Our model system for this study consists of two masses, m1 and m2, each connected to a fixed support by springs of spring constants k1 and k3, and coupled to each other by a third spring of spring constant k2. There is some equilibrium position where the net force on each mass is zero. Measure the locations of each mass from their corresponding equilibrium position, call them x1 and x2. Then the kinetic energy of the system is

T = ½m_{1}x_{1}dot² + ½m_{2}x_{2}dot²

The potential energy of the system comes from stretching or compressing the springs. The change in length of k1 is x1 and the change in length of k3 is x2. The change in length of k2 is (x1 - x2); if both masses are displaced by an equal amount in the same direction, then k2 doesn't change length. The potential energy is

U = ½k_{1}x_{1}² + ½k_{2}(x_{1} - x_{2})² + ½k_{3}x_{2}²

The Lagrangian for the system is

L = T - U = ½m_{1}x_{1}dot² + ½m_{2}x_{2}dot² - ½k_{1}x_{1}² - ½k_{2}(x_{1} - x_{2})² - ½k_{3}x_{2}²

Lagrange's equations yield the following equations of motion

mx_{1}ddot = -k_{1}x_{1} - k_{2}(x_{1} - x_{2}) = -(k_{1} + k_{2})x_{1} + k_{2}x_{2}

mx_{2}ddot = k_{2}(x_{1} - x_{2}) -k_{3}x_{2} = k_{2}x_{1} - (k_{2} + k_{3})x_{2}

This can be written more compactly (and usefully) by introducing matrix notation.
Let __ x__ be the column vector (x

where __ M__ is the diagonal matrix [(m

*Aside 1: Since M is diagonal, it has an inverse, M^{-1}, so we could write xddot = -M^{-1}Kx.
Unfortunately this doesn't bring us closer to a solution and we won't follow this path.*

*Aside 2: M and K are matrices while I is a tensor.
Tensors, like vectors, change in specific ways under coordinate system transformations.
Matrices are a more general mathematical construction, a 2 dimensional array of quantities with rules for addition (of two matrices of equal size) and multiplication (of a matrix with n columns into a matrix with n rows).
A (second order) tensor may be represented as a matrix, and matrices are a nice way to organize and manipulate systems of equations.*

The similarity of the equation does mean that the time dependence in __ x__ should be the same for all elements, that is we can write

where __ A__,

-**M****C**ω²e^{iωt} = -**K****C**e^{iωt}.

Moving all the terms to the same side of the equation, and factoring out the common **C** term and canceling e^{iωt} yields

(**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,

that is, we must find the values of ω² that satisfy the equation.
The trivial solution is the one where __ C__ = 0, but then there is no motion at all in the system and the problem is uninteresting.
For our model system with two masses, the determinant equation is quadratic in ω², yielding two values for ω called normal frequencies (we will take ω>0, since taking ω<0 is equivalent to a change of phase by 180°, not a separate solution).
In general, for a system with N masses, there are N normal frequencies and they are found using the same procedure discussed below.

We will now solve some example coupled oscillator problems.
To keep this first problem simple, we'll look at the model system with both masses equal, m_{1} = m_{2} = m, and idential spring constants k_{1} = k_{2} = k_{3} = k.
The matrix __ M__ becomes m

det(**K** - ω²**M**) = (2k-mω²)² - k² = 3k² - 4kmω² + m²ω^{4} = (3k - mω²)(k - mω²) = 0

The solutions (eigenvalues) are ω² = 3k/m and ω² = k/m.
The normal frequencies are ω_{1} = &radic(k/m) and ω_{2} = &radic(3k/m).
They are the frequencies at which both carts can oscillate simultaneously -- pay attention to the in-class demonstration with coupled pendula.

To understand what this means, let's complete the picture by determining the eigenvectors for these eigenvalues.
The eigenvectors are the values of __ C__ that satisfy the eigenvalue equation (

(2k - k)C_{1} - kC_{2} = C_{1} - C_{2} = 0

-kC_{1} + (2k - k)C_{2} = -C_{1} + C_{2} = 0

where C_{1} and C_{2} are the two components of __ C__.
The solution has C

Now let's take ω = ω_{2} = &radic(3k/m).
The eigenvalue equation reads

(2k - 3k)C_{1} - kC_{2} = -C_{1} - C_{2} = 0

-kC_{1} + (2k - 3k)C_{2} = -C_{1} - C_{2} = 0

The solution has C_{1} = -C_{2}, so a normalized eigenvector is **C**_{2} = √(½) [1, -1].
The position vector is

The 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.