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 $1/e^{2\pi}$.
Related questions
A particle of mass 0.2 kg performs S.H.M. of amplitude 0.05 m with period π s. Its maximumIn damped harmonic motion, the amplitude of oscillationTwo springs of force constants k₁ and k₂ are connected in series and stretched by an exterA particle in S.H.M. of period 2 s starts from rest at the extreme position. Its displacemThe kinetic energy of a particle of mass m in linear S.H.M. of amplitude A at displacementA simple pendulum of length 1 m on the earth's surface (g = π² m/s²) has periodA spring-mass system of mass m and spring constant k oscillates with periodA particle executes S.H.M. with amplitude A and angular frequency ω. Its maximum accelerat