Resonant Frequency of 2 mass spring system 
So the question goes if I has a spring with spring constant $k$ and two masses attached to this spring (one on either side) what is the resonant frequency of the system in terms of $m$ and $k$?
Diagram of system:
[m]-////-[m]

Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation? I know that for a horizontal spring say attached to wall we can take the differential equation
$$m\frac{d^2 x}{d t^2}+kx= 0$$
And then use the equation $Asin(\omega t^2+\phi)$ as a solution and say this is true when $\omega= \sqrt{\frac{k}{m}}$
But I was thinking maybe I could just use the differential equation
$$m\frac{d^2 x}{d t^2}+2kx= 0$$ but I feel like that may be too simple? Is there something I'm missing? Any help would be appreciated! :)
Note: None of this system is undergoing any damping
 A: Assuming the spring is "ideal" (massless) you actually have 2 masses. You can describe your problem as the motion of the center of mass, and either of the masses. And if no external force is exerted on your system, you are only left with the motion of 1 mass relative to the center of the mass of the system.
Let's say your masses are m1, m2, and the spring constant is k. Denote the spring length (at rest) d.
The distance of m1 to center of mass is d * m2 / (m1+m2). Note that if m1 is displaced by x from the center of mass, then the overall spring displacement is given by x * (m1+m2) / m2. In other words, the effective spring constant for the motion of m1 is k * (m1+m2)/m2
So that the oscillation frequency of such a system is given by:
ω = [ k * (m1+m2) / (m1*m2) ] ^ (1/2)
Taking m2 to infinity reduces the formula to ω = [ k/m1 ] ^ (1/2), which is reasonable.
A: To do the equations of motion you need positions from an inertial reference frame. Say a wall far away. Call positions of the two masses $x_1$ and $x_2$. The spring force (tension is positive) is $$ F = k (x_2-x_1 ) $$ and the two equations of motion
$$\begin{align} F & = m_1 \ddot{x}_1 \\ -F & = m_2 \ddot{x}_2 \end{align} $$
All this is trivial. Here comes the trick. Change the coordinates into the separation $x = {x_2-x_1}$ and the center mass position $c = \frac{m_1 x_1+m_2 x_2}{m_1+m_2}$. Substitute in the above $$ \begin{align} x_1 &= c - \frac{m_2}{m_1+m_2} x & \ddot{x}_1 &= \ddot{c} - \frac{m_2}{m_1+m_2} \ddot{x}  \\ x_2 & = c + \frac{m_1}{m_1+m_2} x &  \ddot{x}_2 & = \ddot{c} + \frac{m_1}{m_1+m_2} \ddot{x} \end{align} $$
to get 
$$ \begin{align} \ddot{c} & = 0 \\ \ddot{x} & = - \left( \frac{1}{m_1} + \frac{1}{m_2} \right)^{-1} k x \end{align} $$
The first equation states that the center of mass will not accelerate in the absence of an external force (since $F$ is an internal force only).
The second equation is what you will use for the natural frequency. Your intuition was kind of correct, except you needed to use reduced mass of the system (in parenthesis) to get the frequency.
A: 
Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation?

As there are no external forces  the linear momentum  of the system  will be constant of motion and the constant can be taken as zero as well.
If m1 and m2 are the masses at any time described by position coordinates x1 and x2 -then \
m1. dx1/dt  +m2. dx2/dt =0 
i,e. the velocities of the masses weighted by their mass factor are related.
The Kinetic Energy of the system can be expressed as
T= 1/2 .m1. (dx1/dt)^2 + 1/2 . m2. (dx2/dt)^2
which in terms of new coordinate q 
q= x2-x1-d  will be 
T = 1/2 . m. (dq/dt)^2 
where q is the generalized coordinate  and d is the original length of the spring. The mass factor m can be defined as 1/m = 1/m1 +1/m2 normally called reduced mass of a two body system.
that is the kinetic energy can be written as function of a single generalized velocity say dq/dt.
q which is defined as  (x2-x1-d)  q can be viewed as extension of the spring and the potential energy of the two mass+spring   system  can be expressed as 
V(q) = 1/2. k. q^2   (where k is the spring constant)
Having knowledge of K.E. and P.E.  in terms of generalized coordinate q and dq/dt ...one can write
L , the Lagrangian as L= T- V and one can set up and solve  Lagrange's equation of motion in q
and can arrive at equation of motion
** d^2q/dt^2 = w^2. q**  where  w^2 = k/m
This vibrational frequency w is natural frequency of vibration and can be called resonant frequency.
