The amplitudes of a damped harmonic oscillator after $2 \ s$ and $4 \ s$ are $A_1$ and $A_2$ respectively. If the initial amplitude of the oscillator is $A_0$,then

$A$ particle with a restoring force proportional to displacement and a resisting force proportional to velocity is subjected to a driving force $F \sin \omega t$. If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$,then (where $\omega_0$ is the natural frequency of oscillation of the particle):

The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5 \ s$. In another $10 \ s$ it will decrease to $\alpha$ times its original magnitude,where $\alpha$ equals

The mechanical energy of a damped oscillator becomes half of its initial energy in $4 \ s$. In another $t \ s$,its mechanical energy becomes $12.5 \%$ of its initial mechanical energy. Then $t=$ (in $s$)

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

$A$ $3 \ kg$ 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 $SHM$? ..... $rad/s$