The acceleration of a particle performing $S.H.M.$ at a distance of $3\;cm$ from the mean position is $12\;cm/s^2$. Its time period is ..... $s$.

Under the action of a force $F = -75 y$, where $F$ is in Newton and $y$ is in meters, an object of mass $3 \,kg$ executes simple harmonic motion. If the velocity of the object at the mean position is $2.5 \,ms^{-1}$, the maximum acceleration of the object is (in $\,ms^{-2}$)

$A$ $0.10\, kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x = (10\,cm)\cos [(10\,rad/s)\,t + \pi /2\,rad]$. What is the maximum acceleration experienced by the block?

$A$ particle is performing $S.H.M.$ starting from the extreme position. The graphical representation shows that between displacement and acceleration,there is a phase difference of:

For a particle executing $SHM$, determine the ratio of the average acceleration of the particle between the extreme position and the equilibrium position with respect to the maximum acceleration.

The displacement of a particle varies with time as $x = 12 \sin \omega t - 16 \sin^3 \omega t$ (in $cm$). If its motion is $S.H.M.$,then its maximum acceleration is