In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, the **amplitude decays** as:
ALinearly with time
BExponentially: $A(t) = A_0 e^{-\gamma t}$
CInversely with time
DIndependently of damping
Answer & 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 $\gamma$.
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