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

$A$ simple harmonic oscillator of angular frequency $\omega = 2 \, rad \, s^{-1}$ is acted upon by an external force $F = \sin(t) \, N$. If the oscillator is at rest in its equilibrium position at $t = 0$,its position at later times is proportional to

When an oscillator completes $100$ oscillations,its amplitude is reduced to $\frac{1}{3}$ of its initial value. What will be its amplitude after it completes $200$ oscillations?

The amplitude of a damped harmonic oscillator becomes $50 \%$ of its initial value in a time of $12 \ s$. If the amplitude of the oscillator at a time of $36 \ s$ is $x \%$ of its initial amplitude,then the value of $x$ 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.

$A$ block of mass $200 \, g$ is executing $SHM$ under the influence of a spring with spring constant $K = 90 \, N \, m^{-1}$ and a damping constant $b = 40 \, g \, s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $\ln \frac{1}{2} = -0.693$).