Adventures in Parallel Dimension #1729 II

The speakers are in phase when they emit sound, but since sound takes time to reach its destination, a path difference will cause the sound waves to be out of phase. Let's call the phase difference $\phi$ , and the path difference $\delta$ . Then $\phi=2\pi \frac{\delta}{\lambda}$ .

It is trivial that the point on the couch directly in front of the center of the TV will have maximum constructive interference, and thus is our $0^{th}$ order principal maximum. Using this as our base point, call this $y=0$ , where $y$ is the distance to the right of the $0^{th}$ order maximum while facing the TV. We can then use Pythagorean Theorem to determine the relative phase difference as follows;

$\begin{array}{c}\\ y_{\text{speaker 1}}=50\text{m}, \quad y_{\text{speaker 2}}=-50\text{m} \\ \delta=\sqrt{100+(y+50)^{2}}-\sqrt{100+(y-50)^{2}} \\ \phi=2\pi \left( \sqrt{100+(y+50)^{2}}-\sqrt{100+(y-50)^{2}} \right). \end{array}$

In order to obtain complete destructive interference, the two speakers must have a relative phase difference of $\phi=\pi+2\pi k$ for some $k \in \mathbb{Z}$ . We get;

$\begin{array}{c}\\ \frac{1}{2}+k= \sqrt{100+(y+50)^{2}}-\sqrt{100+(y-50)^{2}}. \end{array}$

Note that $k(y)$ is a strictly increasing function over the range used in the problem; $-500\leq y \leq 500$ ; so we can obtain all possible values of $k$ by looking at the extreme values of $k$ and then counting the number of integers that lie between those extrema.

$\begin{array}{c}\\ k(500)=\sqrt{100+(500+50)^{2}}-\sqrt{100+(500-50)^{2}}-\frac{1}{2}\approx 99.47 \\ k(-500)=\sqrt{100+(-500+50)^{2}}-\sqrt{100+(-500-50)^{2}}-\frac{1}{2}\approx -100.47. \end{array}$

Hence our range of integral $k$ is $-100 \leq k \leq 99$ , and so there are $\boxed{200}$ possible values of $k$ , each of which correspond to a point of complete destructive interference.