$A$ particle performing linear $S.H.M.$ of amplitude $0.1 \,m$ has displacement $0.02 \,m$ and acceleration $0.5 \,m/s^2$. The maximum velocity of the particle in $m/s$ is

The graph in the figure represents:

$A$ 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

$A$ particle executes $S.H.M.$ of period $\frac{2 \pi}{\sqrt{3}} \text{ s}$ along a straight line $4 \text{ cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is (in $\text{ cm}$)

The time period of a particle in simple harmonic motion is $8 \ s$. At $t=0$, it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is

The displacement of a body executing $SHM$ is given by $x = A \sin (2\pi t + \pi /3).$ The first time from $t = 0$ when the velocity is maximum is .... $\sec$