After the beginning of motion,how long will it take for a harmonically oscillating particle to reach a displacement equal to one-half of its amplitude,if the time period is $24 \ sec$ and the particle starts from rest?

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 displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

$A$ particle executes simple harmonic motion represented by the displacement function $x(t) = A \sin (\omega t + \phi)$. If the position and velocity of the particle at $t = 0 \, s$ are $2 \, cm$ and $2 \omega \, cm \, s^{-1}$ respectively,then its amplitude is $x \sqrt{2} \, cm$,where the value of $x$ is ..... .

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:

$A$ 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.