# PHY6200 W07

## Chapter 11: Coupled Oscillators and Normal Modes

### Two Masses and Three Springs

Mxddot = -Kx

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

(K - ω²M)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 - ω²M) = 0,
The solution has C1 = C2, so a normalized eigenvector is C1 = √(½) [1, 1]. The position vector is

x = (A1/√2) [1, 1] e1,t.
Therefore, if the two carts oscillate with frequency ω1 = &radic(k/m) then they move in phase with equal amplitude.

#### The Second Normal Mode

x = (A2/√2) [1, -1] e1,t.
Therefore, if the two carts oscillate with frequency ω2 = &radic(3k/m) then they move 180° out of phase with equal amplitude.

#### 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 C1TC2 = C2TC1 = 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

ξ1 = ½(x1 + x2)

and

ξ2 = ½(x1 - x2).

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

ξ1(t) = A ei(ω1t - δ)
ξ2(t) = 0,

and for the second

ξ1(t) = 0,
ξ2(t) = A ei(ω2t - δ).

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

(k + k2 - mω²)² - k2² = (k - mω²)² + 2k2(k - mω²) = (k - mω²)(k + 2k2 - mω²) = 0

The two normal mode frequencies are

ω1 = √(k/m)
and
ω2 = √((k + 2k2)/m).

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

ω0 = (ω1 + ω2)/2

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

ω1 = ω0 - ε  and  ω2 = ω0 + ε

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

x = C1 [1, 1] ei(ω0 - ε)t  and  x = C2 [1, -1] ei(ω0 + ε)t.
x(t) = {C1 [1, 1] e-iεt + C2 [1, -1] eiεt}e0t.
Let C1 = C2 = A/2, then the position as a function of time becomes
x(t) = A[cosεt, -isinεt] e0t.