The amplitudes of a damped harmonic oscillator after $2 \ s$ and $4 \ s$ are $A_1$ and $A_2$ respectively. If the initial amplitude of the oscillator is $A_0$,then

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 half in $1 \text{ minute}$. The amplitude after $3 \text{ minutes}$ will be $\frac{1}{X}$ times the original,where $X$ is

If the amplitude of a lightly damped oscillator decreases by $1.5 \%$,then the mechanical energy of the oscillator lost in each cycle is: (in $\%$)

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 amplitude of a damped oscillator varies with time as $A(t) = A_0 \exp(-bt / 2m)$, where $b = 70 \text{ g/s}$ and $m = 200 \text{ g}$. How long does it take for the mechanical energy to drop to one-fourth of its initial value (in $s$)? (Take $\ln 2 = 0.7$)