For a simple harmonic motion with amplitude $A$ and time period $T$,what is the velocity at $x = \frac{A}{2}$?

In the figure,what will be the sign of the velocity of the point $P_1$,which is the projection of the velocity of the reference particle $P$? $P$ is moving in a circle of radius $R$ in an anti-clockwise direction.

$A$ particle executing $S.H.M.$ starts from the mean position. Its amplitude is $A$ and time period is $T$. At what displacement is its speed one-fourth of the maximum speed?

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2A}{3}$ from the equilibrium position. The new amplitude of the motion is

Two identical pendulums oscillate with a constant phase difference $\frac{\pi}{4}$ and same amplitude. If the maximum velocity of one is $v$,the maximum velocity of the other will be ........

The piston in the cylinder head of a locomotive has a stroke of $6\,m$ (which is twice the amplitude). If the piston is executing simple harmonic motion with an angular frequency of $200\,rad\,min^{-1}$,its maximum speed is .... $m\,s^{-1}$.