The maximum velocity of a particle performing simple harmonic motion is $6.28 \text{ cm s}^{-1}$. If the length of its path is $8 \text{ cm}$, then what is its period (in $\text{ s}$)?

$Assertion :$ 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.

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 $m/s$?

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

$A$ particle executes $S.H.M.$ according to the equation $x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right] \, cm$,where $t$ is in seconds. The magnitude of the velocity of the particle at $t=\frac{1}{6} \, s$ will be .............. $cm/s$.

Two $S$.$H$.Ms. are represented by equations $y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$ and $y_2 = 0.1 \cos (100 \pi t)$. The phase difference between the speeds of the two particles is