$A$ 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 kinetic energy of a simple harmonic oscillator is oscillating with an angular frequency of $176 \ rad/s$. The frequency of this simple harmonic oscillator is . . . . . . $Hz$. $\left[\text{take } \pi=\frac{22}{7}\right]$

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

$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion is

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).