$A$ body executing simple harmonic motion has a maximum acceleration equal to $24 \, m/s^2$ and a maximum velocity equal to $16 \, m/s$. The amplitude of the simple harmonic motion is:

$A$ horizontal platform with an object placed on it is executing $S.H.M.$ in the vertical direction. The amplitude of oscillation is $3.92 \times 10^{-3} \, m$. What must be the least period of these oscillations,so that the object is not detached from the platform (in $, s$)?

The displacement-time graph of a particle executing $S.H.M.$ is given in the figure: (sketch is schematic and not to scale). Which of the following statements is/are true for this motion?
$(A)$ The force is zero at $t = \frac{3T}{4}$
$(B)$ The acceleration is maximum at $t = T$
$(C)$ The speed is maximum at $t = \frac{T}{4}$
$(D)$ The $P.E.$ is equal to $K.E.$ of the oscillation at $t = \frac{T}{2}$

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.

$Assertion :$ In simple harmonic motion,the velocity is maximum when the acceleration is minimum.
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$.

Two particles $P$ and $Q$ describe simple harmonic motions of same period,same amplitude,along the same line about the same equilibrium position $O.$ When $P$ and $Q$ are on opposite sides of $O$ at the same distance from $O,$ they have the same speed of $1.2 \, m/s$ in the same direction. When their displacements are the same,they have the same speed of $1.6 \, m/s$ in opposite directions. The maximum velocity in $m/s$ of either particle is