\begin{problem*}{14.8}
Two loudspeakers are placed on a wall $2.00\U{m}$ apart.  A listener
stands $3.00\U{m}$ from the wall directly in front of one of the
speakers.  A single oscillator is driving the speakers at a frequency
of $300\U{Hz}$.  \Part{a} What is the phase difference between the two
waves when they reach the observer?  \Part{b} What is the frequency
closest to $300\U{Hz}$ to which the oscillator may be adjusted so that
the observer hears minimal sound?
\end{problem*} % problem 14.8

\begin{solution}
\begin{center}
\begin{asy}
import Mechanics;

real u=1cm;
pair SpeakerA = (0,0);
pair SpeakerB = (0,2u);
pair Listener = (3u,0);

// Extra label text for spacing with inline asymptote
Distance dAB = Distance(SpeakerA, SpeakerB, Label("$L=2\U{m}$", "L=2 m"));
Distance dAL = Distance(Listener, SpeakerA, Label("$d_a=3\U{m}$", "da=3 m"));
Distance dBL = Distance(SpeakerB,Listener,
                        Label("$\qquad\qquad\qquad\qquad\qquad d_b=\sqrt{L^2+d_a^2}=3.606\U{m}$",
                              "$d_b=\sqrt{L^2+d_a^2}=3.606mm$"));

dot(SpeakerA);
label("$S_a$", SpeakerA, W);
dot(SpeakerB);
label("$S_b$", SpeakerB, W);
dot(Listener);
label("Listener", Listener, E);
dAB.draw();
dAL.draw(rotateLabel=false);
dBL.draw(rotateLabel=false);
\end{asy}
\end{center}
\Part{a}
From Table 13.1 we find that the speed of sound in air at $20\dg C$ is
$v=343\U{m/s}$.  The wavelength of this sound is
\begin{equation}
  \lambda = \frac{v}{f} = 1.143\U{m} 
\end{equation}
The phase change from speaker $S_a$ is therefore
\begin{equation}
  \theta_a = k d_a = \frac{2\pi d_a}{\lambda} = 16.49\U{rad} \;,
\end{equation}
and from speaker $S_b$ is
\begin{equation}
  \theta_b = k d_b = \frac{2\pi d_b}{\lambda} = 19.81\U{rad} \;.
\end{equation}
The phase difference is
\begin{equation}
  \Delta\theta = \theta_b - \theta_a = \ans{3.33\U{rad}}
\end{equation}

\Part{b}
For minimal sound, we want the phase difference to be exactly $\pi$
(or some odd multiple of $\pi$).  We see that it's already close to
$\pi$ with our initial frequency of $300\U{Hz}$, only a bit high.
Decreasing the freqency a bit will reduce the rate of dephasing
between the two waves, reducing $\Delta\theta$, so we're looking for a
frequency slightly less than $300\U{Hz}$.
\begin{align}
  \Delta\theta &= \theta_b-\theta_a
    = \frac{2\pi (d_b-d_a)}{\lambda}
    = \frac{2\pi f (d_b-d_a)}{v} \\
  f &= \frac{v\Delta\theta}{2\pi (d_b-d_a)}
    = \frac{v \pi}{2\pi (d_b-d_a)}
    = \frac{v}{2(d_b-d_a)}
    = \frac{343\U{m/s}}{2\cdot 0.606\U{m}} = \ans{283\U{Hz}} \;.
\end{align}
\end{solution}
