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{\bf Question}

A particle moves under the influence of the force $\ds F = -kx +
\frac{kx^3}{\alpha^2}$, where $k$ and $\alpha $ are constants and
$k>0$.
\begin{description}
\item[(a)] Determine the potential $U(x)$ and discuss the motion.
\item[(b)] What happens if the total energy $\ds E = \frac{k
\alpha^2}{4}?$
\end{description}

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{\bf Answer}

\begin{description}
\item[(a)]
$\ds U = - \int F \, dx = -\int \left( -kx + \frac{kx^3}{\alpha
^2} \right)\, dx$

$\ds U(x) = \frac{1}{2}kx^2 - \frac{kx^4}{4\alpha^2}$

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If the particle has total energy less than $U_{\rm max}$ it will
either oscillate about the origin or go off to  $x =\pm \infty$
depending on the initial position.

Otherwise, if $U > U_{\rm max}$, it will move off to $\pm \infty.$

If $U = U_{\rm max}$ see (b) below.


\item[(b)] If $U = U_{\rm max}$.  Either the motion is in $-\alpha < x <
\alpha,$ in which case after infinite time the particle reaches $x
= \alpha$ or $x = -\alpha$; or if $|x|> \alpha$ the particle moves
off to $x = +\infty$ or $ - \infty$ with increasing speed.


\end{description}


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