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 performing $S.H.M.$ about $x = 0$, with an amplitude $a$ and time period $T$. The speed of the particle at $x = \frac{a}{3}$ will be

The velocity of a particle performing simple harmonic motion,when it passes through its mean position is

$A$ particle is executing simple harmonic motion with amplitude $A$. At a distance $x$ from the mean position,when the particle is moving towards the extreme position,it receives a blow in the direction of motion which instantaneously doubles its velocity. What is the new amplitude of the particle? (Frequency remains constant during the motion)

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

The displacement of a particle executing $SHM$ is given by $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$,where $x$ is in $m$ and $t$ is in $s$. The amplitude and maximum speed of the particle are: