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{\bf ADVANCED DYNAMICS --- PHY 4936}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 5}\\%
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Solution for assignment 25: \medskip

Normal Modes of Double Pendulum (continuation of 8).  \bigskip

For small oscillation we have derived the Euler-Lagrange equations,
which read in matrix notation
$$ \left(\matrix{2&1\cr 1&1\cr}\right)\,\left(\matrix{\ddot{\phi}\cr
   \ddot{\psi}\cr}\right) + \left(\matrix{2g/l&0\cr 0&g/l\cr}\right)\,
   \left(\matrix{\phi\cr\psi\cr}\right)\ =\ 0\ . $$
This is solved by the exponential ansatz (physical is the real part
of the solution):
$$ \Phi(t) = \left(\matrix{\phi\cr\psi\cr}\right)\ =\ e^{i\,\omega\,t}
   \left(\matrix{\phi_0\cr\psi_0\cr}\right)~~\Rightarrow~~
   \left(\matrix{\ddot{\phi}\cr\ddot{\psi}\cr}\right)\ =\
   - \omega^2\,e^{i\,\omega\,t} \left(\matrix{\phi_0\cr\psi_0\cr}
   \right) $$
$$ \left(\matrix{-2\omega^2+2g/l&-\omega^2\cr -\omega^2&-\omega^2+g/l
   \cr}\right)\,\left(\matrix{\phi\cr\psi\cr}\right)\ =\ 0\ . $$
$$ 0\ =\ {\rm det}\,\left|\matrix{-2\omega^2+2g/l&-\omega^2\cr -
   \omega^2&-\omega^2+g/l\cr}\right|\ = \omega^4 - \frac{4g}{l}\, 
   \omega^2 + \frac{2g^2}{l^2} $$
with eigenmodes
$$ \omega_{\pm}\ =\ \sqrt{\frac{g}{l}}\,\sqrt{2\pm \sqrt{2}}\ . $$

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