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

The amplitude of a mass-spring system,which is executing simple harmonic motion,decreases with time. If mass $m = 500 \, g$,and the decay constant $b = 20 \, g/s$,then how much time $t$ (in seconds) is required for the amplitude of the system to drop to half of its initial value? (Given $\ln 2 = 0.693$)

$A$ block of mass $100 \,g$ is connected to an elastic spring of spring constant $450 \,N m^{-1}$ and oscillates vertically. The block-spring system is in a viscous surrounding medium with a damping constant $b = 69.3 \,g \,s^{-1}$. Calculate the time in which the amplitude of oscillations drops to half of its initial value. (Take $\ln 2 = 0.693$) (in $\,s$)

$A$ 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

If a damped oscillator is very heavily damped,what is its frequency?

Which of the following quantities does $NOT$ change due to the damping of oscillations?