For a particle executing simple harmonic motion given by $y(cm) = \sin \frac{\pi}{2} \left( \frac{t}{2} + \frac{1}{3} \right)$,what is the maximum acceleration in $cm/sec^2$?

Give the direction of velocity and acceleration of an $SHM$ particle.

$A$ particle is executing simple harmonic motion with an amplitude of $0.1 \, m$. At a certain instant when its displacement is $0.02 \, m$,its acceleration is $0.5 \, m/s^2$. The maximum velocity of the particle is (in $m/s$):

$A$ particle executes $SHM$ on a straight line path. The amplitude of oscillation is $2 \, cm$. When the displacement of the particle from the mean position is $1 \, cm$,the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of velocity. The frequency of $SHM$ (in $s^{-1}$) is:

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 $\,ms^{-2}$)

Two simple harmonic motions of angular frequency $100 \, rad \, s^{-1}$ and $1000 \, rad \, s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is