$A$ particle is performing $S.H.M.$ with maximum velocity $v$. If the amplitude is tripled and periodic time is doubled,then the new maximum velocity will be:

$A$ simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will be

$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

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 .... $m\,s^{-1}$.

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

$A$ simple harmonic oscillation is represented by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. Its speed is maximum when $t$ equals