# PHY6200 W07

## Chapter 11: Coupled Oscillators and Normal Modes

### Recall

#### The Double Pendulum

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 = aeiωt, and get the characteristic equation (K - ω²M)a = 0 which must be solved for ω and a.

#### Equal Lengths and Masses

As a simple example, consider the case when the masses and lengths are equal. This simplifies K and M to

M = mL²[(2, 1), (1, 1)]

and

K = mL²[(2ω0², 0), (0, ω0²)]

where ω0 = √(g/L) is the natural frequency of a pendulum of length L. The determinant is

m²L4[2(ω0² - ω²)² - ω4] = m²L44 - 4ω0²ω² + 2ω0²] = 0

This has solutions

ω1² = (2 - √2)ω0²  and  ω2² = (2 + √2)ω0²

The normal modes are found by substituting back into the characteristic equation. For ω1 we get

(K - ω²M)a = mL²ω0²(√2 - 1)[(2, -√2), (-√2, 1)][a1, a2] = 0,

implying that √2 a1 = a2. For ω2 we get

(K - ω²M)a = mL²ω0²(√2 + 1)[(2, √2), (√2, 1)][a1, a2] = 0,

implying that -√2 a1 = a2.

In the first mode, the masses oscillate together, but the amplitude of the lower mass is √2 times larger than the amplitude of the upper mass. In the second mode, the masses oscillate 180° out of phase, again with the lower mass swinging with an amplitude √2 times greater than the upper mass.

### The General Case

Consider the general case of a system with n degrees of freedom oscillating with small amplitude about a point of stable equilibrium. Assume the system is holonomic so that the n degrees of freedom translats to n generalized coordinates q1, ..., qn. It is convenient to represent these as the column matrix q = [q1, ..., qn]. (Recall my note about the difference between tensors and matrices. The same applies between "true" vectors and column matrices. We often call q a vector, meaning it is a column matrix, not a "true" vector.)

We assume that the system is conservative and has a potential energy function

U(q1, ..., qn) = U(q).

The potential energy is given by

T = ½ ∑a mava²

where the sum is over N particles in the system. The N velocities are related to derivatives of the generalized coordinates. Begin by writing the position of particle a as a function of the generalized coordinates, ra = ra(q1, ..., qn). The time derivative of ra is found using the chain rule

va = dra/dt = (∂ra/∂q1)q1dot + ... + (∂ra/∂qn)qndot = ∑i(∂ra/∂qi)qidot

The square of the velocity is the square of the above expression. We want to be careful to keep track of all the terms. Just writing down the product we have

va² = vava = ∑i,j(∂ra/∂qi)⋅(∂ra/∂qj)qidot qjdot

where we must take the dot product of the vector quantities in parentheses.

Now that we have the most general equations for this situation, we want to specialize to the case of small oscillations about a position of stable equilibrium. Small oscillations imply that any generalized coordinate or its derivatives are of order ε where ε represents a small quantity. We want to keep only the lowest order terms in ε, and we will see that the first order terms in ε are zero, so the lowest order terms are proportional to ε², that is, terms like qi², qiqj, and qidot qjdot.

Let's begin with the potential energy. We assume there's a position of stable equilibrium, and for convenience, adjust the generalized coordinates so that this occurs at q = 0 (this notation means that every qi is zero). Next, we can expand the potential and kinetic energies about this point in a Taylor series. The kinetic energy can depend on the qi through the partial derivative terms. Since each term in the sum for the kinetic energy contains a factor of qidot qjdot, we take only the constant term in the Taylor series expansion of the partial derivatives, so that

T ≈ ½∑a mai,j(∂ra/∂qi)0⋅(∂ra/∂qj)0qidot qjdot

The subscript zero on the partial derivatives indicates that they are evaluated at q = 0. These terms are therefore constants. It is convenient to change the order of the sums

T ≈ ½ ∑i,j [∑a ma(∂ra/∂qi)0⋅(∂ra/∂qj)0]qidot qjdot

Then note that the sum over a just adds up constant partial derivative terms, and these can be moved out of sight by defining the matrix M with components

Mij = ∑a ma(∂ra/∂qi)0⋅(∂ra/∂qj)0

allowing us to write the kinetic energy as

T ≈ ½ ∑i,j Mijqidot qjdot = ½ qdotT M qdot

where I've explicitly shown the matrix expression for the record, though you'll normally use the sum notation.

