In a time of $2 \ s$,the amplitude of a damped oscillator becomes $\frac{1}{e}$ times its initial amplitude $A$. In the next two seconds,the amplitude of the oscillator is

$Assertion :$ Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.
$Reason :$ The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force.

The displacement of a damped oscillator is $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,where $t$ is time in seconds. The time required for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is (in $s$)

$A$ simple pendulum oscillates in air with time period $T$ and amplitude $A$. As the time passes,

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 $\ln 2 = 0.7$)

If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0 \ s$ to $t = \tau \ s$,then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality,the average life time of the pendulum is (assuming damping is small) in seconds: