homework and exercises - Equation of Motion For a Specific Gravitation Question: Is harmonic motion involved? If yes/no, how and why?

This is a question that I have thought about myself, but I'm struggling a bit regarding how to answer it.

This is my question:

Imagine we have a universe with 3 bodies. Out of these 3, 2 are identical and massive, and 1 is small and insignificant. The 2 identical celestial bodies are literally identical in every way possible -- density, mass, volume, etc. They are also fixed in space and unable to move. If their centers are perfectly aligned vertically (although this doesn't really matter since we're in space) and separated by distance $d$, then their Lagrangian point will be at $\frac{d}{2}$ since the bodies are identical.

Now, let's make another axis from this Lagrangian point that is orthogonal to the vertical axis (which $d$ falls on), and call it the horizontal axis. In other words, let's just say this produces some $xy$-plane.

If our third body, which we may assume to be a perfect sphere and insignificant mass, is placed on this horizontal axis $x$ units of distance away from the origin of this $xy$-plane (which is the Lagrangian point), how will it's motion behave? Will it be harmonic and form some sort of space-pendulum or will it eventually fall into the Lagrangian point?

And in either case, what will be its equation of motion? If it falls into the Lagrangian point, depending on $x$ and other parameters, how long will it take to fall in?

I have included a diagram below.

EDITED:

So basically, you can start out with vector summation of the forces and get to here:

$$2\mathbf{F_{G}}\cos(\theta)=m \cdot \mathbf{a}$$

$\theta$ is basically the angle above the horizontal axis.

$$\frac{2GMm}{\frac{d^2}{4}+\Big(x(t)\Big)^2} \cdot \frac{x(t)}{\sqrt{\frac{d^2}{4}+\Big(x(t)\Big)^2}} = m \cdot a(t)$$ $$\frac{2GM\cdot x(t)}{\bigg (\frac{d^2}{4}+\Big(x(t)\Big)^2 \bigg) \sqrt{\frac{d^2}{4}+\Big(x(t)\Big)^2}} = a(t)$$

But now I get kind of stuck.