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:

$A$ block of mass $200 \, g$ is executing $SHM$ under the influence of a spring with spring constant $K = 90 \, N \, m^{-1}$ and a damping constant $b = 40 \, g \, s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $\ln \frac{1}{2} = -0.693$).

In a time of $2 \ s$,the amplitude of a damped oscillator becomes $\frac{1}{e}$ times its initial amplitude $A$. In the next two seconds,the amplitude of the oscillator is

For the damped oscillator shown in the figure,the mass $m$ of the block is $200 \; g$,$k = 90 \; N m^{-1}$,and the damping constant $b$ is $40 \; g s^{-1}$. Calculate:
$(a)$ the period of oscillation,
$(b)$ the time taken for its amplitude of vibrations to drop to half of its initial value,and
$(c)$ the time taken for its mechanical energy to drop to half its initial value.

What is the amplitude of a damped oscillator if time becomes $t = 2m$?

If a damped oscillator is very heavily damped,what is its frequency?