In forced oscillations,a particle oscillates simple harmonically with a frequency equal to

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 $\%$)

The time taken for the amplitude of vibrations of a damped oscillator to drop to $25 \%$ of its initial value is $t$. The time taken for its mechanical energy to drop to $50 \%$ of its initial mechanical energy is

In damped oscillation,the graph between velocity $(V)$ and position $(x)$ will be:

The mechanical energy of a damped oscillator becomes half of its initial energy in $4 \ s$. In another $t \ s$,its mechanical energy becomes $12.5 \%$ of its initial mechanical energy. Then $t=$ (in $s$)

$A$ particle of mass $m$ is attached to a spring (of spring constant $k$) and has a natural angular frequency $\omega_0$. An external force $F(t)$ proportional to $\cos \omega t$ (where $\omega \neq \omega_0$) is applied to the oscillator. The displacement of the oscillator will be proportional to: