\begin{problem*}{12.2}
In an engine, a piston oscillates with simple harmonic motion so that
its position varies according to the expression
\begin{equation}
  x = (5.00\U{cm}) \cos(2t + \pi/6)
\end{equation}
where $x$ is in centimeters and $t$ is in seconds.  At $t = 0$,
find \Part{a} the position of the pistion, \Part{b} its velocity,
and \Part{c} its acceleration.  \Part{d} Find the period and amplitude
of the motion.
\end{problem*}

\begin{solution}
\Part{a}
Simply plugging in $t=0$ into the given position equation we have
\begin{equation}
  x(t=0) = 5.00\U{cm} \cdot \cos(\pi/6) = \ans{4.33\U{cm}}
\end{equation}

\Part{b}
Then we take the derivative to get the velocity
\begin{align}
  v &= \pderiv{t}{x} = -2\U{s$^{-1}$} \cdot 5.00\U{cm} \cdot \sin(2t+\pi/6) \\
  v(t=0) &= -10\U{cm/s} \cdot \sin(\pi/6) = \ans{-5.00\U{cm/s}}
\end{align}

\Part{c}
Taking the derivative again yields the acceleration
\begin{align}
  a &= \pderiv{v}{x} = -2\U{s$^{-1}$}\cdot 10.00\U{cm/s}\cdot \cos(2t+\pi/6) \\
  a(t=0) &= -20\U{cm/s$^2$} \cdot \cos(\pi/6) = \ans{-17.3\U{cm/s$^2$}}
\end{align}

\Part{d}
The period is the time it takes for a complete revolution, i.e.~for a
phase change of $2\pi$, so $\ans{T = \pi\U{s}}$.  The amplitude is the
maximum distance from equilibrium.  Looking at the original equation
for $x(t)$, we realize that $|\cos(\theta)| \le 1$ for all $\theta$,
so the biggest $x$ will ever get is $\ans{A = 5.00\U{cm}}$.
\end{solution}
