In Taylor, read sections 11.1 to 11.2 for today, and 11.3 for Monday.
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
or
or in the generic complex form
where A, B, and C are constant, 2 × 1 column matrices. We will use the complex exponential form of the solution from here on, with the understanding that one must take the real part of the complex function as the actual solution (or the imaginary part, since either is a legitimate solution). 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.
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 m 1, where 1 is the unit matrix, and K becomes k [(2, -1), (-1, 2)]. The determinant expression becomes
The solutions (eigenvalues) are ω² = 3k/m and ω² = k/m. The normal frequencies are ω1 = √(k/m) and ω2 = √(3k/m). They are the frequencies at which both carts can oscillate simultaneously -- pay attention to the in-class demonstration with coupled pendula.
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 - ω² M) C = 0 for a particular eigenvalue. First let's take ω = ω1 = √(k/m). The eigenvalue equation reads
where C1 and C2 are the two components of C. The equation is satisfied for C1 = C2, so a normalized eigenvector is C1= √(½) [1, 1]. The position vector is
Therefore, if the two carts oscillate with frequency ω1 = √(k/m) then they move in phase with equal amplitude.
Now let's take ω = ω2 = √(3k/m). The eigenvalue equation reads
The equation is satisfied for C1 = -C2, so a normalized eigenvector is C2 = √(½) [1, -1]. The position vector is
Therefore, if the two carts oscillate with frequency ω2 = &radic(3k/m) then they move 180° out of phase with equal amplitude.
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.
So we see that it isn't possible for one cart to move and not the other, that is, it isn't possible to vary x1 without also varying x2. Certain combinations of coordinates do have this property, and they are called normal coordinates. Normal coordinates correspond to combinations of the physical coordinates, and always exist in problems such as this.
In this case, we can replace x1 and x2 by
and
These coordinates are orthogonal, and therefore span the space of all possible values for (x1, x2). For this simple example, it is relatively easy to see that ξ1 corresponds to the first eigenvector and ξ2 to the second, so that for the first normal mode
and for the second
Recall that the A's are complex and we must take the real part as the solution. Only ξ1 oscillates in the first normal mode and only ξ2 in the second (of course physically, both masses move, but in the manner dictated by the linear combinations yielding ξ1 and ξ2). The general motion can be written as a linear combination of these two coordinates. One can think of the motion of the system as composed of the motions of each cart individually, or as composed of the motions in each normal mode, where each normal mode represents a particular coordination of the motions of the two carts (a dance if you will).
To complete the loop, let us write the Lagrangian in terms of the normal coordinates.
The normalized coordinates are uncoupled in the Lagrangian --- the Lagrange for ξ1 will not involve ξ2, and vice versa. And one can readily see that this is the Lagrangian for two harmonic oscillators, one of frequency √(k/m) and the other of frequency √(3k/m), agreeing with what we already found. We see that normalized coordinates are a set of coordinates that yield an uncoupled Lagrangian.