$A$ 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 $\,s$)

Write the equation of angular frequency and amplitude for damped oscillation.

What is pure simple harmonic oscillation? Why is it not $100 \%$ possible in practice?

$A$ block of mass $m$ is connected to a light spring of force constant $k$. The system is placed inside a damping medium of damping constant $b$. The instantaneous values of displacement,acceleration and energy of the block are $x, a$ and $E$ respectively. The initial amplitude of oscillation is $A$ and $\omega^{\prime}$ is the angular frequency of oscillations. The incorrect expression related to the damped oscillations is

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 harmonic oscillator becomes half in $3 \ s$ and will become $1/x$ of the initial amplitude in the next $6 \ s$,where $x$ is: