$A$ 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 $\ln \frac{1}{2} = -0.693$).

When an external force with angular frequency $\omega_d$ acts on a system of natural angular frequency $\omega$,the system oscillates with angular frequency $\omega_d$. The condition for the amplitude of oscillations to be maximum is

The displacement of a damped oscillator is $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,where $t$ is time in seconds. The time required for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is (in $s$)

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 $s$)

The angular frequency of the damped oscillator is given by $\omega = \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}}$,where $k$ is the spring constant,$m$ is the mass of the oscillator,and $r$ is the damping constant. If the ratio $\frac{r^2}{mk}$ is $8\%$,the change in time period compared to the undamped oscillator is approximately as follows:

$A$ block of mass $0.1\, kg$ is connected to an elastic spring of spring constant $640\, Nm^{-1}$ and oscillates in a damping medium of damping constant $10^{-2}\, kg\,s^{-1}$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value is closest to ..... $s$.