$A$ 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?

An object undergoing simple harmonic motion takes $0.5 \text{ s}$ to travel from one point of zero velocity to the next such point. The angular frequency of the motion 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 .... $cm$.

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 $\text{ m/s}$)

From the given displacement-time graph of an oscillating particle,find the maximum velocity of the particle.

What is the maximum velocity of a simple harmonic motion with an amplitude of $2 \ m$ and a time period of $2 \ s$?