In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, t…

Question

In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, the **amplitude decays** as:

Accepted Answer

B. Exponentially: $A(t) = A_0 e^{-\gamma t}$

For under-damped case ($\gamma < \omega_0$): $x(t) = A_0 e^{-\gamma t}\cos(\omega' t + \phi)$ where $\omega' = \sqrt{\omega_0^2 - \gamma^2}$.
Amplitude envelope decays exponentially with rate $\gamma$.

Answer

Linearly with time

Answer

Inversely with time

Answer

Independently of damping

In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, the amplitude decays as:

Keep practising Unit 1A: Kinematics: 1D Motion, Position & Velocity