$A$ particle is performing $SHM$ with a path length of $8 \ cm$. If it takes $1 \ s$ to reach from the mean position to the extreme position,what is its velocity at the equilibrium position?

$A$ 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 $0.4 \,m$)

The maximum velocity of a simple harmonic motion represented by $y = 3\sin \left( 100t + \frac{\pi}{6} \right)$ is given by

$A$ 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:

The force ($F$ in newton) acting on a particle of mass $90 \text{ g}$ executing simple harmonic motion is given by $F + 0.04 \pi^2 y = 0$, where $y$ is the displacement of the particle in meters. If the amplitude of the particle is $\frac{6}{\pi} \text{ m}$, then the maximum velocity of the particle is: (in $\text{ m/s}$)

$A$ particle executing $S.H.M.$ starts from the mean position. Its amplitude is $A$ and time period is $T$. At what displacement is its speed one-fourth of the maximum speed?