\begin{problem*}{12.42}
\Part{a} A hanging spring stretches by $35.0\U{cm}$ when an object of
mass $450\U{g}$ is hung on it at rest.  In this situation, we define
its position as $x=0$.  The object is pulled down an additional
$18.0\U{cm}$ and released from rest to oscillate without friction.
What is its position $x$ at a time $84.4\U{s}$ later?  \Part{b} A
hanging spring stretches by $35.5\U{cm}$ when an object of mass
$440\U{g}$ is hung on it at rest.  We define this new position as
$x=0$.  This object is also pulled down an additional $18.0\U{cm}$ and
released from rest to oscillate without friction.  Find its position
$84.4\U{s}$ later.  \Part{c} Why are the answers to parts \Part{a}
and \Part{b} different by such a large percentage when the data are so
similar?  Does this circumstance reveal a fundamental difficulty in
calculating the future?  \Part{d} Find the distance traveled by the
vibrating object in \Part{a}.  \Part{e} Find the distance traveled by
the object in \Part{b}.
\end{problem*}

\begin{solution}
There's a lot of repetition in this problem, so let's do everything
symbolically first.  A stretch of $\Delta x$ due to hanging a mass $m$
implies a spring constant of $k = F/\Delta x = mg/\Delta x$ and an
angular frequency of $\omega = \sqrt{k/m} = \sqrt{g/\Delta{x}}$.
Notice that $\omega$ does not depend on the hanging mass.  At this
point we know everything about how the position changes as a function
of time, namely
\begin{equation}
  x(t) = -A\cos(\omega t) = -A\cos(\sqrt{g/\Delta{x}} t)\;,
\end{equation}
where the $-\cos$ part came from letting up be the positive $x$
direction and noticing that the mass starts at an extreme low point in
its oscillation.  Finally, we note that the total distance traveled as
a function of time is going to be
\begin{equation}
  d = 2A \cdot N_{T/2} + (A+x(t_\text{frac}))
\end{equation}
where $N_{T/2} = floor(2t/T)$ is the number of completed half-periods
and $t_\text{frac} = t - N_{T/2}\cdot T/2$ is the fractional
half-period left over.  The extra $A$ in the right-hand term ensures
that the right-hand term is zero when $t_\text{frac}=0$.  Now we just
have to plug in the two cases\ldots

\Part{a}
\begin{align}
  \Delta x &= 0.350\U{m}  &  A &= 0.180\U{m}      &  t &= 84.4\U{s} \\
  \omega &= 5.29\U{rad/s} &  x(t) &= \ans{-15.8\U{cm}}
\end{align}

\Part{b}
\begin{align}
  \Delta x &= 0.355\U{m}  &  A &= 0.180\U{m}      &  t &= 84.4\U{s} \\
  \omega &= 5.25\U{rad/s} &  x(t) &= \ans{15.9\U{cm}}
\end{align}

\Part{c}
The answers to \Part{a} and \Part{b} are so different because the
angular velocities are slightly different.  The phase starts out the
same but grows slightly faster in \Part{a}.  The difference seems so
large because the phase loops around at $2\pi$, so a small difference
in the ``total phase'' $\omega t$ can produce a large difference in
the ``effective phase'' $\omega t \mod 2\pi$.

This sensisivity to the initial conditions (here to $\Delta x$) is the
same sort of effect that gives rise to chaotic behavior
(see \url{http://en.wikipedia.org/wiki/Chaos_theory#Chaotic_dynamics}).
The simple harmonic oscillator is not chaotic though, because $\Delta
x$ is not a dimension that changes with time.  This means that
predicting the future for a simple harmonic oscillator is not
\emph{that} difficult.  You may have to keep adjusting your model
parameters as you get more data, but with time your model will get
better and better.

Compare this system to one that \emph{is} chaotic, e.g.~the
damped-forced pendulum.  In this situations, the sensitivity to
initial conditions also involves the changing parameters, like the
position of the pendulum.  Not only does your model have to be
perfect, but your understanding of where the pendulum is now must also
be perfect.  This is a much more difficult problem, since the position
of the pendulum is changing with time, so you can't just wait and
aquire a arbitarily precise time average.  This is why it is so hard
to make long term predictions for chaotic systems.

\Part{d}
\begin{align}
  T &= \frac{2\pi}{\omega} = 1.19\U{s}  &  N_{T/2} &= 142
   &  t_\text{frac} &= 93.9\U{ms} \\
  d &= 51.1\U{m} + 2.17\U{cm} = \ans{51.1\U{m}}
\end{align}

\Part{e}
\begin{align}
  T &= \frac{2\pi}{\omega} = 1.20\U{s}  &  N_{T/2} &= 141
   &  t_\text{frac} &= 91.7\U{ms} \\
  d &= 50.8\U{m} + 2.05\U{cm} = \ans{50.8\U{m}}
\end{align}
Notice that the distance $d$ doesn't loop back on itself like the
position $x$, so the small relative difference in ``total phase''
leads to a small relative difference in distance between \Part{d}
and \Part{e}.
\end{solution}
