where __ M__ is the diagonal matrix [(m

(**K** - ω²**M**)**C** = 0.

The two normal mode frequencies are

ω_{1} = √(k/m)

and
ω_{2} = √((k + 2k_{2})/m).

ω_{1} = ω_{0} - ε and ω_{2} = ω_{0} + ε

where some calculation will reveal that ε ≈ k_{2})/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

x_{1}(t) = A cosεt cosωt

x_{2}(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.

U_{1} = m_{1}gL_{1}(1 - cosφ_{1})

U_{2} = m_{2}g[L_{1}(1 - cosφ_{1} + L_{2}(1 - cosφ_{2})]

In the small angle approximation, the total potential energy is

U(φ_{1}, φ_{2}) = ½(m_{1} + m_{2})gL_{1}φ_{1}² + ½m_{2}gL_{2}φ_{2}².

For the first mass, the position is (x_{1}, y_{1}) = L_{1}(sinφ_{1}, 1-cosφ_{1}), and kinetic energy is T_{1} = ½ m_{1}L_{1}²φ_{1}dot².

The position of the second mass is (x_{2}, y_{2}) = (x_{1}, y_{1}) + L_{2}(sinφ_{2}, 1-cosφ_{2}).
The square of the velocity is v_{2}² = x_{2}dot² + y_{2}dot² = L_{1}²φ_{1}dot² + L_{2}²φ_{2}dot² + L_{1}L_{2}φ_{1}dotφ_{2}dot(cosφ_{1}cosφ_{2} + sinφ_{1}sinφ_{2}).
The final trigonometric terms can be written as cos(φ_{1} - φ_{2}) ≈ 1 for small angles.

Putting these together, we can write the kinetic energy as

T = ½(m_{1} + m_{2})L_{1}²φ_{1}dot² + ½m_{2}L_{2}²φ_{2}dot² + m_{2}L_{1}L_{2}φ_{1}dotφ_{2}dot.

And the Lagrangian is T - U.
Lagrange's equations yield, for φ(m_{1} + m_{2})L_{1}²φ_{1}ddot + m_{2}L_{1}L_{2}φ_{2}ddot = -(m_{1} + m_{2})gL_{1}φ_{1}

and for φm_{2}L_{1}L_{2}φ_{1}ddot + m_{2}L_{2}²φ_{2}ddot = -m_{2}gL_{1}φ_{2}.

These can be written in our standard matrix form
where

and

and

Try the same trick again to solve this equation: try the solution __ φ__ =