A simple harmonic oscillator of angular frequency $2 \mathrm{rad} {\mathrm{s}}^{-1}$ 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:

A pendulum with the time period of $1 \mathrm{s}$ is losing energy due to damping. At a certain time, its energy is $45 \mathrm{J}$. If after completing $15$ oscillations its energy has become $15 \mathrm{J}$, then its damping constant (in ${\mathrm{s}}^{-1}$) will be

Match List$‐\mathrm{I}$ (Event) with List$‐\mathrm{II}$ (Order of the time interval for the happening of the event) and select the correct option from the options given below the lists.

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