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).

$A$ $3 \ kg$ sphere dropped through air has a terminal speed of $25 \ m/s$. (Assume that the drag force is $F_d = -bv$). Now suppose the sphere is attached to a spring of force constant $k = 300 \ N/m$,and that it oscillates with an initial amplitude of $20 \ cm$. What is the angular frequency of its damped $SHM$? ..... $rad/s$

$A$ block of mass $1 \, kg$ attached to a spring is made to oscillate with an initial amplitude of $12 \, cm$. After $2 \, minutes$ the amplitude decreases to $6 \, cm$. Determine the value of the damping constant $b$ for this motion. (Take $\ln 2 = 0.693$)

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

The amplitude of vibration of a particle is given by $a_m = \frac{a_0}{a\omega^2 - b\omega + c}$,where $a_0, a, b,$ and $c$ are positive constants. The condition for a single resonant frequency is:

$A$ simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to