The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

Consider a pair of identical pendulums,which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical,the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums?

The equation of $S.H.M.$ is $y = a \sin(2\pi nt + \alpha)$,then its phase at time $t$ is:

Two bodies performing $SHM$ have the same amplitude and frequency. Their positions at a certain instant are as shown in the figure. The phase difference between them is