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

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $x/7 \text{ s}$. The value of $x$ is (Take $\pi = 22/7$)

$A$ 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)

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion 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?