Answer to Question #140294 in Physics for shubham

Question

1.    Calculate the damping constant of an under damped harmonic oscillator, if its amplitude reduces to (1/10)^th of its initial value after 100 oscillations. Given that time period is 2 s.

Answer 1

The amplitude of a under damped harmonic oscillations is proportional to:

$$A\propto \exp(-\omega\zeta t)$$

where $\zeta$ is the damping constant, $t$ is a time and $\omega = \dfrac{2\pi}{2s} = \pi\space rad/s$ is the angular frequency. If the periond of oscillation is 2s then 100 oscillations takes $\Delta t = 100\times 2s = 200s$. In this time the amplitude reduces to (1/10)^th of its initial value:

$$\dfrac{A_2}{A_1} = \exp(-\omega\zeta (t_2-t_1)) = \exp(-\omega\zeta \Delta t) = \dfrac{1}{10}$$

Expressing $\zeta$, obtain:

$$\zeta = \dfrac{\ln10}{\omega \Delta t} = \dfrac{\ln10}{\pi\cdot 200} \approx 0.0037$$

Answer. 0.0037.