Simple harmonic motion

Let the displacement of each particle be $x = A\sin\theta$, where $\theta$ is its phase.
When they cross, $x = \frac{A}{2}$ for both: $\frac{A}{2} = A\sin\theta$, so $\sin\theta = \frac{1}{2}$
$\theta = 30^\circ$ or $150^\circ$
Velocity is $v = A\omega\cos\theta$: at $\theta = 30^\circ$ the velocity is positive and at $\theta = 150^\circ$ it is negative.
Since the particles move in opposite directions, one has phase $30^\circ$ and the other $150^\circ$.
Phase difference = $150^\circ - 30^\circ = 120^\circ = \frac{2\pi}{3}$

The potential energy curve is a parabola, $V(x) \propto x^2$, so the restoring force is $F = -\frac{dV}{dx} \propto -x$.
A force proportional to displacement and directed towards the origin gives simple harmonic motion about the origin.
The particle is released from rest at a point away from the origin, so at t = 0 it is at an extreme position: x is maximum and the velocity (slope of the x–t graph) is zero.
Hence $x(t) = A\cos\omega t$.
The correct graph is the cosine curve that starts from maximum displacement at t = 0 and oscillates about x = 0.

A motion is SHM if the displacement is a single sine or cosine function of time, i.e., if $\frac{d^2y}{dt^2} \propto -y$.
(A) $y = \sin\omega t - \cos\omega t = \sqrt{2}\sin\left(\omega t - \frac{\pi}{4}\right)$: a single sine function, so it is SHM with angular frequency $\omega$.
(B) $y = \sin^3\omega t = \frac{1}{4}(3\sin\omega t - \sin3\omega t)$: a superposition of two SHMs of different frequencies; it is periodic but not SHM.
(C) $y = 5\cos\left(\frac{3\pi}{4} - 3\omega t\right) = 5\cos\left(3\omega t - \frac{3\pi}{4}\right)$: a single cosine function, so it is SHM with angular frequency $3\omega$.
(D) $y = 1 + \omega t + \omega^2t^2$: y keeps increasing with time, so the motion is not even periodic.
Hence only (A) and (C) represent SHM.

$x = a\sin^2\omega t$
Velocity: $\frac{dx}{dt} = 2a\omega\sin\omega t\cos\omega t = a\omega\sin2\omega t$
Acceleration: $\frac{d^2x}{dt^2} = 2a\omega^2\cos2\omega t$
For SHM about the origin we need $\frac{d^2x}{dt^2} = -\omega'^2x$, i.e., acceleration proportional to $-x$.
Here $2a\omega^2\cos2\omega t$ is not proportional to $-a\sin^2\omega t$, so the condition $\frac{d^2x}{dt^2} \propto -x$ is not satisfied for the displacement x as given.
Hence the answer accepted is non simple harmonic motion.
Note: writing $x = \frac{a}{2} - \frac{a}{2}\cos2\omega t$ shows the motion is simple harmonic about the shifted mean position $x = \frac{a}{2}$ with frequency $\frac{\omega}{\pi}$; the key of the source PDF treats it as non-SHM because x is measured from the origin.