The Taylor series expansion of the potential energy is a bit more straightforward. First, note that, at a point of stable equilibrium, all the forces are zero, that is, -∂U/∂qi = 0. Since we're expanding around a point of stable equilibrium, the term that is first order in qi is zero, and we must go to the second order terms, yielding

U ≈ U(0) + ½ ∑i,j (∂²U/∂qi∂qj)0 qi qj = U(0) + ½ ∑i,j Kij qi qj = ½ qT K q,

where I've introduced the matrix K with components Kij = (∂²U/∂qi∂qj)0.

With these approximations for the kinetic and potential energies, the Lagrangian is

L = T - U = ½ ∑i,j Mijqidot qjdot - ½ ∑i,j Kij qi qj = ½ qdotT M qdot - ½ qT K q.

#### The Equation of Motion

We're now ready to look at the equations of motion. There are n of them, corresponding to the n generalized coordinates, each of the form

(d/dt)(∂L/∂qkdot) = ∂L/∂qk,

for k = 1 to n. The double sums present a small complication to the derivatives. Here's a way to get the correct result. First, note that the matrices M and K are symmetric, that is, Mij = Mji and Kij = Kji. Second, notice that there is a term for i=k and one for j=k, so that

L/∂qk = - ½ ∑j Kkj qj - ½ ∑i Kik qi = -∑i Kik qi,

where I've used the fact that i and j are just dummy indices for the summation, and changed j in the first sum to i, then used the symmetry property of K to combine the two terms. Likewise, the other derivative yields

L/∂qkdot = ½ ∑j Mkj qjdot + ½ ∑i Mik qidot = ∑i Mik qidot.
With this simplification the kth equation of motion reads
i Mik qiddot = -∑i Kik qi.

In matrix format this looks like

Mqddot = -Kq

As we did with the pair of oscillators, try a solution of the form q = aeiωt, yielding the eigenvalue equation

(K - ω²M)a = 0

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

det(K - ω²M) = 0.

This is an nth order equation in ω². Aside from a few special cases, it must generally be solved numerically. This type of problem is rather common in science and engineering, so there are numerous computer programs available for solving such systems.

### Three Coupled Pendula

As an example of the above procedure, we'll do an example with three degrees of freedom, three coupled pendula, lined up in and free to swing in one plane. For simplicity, let all the pendula be of length L with mass m, and let both springs have spring constant k. Also, let the springs have an unstretched length equal to the separation of the points of attachement. 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. We want to 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 mL²(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 mgL(φ1² + 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 L(φ2 - φ1) for the first spring and L(φ3 - φ2) for the second, yielding the potential energy

U = ½ kL²((φ2 - φ1)² + (φ3 - φ2)²) = ½ kL²(φ1² - 2φ1φ2 + 2φ2² - 2φ2φ3 + φ3²).

Try using natural units for the rest of this problem. We'll choose units where m = L = 1, in which case the kinetic and potential energies become

T = ½ (φ1dot² + φ2dot² + φ3dot²)

and

U = ½ g(φ1² + φ2² + φ3²) + ½ k(φ1² - 2φ1φ2 + 2φ2² - 2φ2φ3 + φ3²)

We know (from the discussion of the general case) that the equations of motion can be written in the matrix form Mφddot = -Kφ where we now need the 3×3 matrices M and K. By inspection 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 - ω²M)a = 0

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

det(K - ω²M) = 0.
The determinant evaluates to
(g + k - ω²)²(g + 2k -ω²) - 2k²(g + k - ω²) = 0
(g + k - ω²)[(g + k - ω²)(g + 2k -ω²) - 2k²] = (g + k - ω²)[(g - ω²)² + 3k(g - ω²)] = 0
(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.