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 amplitude of a mass-spring system,which is executing simple harmonic motion,decreases with time. If mass $m = 500 \, g$,and the decay constant $b = 20 \, g/s$,then how much time $t$ (in seconds) is required for the amplitude of the system to drop to half of its initial value? (Given $\ln 2 = 0.693$)

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:

$A$ damped harmonic oscillator has a frequency of $5$ oscillations per second. The amplitude drops to half its value for every $10$ oscillations. The time it will take to drop to $\frac{1}{1000}$ of the original amplitude is close to .... $s$

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.

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