The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

$A$ particle is executing $S.H.M.$ If its amplitude is $2 \, m$ and periodic time is $2 \, s$,then the maximum velocity of the particle will be:

$A$ particle is executing $SHM$ along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

$A$ $S.H.M.$ has amplitude $a$ and time period $T$. The maximum velocity 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 .... $m\,s^{-1}$.

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.