Consider two equal bodies (not affected by gravity), each of mass m, attached to three springs, each with spring constant k. They are attached in the following manner, forming a system that is physically symmetric:
where the edge points are fixed and cannot move. Well use x(t) to denote the horizontal displacement of the left mass, and x(t') to denote the displacement of the right mass.
If one denotes acceleration (the second derivative of x(t) with respect to time) as , the equations of motion are:
Since we expect oscillatory motion of a normal mode (where ω is the same for both masses), we try:
Substituting these into the equations of motion gives us:
Since the exponential factor is common to all terms, we omit it and simplify:
And in matrix representation:
If the matrix on the left is invertible, the unique solution is the trivial solution (A, A) = (x, x) = (0,0). The non trivial solutions are to be found for those values of ω whereby the matrix on the left is singular i.e. is not invertible. It follows that the determinant of the matrix must be equal to 0, so:
Solving for , we have two positive solutions:
If we substitute ω into the matrix and solve for (A, A), we get (1, 1).  If we substitute ω, we get (1, −1).  (These vectors are eigenvectors, and the frequencies are eigenvalues.)
The first normal mode is:
Which corresponds to both masses moving in the same direction at the same time. This mode is called antisymmetric.
The second normal mode is:
This corresponds to the masses moving in the opposite directions, while the center of mass remains stationary. This mode is called symmetric.
The general solution is a superposition of the normal modes where c, c, φ, and φ, are determined by the initial conditions of the problem.
The process demonstrated here can be generalized and formulated using the formalism of Lagrangian mechanics or Hamiltonian mechanics.