The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5 \ s$. In another $10 \ s$ it will decrease to $\alpha$ times its original magnitude,where $\alpha$ equals

The mechanical energy of a damped oscillator becomes half of its initial energy in $4 \ s$. In another $t \ s$,its mechanical energy becomes $12.5 \%$ of its initial mechanical energy. Then $t=$ (in $s$)

$A$ 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

In forced oscillations,a particle oscillates simple harmonically with a frequency equal to

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

If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0 \ s$ to $t = \tau \ s$,then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality,the average life time of the pendulum is (assuming damping is small) in seconds: