# PHY6200 W07

## Chapter 11: Coupled Oscillators and Normal Modes

### Recall

### 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 **x**dot is the column vector (x1dot, x2dot), and **x**ddot is the column vector (x1ddot, x2ddot).
The equations of motion can be written in the form

**M****x**ddot = -**K****x**

where **M** is the diagonal matrix [(m1, 0), (0, m2)] and **K** is the matrix [((k1 + k2), -k2), (-k2, (k2 + k3))].

### Three Identical Springs and Two Equal Masses

### Two Weakly Coupled Oscillators: One Spring Different from the other Two

### Lagrangian Approach: The Double Pendulum

### The General Case

### Three Coupled Pendula

### Normal Coordinates*

© 2007 Robert Harr