alleryStart learning →In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, the **amplitude decays** as:
- Linearly with time
- Exponentially: $A(t) = A_0 e^{-\gamma t}$
- Inversely with time
- Independently of damping
Show answer and solution
Correct 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 $\gamma$.