$A$ 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:

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?

$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

$A$ particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $a$ meters. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} \, m$ from its mean position in seconds is

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?