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 . . . . . .

$A$ point mass of $400 \, g$ executes $S.H.M.$ under a force $F = -(10 \, N m^{-1}) x$. If it crosses the centre of oscillation with a speed of $10 \, m s^{-1}$, the amplitude of motion is (in $ \, m$)

$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.

Two $S$.$H$.Ms. are represented by equations $y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$ and $y_2 = 0.1 \cos (100 \pi t)$. The phase difference between the speeds of the two particles is

In case of a simple harmonic motion,if the velocity is plotted along the $X$-axis and the displacement (from the equilibrium position) is plotted along the $Y$-axis,the resultant curve is an ellipse with the ratio:
Major axis (along $X$) $= 20 \pi \times$ Minor axis (along $Y$).
What is the frequency of the simple harmonic motion?