In Taylor, read sections 11.5 for today, and 11.6 for Friday.
For two masses and three springs we found
where M is the diagonal matrix [(m1, 0), (0, m2)] and K is the matrix [((k1 + k2), -k2), (-k2, (k2 + k3))].
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 translates 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
The kinetic energy is given by
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
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
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, qi qj, 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
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
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
allowing us to write the kinetic energy as
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, including generalized forces, 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
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
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
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
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
With this simplification the kth equation of motion reads
In matrix format this looks like
As we did with the pair of oscillators, try a solution of the form q = aeiωt, yielding the eigenvalue equation
which has a non-trivial solution only if the characteristic equation is satisfied
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.