For a particle executing simple harmonic motion $(SHM)$,at its mean position

$A$ particle is performing $S.H.M.$ with a maximum velocity $V$. If the amplitude is doubled and the periodic time is reduced to $\left(\frac{1}{3}\right)^{\text{rd}}$ of its original value,then the new maximum velocity is:

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times 10^{-2} \cos(\pi t) \text{ m}$. The time at which the maximum speed first occurs is (in $\text{ s}$)

$A$ particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium position is at $x = 0$,the force constant is $K$,and the amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be:

If displacement $x$ and velocity $v$ are related as $4v^2 = 16 - x^2$ in a $SHM$,then the time period of the given $SHM$ is (consider $SI$ units).

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .