$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

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

$A$ particle performs simple harmonic motion between $x = -A$ and $x = +A$. How much time does it take to travel from $x = A$ to $x = A/2$?

$A$ particle is executing $S.H.M.$ with time period $T$. Starting from the mean position,the time taken by it to complete $\frac{5}{8}$ oscillations is ..........

Two particles are in $SHM$ on the same straight line with amplitudes $A$ and $2A$ and with the same angular frequency $\omega$. It is observed that when the first particle is at a distance $A/\sqrt{2}$ from the origin and moving toward the mean position,the other particle is at the extreme position on the other side of the mean position. Find the phase difference between the two particles. (in $^o$)

The time taken by a particle executing simple harmonic motion of period $T$ to move from the mean position to half the maximum displacement is