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\newcommand{\ds}{\displaystyle}
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{\bf Question}

A particle of mass $m$ hangs vertically on the end of a spring of
stiffness $k$ and natural length $l$.  The particle is displaced
vertically downwards a distance $\frac{l}{2}$ and released from
rest.
\begin{description}
\item[(a)] What is the maximum height achieved  by the particle in
the subsequence motion?
\item[(b)] Show that the particle oscillates up and down with a
frequency $\ds \sqrt{\frac{k}{m}}.$
\end{description}
\vspace{.25in}

{\bf Answer}

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\bigskip
Using Newton's 2nd law:
\begin{eqnarray*} m \ddot x & = & mg - k(x - l) \\ \ddot x & =
& g + \frac{kl}{m} - {k}{m} x \hspace{.1in}(*) \\ {\rm Energy:\
K.E. + P.E.} & = & {\rm constant} \\ \frac{1}{2} m \dot x^2 +
\frac{1}{2}k(x - l)^2 & = & 0 + \frac{1}{2} k\left(\frac{3}{2}l -
l\right)
\end{eqnarray*}

Since initially $\dot x = 0$ and $x = \frac{3}{2}l$

\begin{description}
\item[(a)]
At the maximum height  $v = 0.$

Therefore $\ds \frac{1}{2} k(x - l)^2 = \frac{1}{2} k \left(
\frac{l}{2} \right) ^2 \Rightarrow x = \frac{l}{2}$

\item[(b)]
Solving (*) gives: $\ds x = l + \frac{mg}{k} + A \cos
\left(\sqrt{\frac{k}{m}}t+ B\right),$ where A and B are constants
and the frequency is $\ds \sqrt{\frac{k}{m}}$
\end{description}



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