The amplitude of a damped oscillator varies with time as $A(t) = A_0 \exp(-bt / 2m)$, where $b = 70 \text{ g/s}$ and $m = 200 \text{ g}$. How long does it take for the mechanical energy to drop to one-fourth of its initial value (in $s$)? (Take $\ln 2 = 0.7$)

For a damped harmonic oscillator governed by the equation $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$, find the time $t$ after which the mechanical energy becomes half of its initial maximum value.

The amplitude of a simple pendulum,oscillating in air with a small spherical bob,decreases from $10 \ cm$ to $8 \ cm$ in $40 \ s$. Assuming that Stokes' law is valid,and the ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$. The time in which the amplitude of this pendulum will reduce from $10 \ cm$ to $5 \ cm$ in carbon dioxide will be close to ..... $s$ $(\ln 5 = 1.601, \ln 2 = 0.693)$

In case of a forced vibration,the resonance wave becomes very sharp when the:

What is the amplitude of a damped oscillator if time becomes $t = 2m$?

In damped oscillation,the graph between velocity $(V)$ and position $(x)$ will be: