The maximum velocity of a body undergoing $S.H.M.$ is $0.2 \ m/s$ and its acceleration at $0.1 \ m$ from the mean position is $0.4 \ m/s^2$. The amplitude of the $S.H.M.$ is .... $m$

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

$A$ particle executes simple harmonic motion with amplitude $A$ and period $T$. If it is halfway between the mean position and the extreme position,then its speed at that point is:

In an angular $SHM$,the angular amplitude of oscillation is $\pi \, rad$ and the time period is $0.4 \, s$. Calculate its angular velocity at an angular displacement of $\pi/2 \, rad$. (Result in $rad/s$)

$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.