The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

$A$ particle executes $S.H.M.$,the graph of velocity as a function of displacement is

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

$A$ particle is executing simple harmonic motion $\text{(S.H.M.).}$ Its acceleration at a distance of $1 \ cm$ from the mean position is $3 \ cm s^{-2}$. If its velocity is $6 \ cm s^{-1}$ when it is at a distance of $2 \ cm$ from its mean position, then the amplitude of $\text{S.H.M.}$ is,

$Assertion :$ 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.

In an angular $SHM$,the angular amplitude of oscillation is $\pi \, rad$ and the time period is $0.4 \, s$. Calculate its angular velocity at an angular displacement of $\pi/2 \, rad$. (Result in $rad/s$)