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

Write the equation of angular frequency and amplitude for damped oscillation.

$A$ 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

The amplitude of a damped oscillator becomes half in $1 \text{ minute}$. The amplitude after $3 \text{ minutes}$ will be $\frac{1}{X}$ times the original,where $X$ is

The amplitude of a damped oscillator is known to decrease to $0.9$ times its original amplitude in $5 \,s$. Approximately, by how many times its original amplitude will it decrease after another $20 \,s$?

The amplitude of a simple pendulum,oscillating in air with a small spherical bob,decreases from $10 \ cm$ to $8 \ cm$ in $40 \ s$. Assuming that Stokes' law is valid,and the ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$. The time in which the amplitude of this pendulum will reduce from $10 \ cm$ to $5 \ cm$ in carbon dioxide will be close to ..... $s$ $(\ln 5 = 1.601, \ln 2 = 0.693)$