The amplitude of a damped oscillator becomes half in $1$ minute. The amplitude after $3$ minutes will be $\frac{1}{x}$ times the original. Then $x$ 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:

What are damped oscillations? Discuss them using the illustration of a spring.

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

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 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