$A$ body is in simple harmonic motion with a time period of $0.5 \ s$ and an amplitude of $1 \ cm$. Find the average velocity in the interval in which it moves from the equilibrium position to half of its amplitude (in $cm/s$).

If $x = a \sin \left( \omega t + \frac{\pi}{6} \right)$ and $x' = a \cos \omega t$,then what is the phase difference between the two waves?

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.

The amplitude and the periodic time of a $S.H.M.$ are $5 \, cm$ and $6 \, s$ respectively. At a distance of $2.5 \, cm$ away from the mean position,the phase will be

$A$ large number of random snapshots using a camera are taken of a particle in simple harmonic motion between $x = -x_0$ and $x = +x_0$,with the origin $x = 0$ as the mean position. $A$ histogram of the total number of times the particle is recorded about a given position (Event no.) would most closely resemble:

Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?