The displacement of a particle moving in $S.H.M.$ at any instant is given by $y = a \sin \omega t$. The acceleration after time $t = \frac{T}{4}$ is (where $T$ is the time period).

$A$ $0.10\, kg$ 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:

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 $\,ms^{-2}$)

$A$ coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

The variation of displacement with time of a simple harmonic motion $(SHM)$ for a particle of mass $m$ is represented by $y = 2 \sin \left(\frac{\pi t}{2} + \phi\right) \text{ cm}$. The maximum acceleration of the particle is