Physics Friday 28

In the previous two posts (here and here), we looked at a collection of masses (2 and 3,respectively) connected in a line by springs of constant k2, and with the ends attached to fixed walls by springs of constant k1. In both cases, we noted that the one-dimensional longitudinal motion could be described via n modes, corresponding to the eigenvalues and eigenvectors of an n×n matrix, where n is the number of masses, and thus position variables, in the system.

In the first problem, we found the matrix to be , where and . Similarly, the next problem had . Now, suppose we let the ending springs have the same constant as the connecting springs: k1=k2=k. Then and , and our matrices become:


We find that we can generalize to a line of n masses with this setup:
for a mass i not on the end (1<i<n), we see:
Similarly, we can examine the end masses, and collectively, we find the resulting n×n matrix to be

or times the n×n symmetric tridiagonal matrix whose diagonal elements are all -2, whose elements one-off from the main diagonal are 1, and all other elements are zero.

Now, we need to do some math. Suppose we have an n×n matrix A, and let v be an eigenvector of A corresponding to eigenvalue λ. Now, for any constant c, we see , so we see that v is an eigenvector of cA with eigenvalue . Thus the eigenvalues of the matrix are simply times the eigenvalues of the tridiagonal matrix ; and both A and B have the same eigenvectors.

Next, suppose we have an n×n matrix B, and let v be an eigenvector of B corresponding to eigenvalue λ. Now, for any constant d, and I the n×n identity matrix, we see
. Thus the eigenvectors of the matrix B are also those of , only the latter has the corresponding eigenvalues increased by d. Thus, let us define , the symmetric tridiangular matrix with zeroes on the diagonal and ones in the near-diagonal terms. If we let λi, i=1,2,…,n be the eigenvalues of C, then the eigenvalues λ’i of A are given by , with the same corresponding eigenvectors.

Now, the eigenvaules of C are the roots of the characteristic equation:
, or
This looks difficult, but there is an important identity involving the Chebyshev polynomials of the second kind Un(x). The identity (see equation (18) here) tells us that
, and thus our characteristic equation becomes

Now, we use the trigonometric definition of Un(x): . Thus, to find the n zeroes xn,k, k=1,2,…,n, of Un(x), we want to find the values of θ, 0≤θ≤2π, for which and . These are just ,
k=1,2,…,n. Taking the cosine to find the zeroes xn,k in decreasing order, , with k=1,2,…,n. Thus, we see that the eigenvalues λi of C are, in decreasing order, , and so the eigenvalues of A are, in decreasing order, .

Now, we see these eigenvalues are all negative: we have -4ω2<λ’i<0, i=1,2,…,n. Thus, our characteristic modes are all oscillatory, with frequencies ωi, i=1,2,…,n, in increasing order, given by

Next week, we examine the eigenvectors, and thus the motion of the masses in the characteristic modes.


Tags: , , , , , ,

2 Responses to “Physics Friday 28”

  1. Physics Friday 29 « Twisted One 151’s Weblog Says:

    […] Friday 29 Last week we found the frequencies for the fundamental modes for the motion of n identical masses, each of […]

  2. Bookmarks about Matrix Says:

    […] – bookmarked by 1 members originally found by xmcrxangelx on 2008-09-10 Physics Friday 28 – bookmarked by 4 members […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: