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

The amplitude of a damped oscillator becomes one third in $2 \, s$. If its amplitude after $6 \, s$ is $1/n$ times the original amplitude,then the value of $n$ is

$A$ 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

In damped $SHM$,the $SI$ unit of damping constant is

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

$A$ particle with a restoring force proportional to displacement and a resisting force proportional to velocity is subjected to a driving force $F \sin \omega t$. If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$,then (where $\omega_0$ is the natural frequency of oscillation of the particle):