$A$ 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 ...... $mm/s$.

For a simple harmonic motion given by the equation $x = 5 \sin (4t - \frac{\pi}{6}) \, m$,what is the velocity when its displacement is $3 \, m$ (in $, m/s$)?

$A$ particle is executing $S.H.M.$ If its amplitude is $2 \, m$ and periodic time is $2 \, s$,then the maximum velocity of the particle will be:

For a particle executing $S.H.M.$ having amplitude $A$,the speed of the particle is $\left(\frac{1}{3}\right)^{rd}$ of its maximum speed when the displacement from the mean position is

$A$ particle performs linear $S.H.M.$ When the displacement of the particle from the mean position is $3 \ cm$ and $4 \ cm$,the corresponding velocities are $8 \ cm/s$ and $6 \ cm/s$ respectively. Its periodic time is

$A$ particle is exhibiting simple harmonic motion and has its displacement $x$ and velocity $v$ related as $4v^2 = 25 - x^2$. The time period of the $SHM$ is