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{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 3}\\%
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Solution for assignment 13: Double Pendulum. \bigskip

\noindent Derivation of the Lagrangian:
\begin{eqnarray} \nonumber
  x_1 &=& l\,\sin\phi \\ \nonumber
  y_1 &=& l\,\cos\phi \\\ \nonumber
  x_2 &=& l\,\sin\phi +l\,\sin\psi \\ \nonumber
  y_2 &=& l\,\cos\phi +l\,\cos\psi \\ \nonumber
  L &=& \frac{1}{2}\,m\,l\,\left(\dot{x_1}^2+\dot{y_1}^2\right)+
        \frac{1}{2}\,m\,l\,\left(\dot{x_2}^2+\dot{y_2}^2\right)-
        m\,g\,l\,(z_1+z_2)+{\rm const} \\ \nonumber &=& 
  \frac{1}{2}\,m\,l\,\dot{\phi}^2+
  \frac{1}{2}\,m\,l\,\left(\dot{\phi}^2+\dot{\psi}^2+
  2\,\dot{\phi}\,\dot{\psi}\,\cos(\phi-\psi)\right)+
  m\,g\,l\,\left(2\,\cos\phi+\cos\psi\right)\\ \nonumber 
  &=& m\,l\,\left(\dot{\phi}^2+\frac{1}{2}\,\dot{\psi}^2+
  \dot{\phi}\,\dot{\psi}\,\cos(\phi-\psi)\right)+
  m\,g\,l\,\left(2\,\cos\phi+\cos\psi\right)\ .
\end{eqnarray}
Taylor expansion to second order in $\phi,\,\psi,\,\dot{\phi}$ and
$\dot{\psi}$ gives
$$ L\ =\ m\,l\,\left(\dot{\phi}^2+\frac{1}{2}\,\dot{\psi}^2+
  \dot{\phi}\,\dot{\psi}\right)-
  m\,g\,l\,\left(\phi^2+\frac{1}{2}\,\psi^2\right)+3\,m\,g\,l\ .$$
The Euler-Lagrange equation are then
\begin{eqnarray} \nonumber
  0\ =\ \frac{d\ }{dt}\,\frac{\partial L}{\partial\dot{\phi}} -
  \frac{\partial L}{\partial\phi} &=& 2\,\ddot{\phi}+\ddot{\psi} 
  + 2\,\frac{g}{l}\,\phi \\ \nonumber
  0\ =\ \frac{d\ }{dt}\,\frac{\partial L}{\partial\dot{\psi}} -
  \frac{\partial L}{\partial\psi} &=&~\,\ddot{\psi}+\ddot{\phi} 
  \,+~\,\frac{g}{l}\,\psi
\end{eqnarray}
and 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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