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

$A$ 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$)

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).

$A$ spring executes $S$.$H$.$M$. with mass $10 \text{ kg}$ attached to it. The force constant of spring is $10 \text{ N/m}$. If at any instant its velocity is $40 \text{ cm/s}$, the displacement at that instant is (Amplitude of $S$.$H$.$M$. is $0.5 \text{ m}$) (in $\text{ m}$)

$A$ particle executes simple harmonic motion with an amplitude of $4 \, cm$. At the mean position,the velocity of the particle is $10 \, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \, cm/s$ is

Show that for a particle executing $SHM$,velocity and displacement have a phase difference of $\frac{\pi}{2}$.