If a particle under $S.H.M.$ has a time period of $0.1 \ s$ and an amplitude of $2 \times 10^{-3} \ m$,what is its maximum velocity?

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

In an angular $SHM$,the angular amplitude of oscillation is $\pi \, rad$ and the time period is $0.4 \, s$. Calculate its angular velocity at an angular displacement of $\pi/2 \, rad$. (Result in $rad/s$)

$A$ particle is executing simple harmonic motion $\text{(S.H.M.).}$ Its acceleration at a distance of $1 \ cm$ from the mean position is $3 \ cm s^{-2}$. If its velocity is $6 \ cm s^{-1}$ when it is at a distance of $2 \ cm$ from its mean position, then the amplitude of $\text{S.H.M.}$ is,

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

$A$ particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium position is at $x = 0$,the force constant is $K$,and the amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be: