss, which is the absolute value of the energy change.

Your calculation of the times at which the energy is all potential is overlooking the fact that the peaks in x do not correspond to the points where the cosine has value 1 because the exponential is also time dependent. However, the times between peaks still satisfy the condition you used, so the result follows subject to the other approximations you made.