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The chain has
particles of unit mass with displacement
for
. The ends of the chain are attached to a rigid
wall. The equation of motion then reads
 |
(1) |
for
and at the ends we have
This system of equations can be rewritten in matrix form as
 |
(4) |
where
is an
tridiagonal matrix with 2's on the diagonal
and
's on the super- and subdiagonals.
The matrix equation is solved by first diagonalizing
. The
eigenvectors and eigenvalues (diagonal values) satisfy
 |
(5) |
for
. These eigenvectors define the normal modes of
vibration. The exact analytic solution gives the eigenvalues
![\begin{displaymath}
\omega^2_k = 2 - 2 \cos[k\pi/(N+1)]
\end{displaymath}](img19.gif) |
(6) |
with eigenvectors proportional to
![\begin{displaymath}
{\bf u}^{(k)}_i \propto \sin[(i+1)k\pi/(N+1)].
\end{displaymath}](img20.gif) |
(7) |
You should normalize each eigenvector so the sum of the squares of the
components is 1. For a given normal mode
we can plot the
components of the eigenvector
vs
. For the
lowest mode
, there are no nodes, for
, there is one
node, etc. You are asked to plot these.
The general solution to the linear system can then be written as a
linear combination of the eigenvectors
 |
(8) |
where the coefficient functions
for the linear system,
only are just
 |
(9) |
The initial values
and
can be obtained from the
initial displacement vector
and initial velocity
from
where we have assumed that the eigenvectors are normalized to 1 and
are real.
The energy of the system is easily found to be
 |
(12) |
and using the orthonormal properties of the eigenvectors, we can
easily show that
 |
(13) |
where
 |
(14) |
The normal mode velocities and amplitudes can be derived from the
diplacement velocities and amplitudes using
Next: Nonlinear Problem
Up: Details
Previous: Details
Carleton DeTar
2005-11-07