Linear Problem

The chain has $N$ particles of unit mass with displacement $x_i$ for $i = 0,\ldots{},N-1$. The ends of the chain are attached to a rigid wall. The equation of motion then reads

\frac{d^2x_i(t)}{dt^2} = x_{i+1}(t) + x_{i-1}(t) - 2x_i(t)
\end{displaymath} (1)

for $i = 1,\ldots{},N-2$ and at the ends we have
$\displaystyle \frac{d^2x_0(t)}{dt^2}$ $\textstyle =$ $\displaystyle x_1(t) - 2x_0(t)$ (2)
$\displaystyle \frac{d^2x_{N-1}(t)}{dt^2}$ $\textstyle =$ $\displaystyle x_{N-2}(t) - 2x_{N-1}(t)$ (3)

This system of equations can be rewritten in matrix form as
\frac{d^2{\bf x}(t)}{dt^2} = -A{\bf x}(t)
\end{displaymath} (4)

where $A$ is an $N\times N$ tridiagonal matrix with 2's on the diagonal and $-1$'s on the super- and subdiagonals.

The matrix equation is solved by first diagonalizing $A$. The eigenvectors and eigenvalues (diagonal values) satisfy

A{\bf u}^{(k)} = \omega^2_k {\bf u}^{(k)}
\end{displaymath} (5)

for $k = 1,\ldots{},N$. These eigenvectors define the normal modes of vibration. The exact analytic solution gives the eigenvalues
\omega^2_k = 2 - 2 \cos[k\pi/(N+1)]
\end{displaymath} (6)

with eigenvectors proportional to
{\bf u}^{(k)}_i \propto \sin[(i+1)k\pi/(N+1)].
\end{displaymath} (7)

You should normalize each eigenvector so the sum of the squares of the components is 1. For a given normal mode $(k)$ we can plot the components of the eigenvector ${\bf u}^{(k)}_i$ vs $i$. For the lowest mode $k = 1$, there are no nodes, for $k = 2$, 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

{\bf x}(t) = \sum_{k = 1}^{N}a_k(t){\bf u}^{(k)}
\end{displaymath} (8)

where the coefficient functions $a_k(t)$ for the linear system, only are just
a_k(t) = a_k(0)\cos(\omega_k t) + \dot a_k(0)\sin(\omega_k t)/\omega_k.
\end{displaymath} (9)

The initial values $a_k(0)$ and $\dot a_k(0)$ can be obtained from the initial displacement vector ${\bf x}(0)$ and initial velocity ${\bf v}(0)$ from
$\displaystyle a_k(0)$ $\textstyle =$ $\displaystyle {\bf u}^{(k)} \cdot {\bf x}(0)$ (10)
$\displaystyle \dot a_k(0)$ $\textstyle =$ $\displaystyle {\bf u}^{(k)} \cdot {\bf v}(0)$ (11)

where we have assumed that the eigenvectors are normalized to 1 and are real.

The energy of the system is easily found to be

E = \frac{1}{2}{\bf v}\cdot{\bf v} + \frac{1}{2}{\bf x}\cdot A{\bf x}
\end{displaymath} (12)

and using the orthonormal properties of the eigenvectors, we can easily show that
E = \sum_{k=1}^{N} E_k
\end{displaymath} (13)

E_k = \frac{1}{2}\dot a_k(t)^2 + \frac{1}{2}\omega_k^2    a_k(t)^2.
\end{displaymath} (14)

The normal mode velocities and amplitudes can be derived from the diplacement velocities and amplitudes using
$\displaystyle a_k(t)$ $\textstyle =$ $\displaystyle {\bf u}^{(k)} \cdot {\bf x}(t)$  
$\displaystyle \dot a_k(t)$ $\textstyle =$ $\displaystyle {\bf u}^{(k)} \cdot {\bf v}(t)$ (15)