The acceleration of a particle executing $SHM$ is

For a particle in $SHM$,if the amplitude of the displacement is $a$ and the amplitude of velocity is $v$,the amplitude of acceleration is

$A$ point mass oscillates along the $X$-axis according to the law $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$. If the acceleration of the particle is written as $a=A \cos (\omega t-\delta)$, then

What is the maximum acceleration of a particle performing $SHM$ given by the equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$,where $y$ is in $cm$?

The displacement of a particle performing $S.H.M.$ is given by $x=5 \sin (3 t+3)$,where $x$ is in $cm$ and $t$ is in $s$. The maximum acceleration of the particle will be (in $cm \ s^{-2}$)

$A$ 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: