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{\bf Question}

A model of a linear molecule consists of four masses $m$, $2m$,
$2m$, $m$ connected by springs with spring constant $k$.  Denoting
the displacement of the masses from their equlibrium positions by
$x_1, x_2, x_3, x_4$ respectivily, the equations can be written in
the matrix form as:

$$\frac{d}{dt^2}{\bf x} = A{\bf x}$$

where

$$ {\bf x} = \left( \begin{array}{r} x_1  \\ x_2 \\x_3
\\ x_4 \end{array} \right),\
A = \left(\begin{array}{rrrr} {-2b} & {2b} & 0 & 0 \\ b & {-2b}& b
& 0 \\ 0 & 0 & {2b} & {-2b} \\ \end{array} \right), \   {\rm and}\
b = \frac{k}{2m} $$

The matrix A has eigenvalues $0, -b, -3b$ and $-4b$.  Find all the
corresponding eigenvectors of A, and hence show that the equations
have three types of oscillatory solution.

\begin{description}
\item[(i)]
oscillations of frequency $\frac{1}{2\pi} \sqrt {\frac{k}{2m}}$
when $x_2 = \frac {1}{2}x_1$, $x_3 = -\frac {1}{2}x_1$, $x_4 =
-x_1$,
\item[(ii)]
oscillations of frequency $\frac{1}{2\pi} \sqrt {\frac{3k}{2m}}$
when $x_2 = x_3 = -\frac {1}{2}x_1$, $x_4 = x_1$,
\item[(iii)]
oscillations of frequency $\frac{1}{\pi} \sqrt {\frac{k}{2m}}$
when $x_2 = -x_1$, $x_3 =x_1$, $x_4 = -x_1$,
\end{description}

{\bf Answer} \vspace{.2in}

$\displaystyle \left( \begin{array}{c} \frac{d^2x_1}{dt^2} \\
\frac{d^2x_2}{dt^2} \\\frac{d^2x_3}{dt^2}
\\ \frac{d^2x_4}{dt^2}
\end{array} \right) = \left( \begin{array}{cccc} -2b & 2b & 0 & 0
\\ b & -2b & b & 0 \\ 0 & b & -2b & b \\ 0 & 0 & 2b & -2b \end{array} \right)
\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}
\right) $

Eigenvalues are: 0, -b, -3b, -4b.

To find eigenvectors:

$ \underline {\lambda = 0}$ \hspace{.2in} Solve $\left(
\begin{array}{cccc} -2b & 2b & 0 & 0
\\ b & -2b & b & 0 \\ 0 & b & -2b & b \\ 0 & 0 & 2b & -2b \end{array} \right)
\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}
\right) = \left(
\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$

or $\left. \begin{array}{rcl} {-2bx_1 + 2bx_2} & = & 0 \\ {bx_1 -
2bx_2 + bx_3} & = & 0 \\ {bx_2 - 2bx_3 + bx_4} & = & 0 \\ {2bx_3 -
2bx_4 } & = & 0 \end{array} \right\}
\begin{array}{lrcl} {\rm let\ \ } & x_1 & = & \alpha \\ {\rm so\ \ }
& x_2 & = & \alpha \\{\rm then\ } & x_3 = 2x_2 - x_1 = \alpha \\
{\rm and\ } & x_4 = \alpha \\ \end{array}$

Suitable eigenvector $\alpha \left( \begin{array}{c} 1 \\ 1 \\ 1
\\ 1 \end{array} \right)$

$ \underline {\lambda = -b}$ \hspace{.2in} Solve $\left(
\begin{array}{cccc} -b & 2b & 0 & 0
\\ b & -b & b & 0 \\ 0 & b & -b & b \\ 0 & 0 & 2b & -b \end{array} \right)
\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}
\right) = \left(
\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$

or $\left. \begin{array}{rcl} {-bx_1 + 2bx_2} & = & 0 \\ {bx_1 -
bx_2 + bx_3} & = & 0 \\ {bx_2 - bx_3 + bx_4} & = & 0 \\ {2bx_3 -
bx_4 } & = & 0 \end{array} \right\}
\begin{array}{lrcl} {\rm let\ \ } & x_2 & = & \beta \\ {\rm so\ \ }
& x_1 & = & 2\beta \\{\rm then\ } & x_3 = x_2 - x_1 = -\beta \\
{\rm and\ } & x_4 = -2\beta \\ \end{array}$

Suitable eigenvector $\beta \left( \begin{array}{c} 2 \\ 1 \\ -1
\\ -2 \end{array} \right)$


$ \underline {\lambda = -3b}$ \hspace{.2in} Solve $\left(
\begin{array}{cccc} b & 2b & 0 & 0
\\ b & b & b & 0 \\ 0 & b & b & b \\ 0 & 0 & 2b & b \end{array} \right)
\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}
\right) = \left(
\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$

or $\left. \begin{array}{rcl} {bx_1 + 2bx_2} & = & 0 \\ {bx_1 +
bx_2 + bx_3} & = & 0 \\ {bx_2 + bx_3 + bx_4} & = & 0 \\ {2bx_3 -
bx_4 } & = & 0 \end{array} \right\}
\begin{array}{lrcl} {\rm let\ \ } & x_2 & = & \gamma \\ {\rm so\ \ }
& x_1 & = & -2\gamma \\{\rm then\ } & x_3 = -x_1 - x_2 = \gamma
\\ {\rm and\ } & x_4 = -2\gamma \\ \end{array}$

Suitable eigenvector $\gamma \left( \begin{array}{c} -2 \\ 1 \\ 1
\\ -2 \end{array} \right)$


$ \underline {\lambda = -4b}$ \hspace{.2in} Solve $\left(
\begin{array}{cccc} 2b & 2b & 0 & 0
\\ b & 2b & b & 0 \\ 0 & b & 2b & b \\ 0 & 0 & 2b & 2b \end{array} \right)
\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}
\right) = \left(
\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$

or $\left. \begin{array}{rcl} {2bx_1 + 2bx_2} & = & 0 \\ {bx_1 +
2bx_2 + bx_3} & = & 0 \\ {bx_2 + 2bx_3 + bx_4} & = & 0 \\ {2bx_3 +
2bx_4 } & = & 0 \end{array} \right\}
\begin{array}{lrcl} {\rm let\ \ } & x_1 & = & \delta \\ {\rm so\ \ }
& x_2 & = & -\delta \\{\rm then\ } & x_3 = -x_1 - 2x_2 = \delta \\
{\rm and\ } & x_4 = -\delta \\ \end{array}$

Suitable eigenvector $\delta \left( \begin{array}{c} 1 \\ -1 \\ 1
\\ -1 \end{array} \right)$

The solutions are therefore:
\begin{enumerate}
\item
Eigenvalue $\lambda = 0$ and eigenvector ${\bf x} = \left(
\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \right)$ gives a
solution $${\bf x} = \left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1
\end{array} \right)(B + Ct)$$ (This solution is non - oscillatory)
\item
Eigenvalue $\lambda = -b$ and eigenvector ${\bf x} = \left(
\begin{array}{c} 2 \\ 1 \\ -1 \\ -2 \end{array} \right)$ gives a
solution $${\bf x} = \left( \begin{array}{c} 2 \\ 1 \\ -1 \\ -2
\end{array} \right)(D\cos \sqrt b t + E\sin \sqrt b t)$$

which oscillates with frequency $\displaystyle
\frac{\sqrt{b}}{2\pi}=\frac{1}{2\pi} \sqrt{\frac{k}{2m}}.$

(Recall: A solution of the form (${\bf x} = D\cos \sqrt b t +
E\sin \sqrt b t$ oscillates with time period $T =
\frac{2\pi}{\omega}$ and frequency $ f = \frac{1}{T} =
\frac{\omega}{2\pi}$ )
\item
Eigenvalue $\lambda = -3b$ and eigenvector ${\bf x} = \left(
\begin{array}{c} -2 \\ 1 \\ 1 \\ -2 \end{array} \right)$ gives a
solution $${\bf x} = \left( \begin{array}{c} -2 \\ 1 \\ 1 \\ -2
\end{array} \right)(F\cos \sqrt {3b} t + G\sin \sqrt {3b} t)$$
which oscillates with frequency $\displaystyle
\frac{\sqrt{3b}}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{3k}{2m}}$
\item
Eigenvalue $\lambda = -4b$ and eigenvector ${\bf x} = \left(
\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array} \right)$ gives a
solution $${\bf x} = \left( \begin{array}{c} 1 \\ -1 \\ 1 \\ -1
\end{array} \right)(H\cos 2 \sqrt {b} t + J\sin 2 \sqrt {b} t)$$
which oscillates with frequency $\displaystyle
\frac{2\sqrt{b}}{2\pi} = \frac{1}{\pi}\sqrt{\frac{k}{2m}}.$
\end{enumerate}

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