The average acceleration of a particle performing $SHM$ over one complete oscillation is:

$A$ point mass oscillates along the $x$-axis according to the law $x = x_0 \cos(\omega t + \pi/4)$. If the acceleration of the particle is written as $a = A \cos(\omega t + \delta)$,then:

$A$ 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 $m/s$):

$A$ 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

If a particle is executing simple harmonic motion,then the acceleration of the particle:

$A$ particle executes harmonic motion with an angular velocity and maximum acceleration of $3.5\, rad/s$ and $7.5\, m/s^2$ respectively. The amplitude of oscillation is .... $m$