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{\bf ADVANCED DYNAMICS --- PHY--4241/5227}\\\vspace {2mm} 
{\bf Midterm Exam Solutions}\\%
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\centerline{\bf PROBLEM 1}
\noindent
From all possible paths that a physical system can take in going from 
point $x_1$ at time $t_1$ to point $x_2$ at time $t_2$ it will select
the one that minimizes the action 
$$ S = \int_{t_1}^{t_2} dt\,L = \int_{t_1}^{t_2} dt\,(T-V) $$
under local variations of the path. The Euler-Lagrange equations 
follow. In our case:
$$
  \frac{d~}{dt} \frac{\partial L}{\partial\dot x} 
  = \frac{\partial L}{\partial x}\ \Rightarrow\ 
  m\ddot x = - V'(x) = F(x)
$$
in agreement with Newton's second law.
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\centerline{\bf PROBLEM 2}
\noindent
(1) The momentum is
$$ p = \frac{\partial L}{\partial\dot x} = m \dot x$$
\noindent
(2) Eliminating $\dot x$ in favor of $p$ we obtain the Hamiltonian
$$ H = T + V = \frac{p^2}{2m} + \frac{1}{2} k x^2 $$
\noindent
(3) Hamilton's equations are
$$ \frac{\partial H}{\partial p} = \frac{p}{m} = \dot x~~{\rm and}~~ 
   \frac{\partial H}{\partial x} = k x = - \dot p $$
\noindent
(4) Newton's force law follows
$$ k x = - \dot p = -m \ddot x~~{\rm or}~~ m \ddot x = - k x $$
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\centerline{\bf PROBLEM 3}
\noindent
The given Lagrangian is independent of $t$ and $\phi$. Thus,
$$
  \frac{\partial L}{\partial t} = 0\ \Rightarrow\ E=T+V
  ~~{\rm conserved.}
$$
and
$$
  0 = \frac{\partial L}{\partial \phi} = \frac{d~}{dt}
  \frac{\partial L}{\partial\dot\phi} = 
  m r^2 \sin^2(\theta) \dot\phi = p_{\phi}
  ~~{\rm conserved.}
$$
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