The maximum velocity of a simple harmonic motion represented by $y = 3\sin \left( 100t + \frac{\pi}{6} \right)$ is given by

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

If a particle under $S.H.M.$ has a time period of $0.1 \ s$ and an amplitude of $2 \times 10^{-3} \ m$,what is its maximum velocity?

$A$ simple harmonic oscillator has a period of $0.01 \, s$ and an amplitude of $0.2 \, m$. The magnitude of the velocity in $m \cdot s^{-1}$ at the centre of oscillation 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

$A$ particle is performing $S.H.M.$ with a maximum velocity $V$. If the amplitude is doubled and the periodic time is reduced to $\left(\frac{1}{3}\right)^{\text{rd}}$ of its original value,then the new maximum velocity is: