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%%%%%%%%%%%%%%%%% ADVANCED DYNAMICS %%%%%%%%%%%%%%%%
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{\bf SOLUTIONS MIDTERM ADVANCED DYNAMICS --- PHY--4241/5227}\\
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\centerline{\bf PROBLEM 1 (20 points)}
\begin{enumerate}
\item Yes, because $L$ does not depend explicitly on time.
\item $$ \vec{R} = \frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}\,.$$
The center of mass frame is defined by $\vec{R}=0$.
\item $$\vec{r}_1-\vec{r}_2+\left(\frac{m_1+m_2}{m_2}\right)\vec{R}=+
\vec{r}_1+\frac{m_1}{m_2}\vec{r}_1=\frac{m_1+m_2}{m_2}\vec{r}_1\,,$$
$$ \vec{r}_1-\vec{r}_2-\left(\frac{m_1+m_2}{m_1}\right)\vec{R}= -
\vec{r}_2-\frac{m_2}{m_1}\vec{r}_2=\frac{m_1+m_2}{m_1}\vec{r}_2\,,$$
Therefore, in the center of mass frame
$$ \vec{r}_1 = \frac{m_2\,\vec{r}}{m_1+m_2}~~~{\rm and}~~~
   \vec{r}_2 =-\frac{m_1\,\vec{r}}{m_1+m_2}\,.$$
The Lagrangian becomes
$$ L = \mu\,\dot{\vec{r}}^{\,2} - \frac{k}{2}\,\vec{r}^{\,2}
   ~~{\rm with}~~\mu=\frac{m_1m_2}{m_1+m_2}~~{\rm reduced\ mass}\,.$$
\item This is the 3D harmonic oscillator, which reads in Cartesian
coordinates.
$$L=\sum_{i=1}^3\left(\mu\,\dot{x_i}^{\,2}-\frac{k}{2}\,x_i^2\right)$$
and the Euler-Lagrange euations read
$$ 0 = \frac{d~}{dt} \frac{\partial L}{\partial\dot{x}_j} - 
   \frac{\partial L}{\partial x_j} = \mu\,\ddot{x}_j - k\,x_j\,,
   ~~~j=1,2,3\,.  $$
\end{enumerate}
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\centerline{\bf PROBLEM 2 (20 points)}

See posted midterm solutions.
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\centerline{\bf PROBLEM 3 (10 points)}

See solution for assignment~30 (set 9).
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