In a time of $2 \ s$,the amplitude of a damped oscillator becomes $\frac{1}{e}$ times its initial amplitude $A$. In the next two seconds,the amplitude of the oscillator is

For a damped harmonic oscillator governed by the equation $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$, find the time $t$ after which the mechanical energy becomes half of its initial maximum value.

The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)$ of its original amplitude in $2 \ s$. If its amplitude after $6 \ s$ becomes $\left(\frac{1}{n}\right)$ times the original amplitude,the value of $n$ is ($n$ is a non-zero integer).

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 $s$)

Write the expression for the mechanical energy of a damped oscillator.

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