where __ M__ is the diagonal matrix [(m

(**K** - ω²**M**)**C** = 0.

This is an eigenvalue equation. In mathematics it is shown that such equations have a non-trivial solution if and only if the determinant of the quantity in parentheses is zero

det(**K** - ω²**M**) = 0,

The solution has CThe most general motion is a linear combination of these two solutions.
Notice that **C**_{1} and **C**_{2} are orthoganol column matrices ("vectors"), that is **C**_{1}^{T}**C**_{2} = **C**_{2}^{T}**C**_{1} = ½ - ½ = 0.
Therefore, they can be thought of as basis vectors in the two dimensional space of the coordinates x_{1} and x_{2}.
Any possible solution can be expressed as a linear combination of our two eigenvector solutions.

So we see that it isn't possible for one cart to move and not the other, that is, it isn't possible to vary x_{1} without also varying x_{2}.
Certain combinations of coordinates do have this property, and they are called normal coordinates.
Normal coordinates correspond to combinations of the physical coordinates, and always exist in problems such as this.

In this case, we can replace x_{1} and x_{2} by

ξ_{1} = ½(x_{1} + x_{2})

and

ξ_{2} = ½(x_{1} - x_{2}).

These coordinates are orthogonal, and therefore span the space of all possible values for x_{1} and x_{2}.
For this simple example, it is relatively easy to see that ξ_{1} corresponds to the first eigenvector and ξ_{2} to the second, so that for the first normal mode

ξ_{1}(t) = A e^{i(ω1t - δ)}

ξ_{2}(t) = 0,

and for the second

ξ_{1}(t) = 0,

ξ_{2}(t) = A e^{i(ω2t - δ).}

Only ξ_{1} oscillates in the first normal mode and only ξ_{2} in the second (of course physically, both masses move, but in the manner dictated by the linear combinations yielding ξ_{1} and ξ_{2}).
The general motion can be written as a linear combination of these two coordinates.
One can think of the motion of the system as composed of the motions of each cart individually, or as composed of the motions in each normal mode, where each normal mode represents a particular coordination of the motions of the two carts (a dance if you will).

Another situation of particular interest occurs when a set of (identical) harmonic oscillators are coupled, but only weakly.
This is a technique commonly applied to solving problem:
if you have a difficult problem that involves two simpler situations, that you do know how to solve, coupled together, first consider the problem with weak coupling.
In our simple model, this situation would correspond to the case of two equal masses, and identical springs 1 and 3, k_{1} = k_{3} = k, and a much weaker spring coupling them, k_{2} << k.
This implies that at some point we will make an approximation and drop terms that are small.

Before beginning, we expect to find solutions that are close to two uncoupled oscillators. There will be some small effect due to the weak coupling.

The matrix __ M__ is unchanged from before, and we can now write

(k + k_{2} - mω²)² - k_{2}² = (k - mω²)² + 2k_{2}(k - mω²) = (k - mω²)(k +
2k_{2} - mω²) = 0

The two normal mode frequencies are

ω_{1} = √(k/m)

and
ω_{2} = √((k + 2k_{2})/m).

Thus far we've made no approximations.
In the weak coupling limit (k_{2}<_{1} and ω_{2}.
Start by taking the average of ω_{1} and ω_{2}

ω_{0} = (ω_{1} + ω_{2})/2

Now we can write the two normal frequencies as being a little below and a little above the average frequency ω_{0}

ω_{1} = ω_{0} - ε and ω_{2} = ω_{0} + ε

where some calculation will reveal that ε ≈ k_{2})/2m.