The amplitude of a damped oscillator is known to decrease to $0.9$ times its original amplitude in $5 \,s$. Approximately, by how many times its original amplitude will it decrease after another $20 \,s$?

$A$ particle of mass $m$ is attached to a spring (of spring constant $k$) and has a natural angular frequency $\omega_0$. An external force $F(t)$ proportional to $\cos \omega t$ (where $\omega \neq \omega_0$) is applied to the oscillator. The displacement of the oscillator will be proportional to:

Resonance is an example of

The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)$ of its original amplitude in $2 \ s$. If its amplitude after $6 \ s$ becomes $\left(\frac{1}{n}\right)$ times the original amplitude,the value of $n$ is ($n$ is a non-zero integer).

$A$ $3 \ kg$ sphere dropped through air has a terminal speed of $25 \ m/s$. (Assume that the drag force is $F_d = -bv$). Now suppose the sphere is attached to a spring of force constant $k = 300 \ N/m$,and that it oscillates with an initial amplitude of $20 \ cm$. What is the angular frequency of its damped $SHM$? ..... $rad/s$

$A$ 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