Q.A particle performing linear S.H.M. has a period of 6.28 seconds and a path length of 20 cm. What is the velocity when its displacement is 6 cm from mean position?

Use $v=\omega\sqrt{A^2-x^2}$ for linear S.H.M., with $\omega$ from the period and $A$ from the path length.

Period $T=6.28\text{ s}\approx 2\pi\text{ s}$, so angular frequency:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{6.28}\approx 1\text{ rad/s}$

Path length (i.e. the full length traversed, from one extreme to the other) $=20\text{ cm}$, so the amplitude is half of this:

$A=\frac{20}{2}=10\text{ cm}$

For linear S.H.M., the velocity at displacement $x$ from the mean position is …