$A$ particle executes $S.H.M.$ with a period of $6 \, s$ and amplitude of $3 \, cm$. Its maximum speed in $cm/s$ is

$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation is

$A$ particle is executing a linear simple harmonic motion. Let $V_1$ and $V_2$ be its speeds at distances $x_1$ and $x_2$ from the equilibrium position,respectively. The amplitude of oscillation is

The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position,the speed of the particle is $16 \,cm/s$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3} \,cm/s$ will be .... $cm$.

$A$ spring executes $S.H.M.$ with a mass of $1 \,kg$ attached to it. The force constant of the spring is $4 \,N/m$. If at any instant its velocity is $20 \,cm/s$, the displacement at that instant is (Amplitude of $S.H.M.$ is $0.4 \,m$)