In a damped oscillation, the amplitude decreases exponentially as $A(t) = A_0 e^{-\gamma t}$. The time after which the amplitude becomes $\dfrac{A_0}{e}$ is:
A$\gamma$
B$1/\gamma$
C$\gamma^2$
D$e^{\gamma}$
Answer & Solution
Correct answer: B. $1/\gamma$
Setting $A(t) = A_0/e$ in $A(t) = A_0 e^{-\gamma t}$:
$\dfrac{A_0}{e} = A_0 e^{-\gamma t} \Rightarrow e^{-1} = e^{-\gamma t} \Rightarrow \gamma t = 1 \Rightarrow t = \dfrac{1}{\gamma}$.
$1/\gamma$ is called the **characteristic decay time** of the damping. Energy decays at twice that rate ($E \propto A^2$), so energy falls to $1/e$ in time $1/(2\gamma)$.
The quality factor $Q = \omega_0/(2\gamma)$ ties this damping rate to the number of oscillations before energy drops to $1/e^{2\pi}$.
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