If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

$A$ body executing $S.H.M.$ along a straight line has a velocity of $3 \, m/s$ when it is at a distance of $4 \, m$ from its mean position and $4 \, m/s$ when it is at a distance of $3 \, m$ from its mean position. Its angular frequency and amplitude are:

$A$ particle is executing simple harmonic motion with amplitude $A$. At a distance $x$ from the mean position,when the particle is moving towards the extreme position,it receives a blow in the direction of motion which instantaneously doubles its velocity. What is the new amplitude of the particle? (Frequency remains constant during the motion)

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

The maximum speed of a particle in $S.H.M.$ is $V$. The average speed is

The displacements of two particles executing simple harmonic motion are represented as $y_{1} = 2 \sin (10 t + \theta)$ and $y_{2} = 3 \cos 10 t$. The phase difference between the velocities of these waves is