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

$A$ $S.H.M.$ has amplitude $a$ and time period $T$. The maximum velocity will be

$A$ particle performing $SHM$ is found at its equilibrium position at $t = 1 \, s$. It is found to have a speed of $0.25 \, m/s$ at $t = 2 \, s$. If the period of oscillation is $6 \, s$,calculate the amplitude of oscillation.

$A$ particle is making simple harmonic motion along the $x$-axis. If at a distance $x_{1}$ and $x_{2}$ from the mean position the velocities of the particle are $v_{1}$ and $v_{2}$ respectively,the time period of its oscillation is given as -

The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position,the speed of the particle is $16 \,cm/s$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3} \,cm/s$ will be .... $cm$.

Two simple harmonic motions are represented by equations $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$. What is the phase difference between their velocities?