$A$ particle is performing simple harmonic motion. If the oscillations are damped oscillations,then the angular frequency is given by:

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

The displacement of a damped harmonic oscillator is given by $x(t) = e^{-0.1t} \cos(10\pi t + \varphi)$. The time taken for its amplitude of vibration to drop to half of its initial value is close to .... $s$

$A$ 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:

The displacement of a damped oscillator is $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,where $t$ is time in seconds. The time required for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is (in $s$)

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: