The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $x/7 \text{ s}$. The value of $x$ is (Take $\pi = 22/7$)

$A$ particle is exhibiting simple harmonic motion and has its displacement $x$ and velocity $v$ related as $4v^2 = 25 - x^2$. The time period of the $SHM$ is

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 . . . . . .

$A$ particle performing $SHM$ has a time period of $\frac{2 \pi}{\sqrt{3}} \,s$ and a path length of $4 \,cm$. The displacement from the mean position at which the magnitude of acceleration is equal to the magnitude of velocity is (in $\,cm$)

$A$ particle is executing a linear simple harmonic motion. Let $V_1$ and $V_2$ be its speeds at distances $x_1$ and $x_2$ from the equilibrium position,respectively. The amplitude of oscillation is

The kinetic energy of a particle executing $S.H.M.$ is $16 \, J$ when it is at its mean position. If the mass of the particle is $0.32 \, kg$,then what is the maximum velocity of the particle in $m/s$?