$A$ simple pendulum oscillates in air with time period $T$ and amplitude $A$. As the time passes,

The amplitude of a damped oscillator becomes one third in $2 \, s$. If its amplitude after $6 \, s$ is $1/n$ times the original amplitude,then the value of $n$ is

Two damped spring-mass oscillating systems have identical spring constants and decay times. However,system $A$'s mass $m_A$ is twice system $B$'s mass $m_B$. How do their damping constants,$b$,compare?

$A$ block of mass $m$ is connected to a light spring of force constant $k$. The system is placed inside a damping medium of damping constant $b$. The instantaneous values of displacement,acceleration and energy of the block are $x, a$ and $E$ respectively. The initial amplitude of oscillation is $A$ and $\omega^{\prime}$ is the angular frequency of oscillations. The incorrect expression related to the damped oscillations is

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:

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