The graph between velocity and position for a damped oscillation will be:

$A$ 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 ..... $kg/s$.

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

The displacement of a damped harmonic oscillator is given by $x(t) = e^{-0.1 t} \cos(10 \pi t + \varphi)$. Here $t$ is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to: (in $s$)

Explain the behaviour of the oscillator when the driving frequency is far from the natural frequency in small damped oscillations.

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