PHY6200 W07

Chapter 11: Coupled Oscillators and Normal Modes


Two Masses and Three Springs

Our model system for this study consists of two masses, m1 and m2, each connected to a fixed support by springs of spring constants k1 and k3, and coupled to each other by a third spring of spring constant k2. There is some equilibrium position where the net force on each mass is zero. Measure the locations of each mass from their corresponding equilibrium position, call them x1 and x2. Then the kinetic energy of the system is

T = ½m1x1dot² + ½m2x2dot²

The potential energy of the system comes from stretching or compressing the springs. The change in length of k1 is x1 and the change in length of k3 is x2. The change in length of k2 is (x1 - x2); if both masses are displaced by an equal amount in the same direction, then k2 doesn't change length. The potential energy is

U = ½k1x1² + ½k2(x1 - x2)² + ½k3x2²

The Lagrangian for the system is

L = T - U = ½m1x1dot² + ½m2x2dot² - ½k1x1² - ½k2(x1 - x2)² - ½k3x2²

Lagrange's equations yield the following equations of motion

mx1ddot = -k1x1 - k2(x1 - x2) = -(k1 + k2)x1 + k2x2
mx2ddot = k2(x1 - x2) -k3x2 = k2x1 - (k2 + k3)x2

This can be written more compactly (and usefully) by introducing matrix notation. Let x be the column vector (x1, x2), then xdot is the column vector (x1dot, x2dot), and xddot is the column vector (x1ddot, x2ddot). The equations of motion can be written in the form

Mxddot = -Kx

where M is the diagonal matrix [(m1, 0), (0, m2)] and K is the matrix [((k1 + k2), -k2), (-k2, (k2 + k3))]. This equation looks similar to the simple harmonic oscillator eqution mxddot = -kx. Sometimes looks can be deceiving, but the general aim in mathematics is to simplify the notation so that things with similar properties are more readily apparent. However, the equations are not so similar that we can write xddot = -(K/M)x. M is a matrix, and matrix division is not a defined operation.

Aside 1: Since M is diagonal, it has an inverse, M-1, so we could write xddot = -M-1Kx. Unfortunately this doesn't bring us closer to a solution and we won't follow this path.

Aside 2: M and K are matrices while I is a tensor. Tensors, like vectors, change in specific ways under coordinate system transformations. Matrices are a more general mathematical construction, a 2 dimensional array of quantities with rules for addition (of two matrices of equal size) and multiplication (of a matrix with n columns into a matrix with n rows). A (second order) tensor may be represented as a matrix, and matrices are a nice way to organize and manipulate systems of equations.

The similarity of the equation does mean that the time dependence in x should be the same for all elements, that is we can write

x = Acosωt,
x = Bsinωt,
or in the generic complex form
x = Acosωt + iBsinωt = Ceiωt,

where A, B, and C are constant, 2×1 column matrices. Inserting this form for a trial solution into the matrix equation yields

-MCω²eiωt = -KCeiωt.

Moving all the terms to the same side of the equation, and factoring out the common C term and canceling eiωt yields

(K - ω²M)C = 0.

This is an eigenvalue equation. In mathematics it is shown that such equations have a non-trivial solution if and only if the determinant of the quantity in parentheses is zero

det(K - ω²M) = 0,

that is, we must find the values of ω² that satisfy the equation. The trivial solution is the one where C = 0, but then there is no motion at all in the system and the problem is uninteresting. For our model system with two masses, the determinant equation is quadratic in ω², yielding two values for ω called normal frequencies (we will take ω>0, since taking ω<0 is equivalent to a change of phase by 180°, not a separate solution). In general, for a system with N masses, there are N normal frequencies and they are found using the same procedure discussed below.

Three Identical Springs and Two Equal Masses

We will now solve some example coupled oscillator problems. To keep this first problem simple, we'll look at the model system with both masses equal, m1 = m2 = m, and idential spring constants k1 = k2 = k3 = k. The matrix M becomes m1, where 1 is the unit matrix, and K becomes k[(2, -1), (-1, 2)]. The determinant expression becomes

det(K - ω²M) = (2k-mω²)² - k² = 3k² - 4kmω² + m²ω4 = (3k - mω²)(k - mω²) = 0

The solutions (eigenvalues) are ω² = 3k/m and ω² = k/m. The normal frequencies are ω1 = &radic(k/m) and ω2 = &radic(3k/m). They are the frequencies at which both carts can oscillate simultaneously -- pay attention to the in-class demonstration with coupled pendula.

The First Normal Mode

To understand what this means, let's complete the picture by determining the eigenvectors for these eigenvalues. The eigenvectors are the values of C that satisfy the eigenvalue equation (K - ω²M)C = 0 for a particular eigenvalue. First let's take ω = ω1 = &radic(k/m). The eigenvalue equation reads

(2k - k)C1 - kC2 = C1 - C2 = 0
-kC1 + (2k - k)C2 = -C1 + C2 = 0

where C1 and C2 are the two components of C. The solution has C1 = C2, so a normalized eigenvector is C1 = √(½) [1, 1]. The position vector is

x = (A1/√2) [1, 1] e1,t.
Therefore, if the two carts oscillate with frequency ω1 = &radic(k/m) then they move in phase with equal amplitude.

The Second Normal Mode

Now let's take ω = ω2 = &radic(3k/m). The eigenvalue equation reads

(2k - 3k)C1 - kC2 = -C1 - C2 = 0
-kC1 + (2k - 3k)C2 = -C1 - C2 = 0

The solution has C1 = -C2, so a normalized eigenvector is C2 = √(½) [1, -1]. The position vector is

x = (A2/√2) [1, -1] e1,t.
Therefore, if the two carts oscillate with frequency ω2 = &radic(3k/m) then they move 180° out of phase with equal amplitude.

The General Solution

The most general motion is a linear combination of these two solutions. Notice that C1 and C2 are orthoganol column matrices ("vectors"), that is C1TC2 = C2TC1 = ½ - ½ = 0. Therefore, they can be thought of as basis vectors in the two dimensional space of the coordinates x1 and x2. Any possible solution can be expressed as a linear combination of our two eigenvector solutions.

© 2007 Robert Harr