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\begin{document}

\begin{center}
{\bf ADVANCED DYNAMICS --- PHY 4936}\\\vspace {2mm} 
\end{center}
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Solution 27: (A) Let $\lambda = \omega^2$. We have to solve
$$ {\rm det}\, \left|\matrix{5-\lambda&amp;1-\lambda\cr 
               1-\lambda&amp;2(1-\lambda)\cr}\right|\ =\ 0\,,$$
which leads to the quadratic equation
\begin{eqnarray} \nonumber
 0 &amp;=&amp; (5-\lambda)\,2\,(1-\lambda)-(1-\lambda)\,(1-\lambda)\\ \nonumber
   &amp;=&amp; 10-2\,\lambda-10\,\lambda+2\,\lambda^2-1+2\,\lambda-\lambda^2\\
   &amp;=&amp; \lambda^2 - 10\,\lambda + 9 \nonumber
\end{eqnarray}
with the solutions
\begin{eqnarray} \nonumber
  \omega^2_{1,2} &amp;=&amp; \lambda_{1,2}\ =\ 5 \pm\sqrt{5^2-9} = 5\pm 4\\ \nonumber
  \omega^2_1 &amp;=&amp; \lambda_1\ =\ 9\,,~~~ \omega^2_2\ =\ \lambda_2\ =\ 1\,.
\end{eqnarray}
(B) Using normal coordinates 
$\Theta_1={\rm Re}\,\left[C_1\,\exp(i\omega_1t)\right]$ and
$\Theta_2={\rm Re}\,\left[C_2\,\exp(i\omega_2t)\right]$, the 
general solutions are the superpositions
\begin{eqnarray} \nonumber
  x_1&amp;=&amp;\triangle_{11}\,\Theta_1+\triangle_{12}\,\Theta_2\\ \nonumber
  x_2&amp;=&amp;\triangle_{21}\,\Theta_1+\triangle_{22}\,\Theta_2
\end{eqnarray}
It turns out that we have to take the minor of the second row, because
for the first row the $\triangle_{12}$ and $\triangle_{22}$ minors are 
both zero. Solutions are then
\begin{eqnarray} \nonumber
  x_1 &amp;=&amp;  8\,\Theta_1 + 0 \ =\  8\,\Theta_1 \\ \nonumber
  x_2 &amp;=&amp; -4\,\Theta_1 + 4\,\Theta_2\,.
\end{eqnarray}
(C) Substituting these equations into the Lagrangian 
\begin{eqnarray} \nonumber
  L &amp;=&amp; \frac{1}{2}\,\left( \dot x_1\,\dot x_1 + \dot x_1\,\dot x_2 
  + \dot x_2\,\dot x_1 + 2\,\dot x_2\,\dot x_2  
  - 5\,x_1\,x_1 - x_1\,x_2 - x_2\,x_1 - 2\,x_2\,x_2 \right) 
\end{eqnarray}
diagonalizes simultaneously both terms and gives
\begin{eqnarray} \nonumber
  L &amp;=&amp; 16\,\left(\dot\Theta_1\right)^2+16\,\left(\dot\Theta_2\right)^2 
     - 144\,\left(\Theta_1\right)^2-16\,\left(\Theta_2\right)^2\ .
\end{eqnarray}
(D) The initial condition $\Theta_1(0)=1$ and $\dot\Theta_1(0)=0$ implies
$\Theta_1(t)=\cos(\omega_1 t)=\cos(3t)$.

and initial condition $\Theta_2(0)=0$ and $\dot\Theta_1(0)=1$ implies
$\Theta_2(t)=\sin(\omega_2 t)=\sin(t)$.

Plots in the $\Theta_1$, $\Theta_2$ and $x_1$, $x_2$ planes follow.

\begin{figure}[tb]
\vspace{5pc}
 \centerline{\hbox{ \psfig{figure=O2D.pdf,width=18.0cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader}
\end{figure}

\begin{figure}[tb]
\vspace{5pc}
 \centerline{\hbox{ \psfig{figure=O2Dx.pdf,width=18.0cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader}
\end{figure}
 
\end{document}

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