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 $\%$)

$A$ spring-mass system executes damped harmonic oscillations given by the equation $y = A e^{-\frac{bt}{2m}} \sin(\omega' t + \phi)$,where the symbols have their usual meanings. If a $2 \ kg$ mass $(m)$ is attached to a spring of force constant $(K) = 1250 \ N/m$,the period of the oscillation is $(\pi / 12) \ s$. The damping constant $b$ has the value ..... $kg/s$.

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

Which of the diagrams shown in the figure represents the variation of the total mechanical energy of a pendulum oscillating in water as a function of time?

The amplitude of a damped harmonic oscillator becomes $50 \%$ of its initial value in a time of $12 \ s$. If the amplitude of the oscillator at a time of $36 \ s$ is $x \%$ of its initial amplitude,then the value of $x$ 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.