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.

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

The two normal mode frequencies are

ω1 = √(k/m)
ω2 = √((k + 2k2)/m).
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.

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


φ = [φ1, φ2]


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


K = [((m1 + m2)gL1, 0), (0, m2gL1)].

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

© 2007 Robert Harr