where M is the diagonal matrix [(m1, 0), (0, m2)] and K is the matrix [((k1 + k2), -k2), (-k2, (k2 + k3))].
The two normal mode frequencies are
where some calculation will reveal that ε ≈ k2)/2m.
Let C1 = C2 = A/2, then the position as a function of time becomes
The actual motion is the real part of x. This gives
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.
In the small angle approximation, the total potential energy is
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
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