PHY6200 W07

Chapter 11: Coupled Oscillators and Normal Modes

Recall

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.

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

The two normal mode frequencies are

ω₁ = √(k/m)

and

ω₂ = √((k + 2k₂)/m).

ω₁ = ω₀ - ε and ω₂ = ω₀ + ε

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

x = C₁ [1, 1] e^{i(ω₀ - ε)t} and x = C₂ [1, -1] e^{i(ω₀ + ε)t}.

x(t) = {C₁ [1, 1] e^-iεt + C₂ [1, -1] e^iεt}e^iω₀t.

Let C1 = C2 = A/2, then the position as a function of time becomes

x(t) = A[cosεt, -isinεt] e^iω₀t.

The actual motion is the real part of x. This gives

x₁(t) = A cosεt cosωt

x₂(t) = A sinεt sinωt

Let's look at this result in detail. Since ε << ω we can consider the factors A cosεt and A sinεt as a slowly changing amplitude for the cosωt and sinωt oscillations. That is, we can approximate the overall motion as (rapid) oscillations at the uncoupled frequency occurring with an envelope of slowly varying amplitude. The amplitude envelopes vary out of phase, so that first mass 1 swings at full amplitude while mass 2 is at rest; then the energy in mass 1 is transferred to mass 2, with a corresponding increase of the amplitude of mass 2 and reduction of amplitude of mass 1; until finally mass 2 swings at full amplitude and mass 1 comes to rest; then the whole process repeats but with energy being transferred from mass 2 to mass 1.

This phenomenon is known as beats. It arises when two modes exist that are close in frequency. It occurs with sound waves, electrical waves, and quantum mechanical probability waves.

Lagrangian Approach: The Double Pendulum

Lagrangian Approach for Two Carts and Three Springs

Already done in lecture.

The Double Pendulum

U₁ = m₁gL₁(1 - cosφ₁)

U₂ = m₂g[L₁(1 - cosφ₁ + L₂(1 - cosφ₂)]

In the small angle approximation, the total potential energy is

U(φ₁, φ₂) = ½(m₁ + m₂)gL₁φ₁² + ½m₂gL₂φ₂².

For the first mass, the position is (x₁, y₁) = L₁(sinφ₁, 1-cosφ₁), and kinetic energy is T₁ = ½ m₁L₁²φ₁dot².

The position of the second mass is (x₂, y₂) = (x₁, y₁) + L₂(sinφ₂, 1-cosφ₂). The square of the velocity is v₂² = x₂dot² + y₂dot² = L₁²φ₁dot² + L₂²φ₂dot² + L₁L₂φ₁dotφ₂dot(cosφ₁cosφ₂ + sinφ₁sinφ₂). The final trigonometric terms can be written as cos(φ₁ - φ₂) ≈ 1 for small angles.

Putting these together, we can write the kinetic energy as

T = ½(m₁ + m₂)L₁²φ₁dot² + ½m₂L₂²φ₂dot² + m₂L₁L₂φ₁dotφ₂dot.

And the Lagrangian is T - U. Lagrange's equations yield, for φ₁

(m₁ + m₂)L₁²φ₁ddot + m₂L₁L₂φ₂ddot = -(m₁ + m₂)gL₁φ₁

and for φ₂

m₂L₁L₂φ₁ddot + m₂L₂²φ₂ddot = -m₂gL₁φ₂.

These can be written in our standard matrix form

Mφddot = -Kφ

where

φ = [φ₁, φ₂]

and

M = [((m₁ + m₂)L₁², m₂L₁L₂), (m₂L₁L₂, m₂L₂²)]

and

K = [((m₁ + m₂)gL₁, 0), (0, m₂gL₁)].

Try the same trick again to solve this equation: try the solution φ = ae^iωt, and get the characteristic equation (K - ω²M)a = 0 which must be solved for ω and a.