$A$ spring executes $SHM$ with a mass of $10\,kg$ attached to it. The force constant of the spring is $10\,N/m$. If at any instant its velocity is $40\,cm/s$,the displacement will be .... $m$ (where amplitude is $0.5\,m$).

$A$ particle executes $S.H.M.$ according to the equation $x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right] \, cm$,where $t$ is in seconds. The magnitude of the velocity of the particle at $t=\frac{1}{6} \, s$ will be .............. $cm/s$.

$A$ 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)

The velocity of a particle in simple harmonic motion at displacement $y$ from the mean position is

When a particle executes $SHM$,the nature of the graphical representation of velocity as a function of displacement is:

$A$ 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: