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% BB Mar 10 2010 \documentclass[12pt]{article} \textheight 9.5in \textwidth 6.0in \topmargin -0.7in \headsep 0.4in \oddsidemargin 0.25in \evensidemargin 0.25in \def\pmbf#1{{\mbox{\boldmath${#1}$}}} \def\undertext#1{$\underline{\smash{\hbox{#1}}}$} \pagestyle{empty} \begin{document} %----------------------------------------------------% %%%%%%%%%%%%%%%%% ADVANCED DYNAMICS %%%%%%%%%%%%%%%% %----------------------------------------------------% \begin{center} {\bf SOLUTIONS MIDTERM ADVANCED DYNAMICS --- PHY--4241/5227}\\ \end{center} \setcounter{equation}{0} \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%% \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 $$\vec{r}_1-\vec{r}_2+\left(\frac{m_1+m_2}{m_2}\right)\vec{R}=+ \vec{r}_1+\frac{m_1}{m_2}\vec{r}_1=\frac{m_1+m_2}{m_2}\vec{r}_1\,,$$ $$ \vec{r}_1-\vec{r}_2-\left(\frac{m_1+m_2}{m_1}\right)\vec{R}= -
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Unformatted text preview: \vec{r}_2-\frac{m_2}{m_1}\vec{r}_2=\frac{m_1+m_2}{m_1}\vec{r}_2\,,$$ 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} %%%%%%%%%%%%%%%%%%%%%%%%%% \centerline{\bf PROBLEM 2 (20 points)} See posted midterm solutions. \hfil\break %%%%%%%%%%%%%%%%%%%%%%%%%% \centerline{\bf PROBLEM 3 (10 points)} See solution for assignment~30 (set 9). %%%%%%%%%%%%%%%%%%%%%%%%%% \vfill \end{document}...
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