$A$ 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 $......... \, s$.

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 .........

In the following table,time is in column-$I$ and the phase of an oscillator starting from the mean position is in column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $t = \frac{T}{8}$	$(i)$ $\theta = \frac{5\pi}{4}$
$(b)$ $t = \frac{5T}{8}$	$(ii)$ $\theta = \frac{3\pi}{2}$
	$(iii)$ $\theta = \frac{\pi}{4}$

$A$ particle performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is:

The displacement of a particle executing simple harmonic motion is $y = A \sin (2t + \phi) \ m$,where $t$ is time in seconds and $\phi$ is the phase angle. At time $t = 0$,the displacement and velocity of the particle are $2 \ m$ and $4 \ ms^{-1}$ respectively. The phase angle $\phi$ is: (in $^{\circ}$)

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