Reviewed solutions to problem set #9.
Taylor 11.1 for today and Taylor 11.2-11.4 for Wednesday.
In chapter 5 we studied systems with a single oscillator, subject to damping and driving forces, learning about the behavior of resonance. In this chapter we will consider systems of coupled oscillators and examine the behavior possible in such systems.
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
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
The Lagrangian for the system is
Lagrange's equations yield the following equations of motion
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
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
where A, B, and C are constant, 2×1 column matrices. Inserting this form for a trial solution into the matrix equation yields
Moving all the terms to the same side of the equation, and factoring out the common C term and canceling eiωt yields
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
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.
© 2008 Robert Harr