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 equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

$A$ spring executes $S.H.M.$ with a mass of $1 \,kg$ attached to it. The force constant of the spring is $4 \,N/m$. If at any instant its velocity is $20 \,cm/s$, the displacement at that instant is (Amplitude of $S.H.M.$ is $0.4 \,m$)

The velocity of a particle performing simple harmonic motion,when it passes through its mean position is

$A$ simple pendulum performs $S.H.M$ about $x = 0$ with an amplitude $A$ and time-period $T$. The speed of the pendulum at $x = \frac{A}{2}$ will be

The displacement of a particle executing $SHM$ is given by $y = 0.25 \sin(200t) \ cm$. The maximum speed of the particle is $......... \ cm \ s^{-1}$.