The amplitude of vibration of a particle is given by $a_m = \frac{a_0}{a\omega^2 - b\omega + c}$,where $a_0, a, b,$ and $c$ are positive constants. The condition for a single resonant frequency is:

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).

The value of maximum possible amplitude in the case of forced oscillations when the driving frequency is close to the natural frequency is:

Explain the behaviour of the oscillator when the driving frequency is far from the natural frequency in small damped oscillations.

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

$A$ block of mass $100 \,g$ is connected to an elastic spring of spring constant $450 \,N m^{-1}$ and oscillates vertically. The block-spring system is in a viscous surrounding medium with a damping constant $b = 69.3 \,g \,s^{-1}$. Calculate the time in which the amplitude of oscillations drops to half of its initial value. (Take $\ln 2 = 0.693$) (in $\,s$)