For a particle performing $S.H.M.$ given by the equation $y = 2 \sin \left( \frac{\pi t}{2} + \phi \right) \, (cm)$,what is the maximum acceleration of the particle?

$A$ point mass oscillates along the $x$-axis according to $x = x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$. If the acceleration of the point mass is written as $a = A \sin (\omega t + \delta)$,then:

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 $\,ms^{-2}$)

$A$ piston is performing $S.H.M.$ in the vertical direction with a frequency of $0.5 \,Hz$. $A$ block of $10 \,kg$ is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is: (in $\,m$)

The maximum force acting on a particle executing simple harmonic motion is $10 \,N$. The force on the particle when it is midway between mean and extreme positions will be

$A$ horizontal platform with a small object placed on it executes a linear $S.H.M.$ in the vertical direction. The amplitude of oscillation is $40 \text{ cm}$. What should be the least period of these oscillations, so that the object is not detached from the platform (in $\pi \text{ s}$)? [Take $g = 10 \text{ m/s}^2$]