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?

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 $3 \ s$. The time taken by it to cover a distance equal to half the amplitude from the mean position is $\left[\sin 30^{\circ}=0.5\right]$.

The phase of a particle executing simple harmonic motion is $\frac{\pi}{2}$ when it has:

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

The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in the $1^{\text{st}}$ second to the $2^{\text{nd}}$ second is: