The amplitude of a mass-spring system,which is executing simple harmonic motion,decreases with time. If mass $m = 500 \, g$,and the decay constant $b = 20 \, g/s$,then how much time $t$ (in seconds) is required for the amplitude of the system to drop to half of its initial value? (Given $\ln 2 = 0.693$)

If the amplitudes of a damped harmonic oscillator at times $t=0, t_1$ and $t_2$ are $A_0, A_1$ and $A_2$ respectively, then the amplitude of the oscillator at a time of $(t_1+t_2)$ is

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

$A$ body of mass $0.3 \ kg$ hangs by a spring with a force constant of $50 \ N/m$. The amplitude of oscillations is damped and reaches $1/e$ of its original value in about $100$ oscillations. If $\omega$ and $\omega^{\prime}$ are the angular frequencies of undamped and damped oscillations respectively, then the percentage value of $\left(\frac{\omega-\omega^{\prime}}{\omega}\right) \times 100$ is:

Define: Natural oscillations,free oscillations,and forced oscillations.

Obtain the differential equation of forced oscillation.