$A$ particle executes $SHM$ with a period of $1.2 \, s$ and an amplitude of $8 \, cm$. Find the time it takes to travel $3 \, cm$ from the positive extremity of its oscillation.

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

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

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?

$A$ body is in simple harmonic motion with a time period of $0.5 \ s$ and an amplitude of $1 \ cm$. Find the average velocity in the interval in which it moves from the equilibrium position to half of its amplitude (in $cm/s$).

$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?