PHY6200 W07

Chapter 11: Coupled Oscillators and Normal Modes

Recall

Two Masses and Three Springs

where M is the diagonal matrix [(m₁, 0), (0, m₂)] and K is the matrix [((k₁ + k₂), -k₂), (-k₂, (k₂ + k₃))].

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

The Second Normal Mode

The General Solution

The most general motion is a linear combination of these two solutions. Notice that C₁ and C₂ are orthoganol column matrices ("vectors"), that is C₁^TC₂ = C₂^TC₁ = ½ - ½ = 0. Therefore, they can be thought of as basis vectors in the two dimensional space of the coordinates x₁ and x₂. Any possible solution can be expressed as a linear combination of our two eigenvector solutions.

Normal Coordinates

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 x₁ without also varying x₂. 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 x₁ and x₂ by

and

These coordinates are orthogonal, and therefore span the space of all possible values for x₁ and x₂. For this simple example, it is relatively easy to see that ξ₁ corresponds to the first eigenvector and ξ₂ to the second, so that for the first normal mode

and for the second

Only ξ₁ oscillates in the first normal mode and only ξ₂ in the second (of course physically, both masses move, but in the manner dictated by the linear combinations yielding ξ₁ and ξ₂). 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).

Two Weakly Coupled Oscillators: One Spring Different from the Other Two

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, k₁ = k₃ = k, and a much weaker spring coupling them, k₂ << 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 as K = [((k + k₂), -k₂), (-k₂, (k + k₂))]. The determinant equation becomes

The two normal mode frequencies are

Thus far we've made no approximations. In the weak coupling limit (k₂<2 being slightly higher than the uncoupled frequency. It is convenient to express this result in a different way that emphasizes the closeness of ω₁ and ω₂. Start by taking the average of ω₁ and ω₂

Now we can write the two normal frequencies as being a little below and a little above the average frequency ω₀

where some calculation will reveal that ε ≈ k₂)/2m.