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 *k*_{2}, and with the ends attached to fixed walls by springs of constant *k*_{1}. 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: *k*_{1}=*k*_{2}=*k*. Then and , and our matrices become:

and

.

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 *cλ*. 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 *λ _{i}*,

*i*=1,2,…,

*n*be the eigenvalues of

*C*, then the eigenvalues

*λ’*of A are given by , with the same corresponding eigenvectors.

_{i}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

*U*(

_{n}*x*). The identity (see equation (18) here) tells us that

, and thus our characteristic equation becomes

Now, we use the trigonometric definition of

*U*(

_{n}*x*): . Thus, to find the n zeroes

*x*,

_{n,k}*k*=1,2,…,

*n*, of

*U*(

_{n}*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

*x*in decreasing order, , with

_{n,k}*k*=1,2,…,

*n*. Thus, we see that the eigenvalues

*λ*of

_{i}*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 -4

*ω*

^{2}<

*λ’*<0,

_{i}*i*=1,2,…,

*n*. Thus, our characteristic modes are all oscillatory, with frequencies

*ω*,

_{i}*i*=1,2,…,

*n*, in increasing order, given by

,

*i*=1,2,…,

*n*.

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

Tags: Chebyshev Polynomials, Coupled Oscillator, Eigenvalues, Eigenvectors, Friday Physics, physics, Vibrational Modes

July 18, 2008 at 12:07 am |

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

September 14, 2008 at 11:00 pm |

[…] – bookmarked by 1 members originally found by xmcrxangelx on 2008-09-10 Physics Friday 28 https://twistedone151.wordpress.com/2008/07/11/physics-friday-28/ – bookmarked by 4 members […]