PHY5210 W15

Chapter 11: Coupled Oscillators and Normal Modes

Reading:

In Taylor, read sections 11.6 for today, and 11.7 for Monday.

Recall

The general equation of motion for the small oscillation of n oscillators is written

Mqddot = -Kq

where the matrices are given by

Mij = ∑a ma (∂ra/∂qi)0 ⋅ (∂ra/∂qj)0

and

Kij = (∂²U/∂qiqj)0

Assuming all oscillators move at the same frequency yields the eigenvalue problem to solve for the normal modes and frequencies

(K - ω² M) a = 0

Three Coupled Pendula

As an example of the general procedure, we'll work an example with three degrees of freedom, three coupled pendula, lined up in and free to swing in one plane. The first and second pendulum masses and the second and third are connected with springs. For simplicity, let all the pendula be of length L with mass m, and let both coupling springs have spring constant k. Also, let the springs have an unstretched length equal to the separation of the points of attachement. Then the equilibrium position is when all three pendula hang vertically. We'll measure their displacements from equilibrium by the angles φ1, φ2, and φ3 that each pendulum makes with vertical. Next write down the kinetic and potential energies in the small angle approximation.

The kinetic energy is easy and doesn't require any approximations,

T = ½ m L² (φ1dot² + φ2dot² + φ3dot²)

The potential energy is composed of two parts, gravitational potential energy and elastic potential energy. The gravitational potential energy depends on how high each pendulum has moved vertically from the equilibrium position. In the small angle approximation and neglecting constant terms, it is

U = ½ m g L1² + φ2² + φ3²)

The elastic potential energy depends on the change in length of each spring. In the small angle approximation, the horizontal motion is first order in the angle while the vertical motion is second order, so we neglect the vertical motion. The change in length is then L2 - φ1) for the first spring and L3 - φ2) for the second, yielding the potential energy

U = ½ k L² ((φ2 - φ1)² + (φ3 - φ2)²) = ½ k L² (φ1² - 2φ1 φ2 + 2φ2² - 2φ2φ 3 + φ3²).

Try using natural units for the rest of this problem. Natural units are commonly used in advanced physics (quantum mechanics problems for example) to simplify the equations during derivations. The idea is to set the units to make constants that appear in the equations equal to 1. At the end of the problem, the constants are restored by determining the combination needed to yield the correct units on the result. We'll choose units where m = L = 1, in which case the kinetic and potential energies simplify to

T = ½ (φ1dot² + φ2dot² + φ3dot²)

and

U = ½ g1² + φ2² + φ3²) + ½ k1² - 2φ1 φ2 + 2φ2² - 2φ2 φ3 + φ3²)

We know (from the discussion of the general case) that the equations of motion can be written in the matrix form M φddot = -K φ where we now need the 3 × 3 matrices M and K. Taking the derivatives of T we find that M is the diagonal matrix

M = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]

and K has elements

K = [(g + k, -k, 0), (-k, g + 2k, -k), (0, -k, g + k)]

By assuming a solution of the form φ = aeiωt we get the eigenvalue equation

(K - ω² M) a = 0

which has a non-trivial solution only if the characteristic equation is satisfied

det(K - ω² M) = 0.

The determinant evaluates to

(g + k - ω²)² (g + 2k -ω²) - 2k² (g + k - ω²) = 0

Expanding and factoring yields

(g + k - ω²)[(g + k - ω²)(g + 2k -ω²) - 2k²] = (g + k - ω²)[(g - ω²)² + 3k (g - ω²)] = 0

and finally

(g + k - ω²)(g + 3k - ω²)(g - ω²) = 0

There are three normal frequencies

ω1² = g ,    ω2² = g + k ,  and   ω3² = g + 3k.

The corresponding normal modes are,

a1 = [1, 1, 1],    a2 = [1, 0, -1]   and   a3 = [1, -2, 1].

Since we used "natural units", the results must be adjusted to get calculable quantities. In particular, the normal frequencies need to be adjusted. This is done by reinserting the missing factors of m and L so that units come out correctly. For example, ω1² = g in natural units, but since frequency squared has units of [1/s²], and g has units of [m/s²] we must divide g by a length to match. Therefore ω1² = g/L is the standard result.

The second normal frequency needs to have a factor of L divided into g, and a factor of m divided into k, yielding ω2² = g/L + k/m, and similarly the third normal frequency becomes ω3² = g/L + 3k/m.

Normal Coordinates*

Please read this section. I will not cover it in lecture.

The General Case

© 2015 Robert Harr