$A$ particle executes $SHM$ with an amplitude of $20 \, cm$ and a time period of $12 \, s$. What is the minimum time required for it to move between two points $10 \, cm$ on either side of the mean position?

Two particles executing $S.H.M.$ of same frequency,meet at $x=+A/2$,while moving in opposite directions. Phase difference between the particles is .........

Two particles $A$ and $B$ execute $SHMs$ of periods $T$ and $\frac{3T}{2}$. If they start from the mean position,then the phase difference between them,when the particle $A$ completes two oscillations,will be

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

$A$ 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:

$A$ particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$,and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$,then