Problem 13.13

13.13. a)
To find the phase difference, plug in the values given.
$\phi_{1}(5,2) = (20)(5) - (30)(2) ~=~ 40 rad$
$\phi_{2}(5,2) = (25)(5) - (40)(2) ~=~ 45 rad$
Therefore, we can find the phase difference.
$\Delta \phi = \phi_{2} - \phi_{1}$
$\Delta \phi = 5 rad = -73.5{}^{\circ}$

b)
We want to know where $\Delta \phi$ is an odd multiple of $\pi$. Subtracting the two equations for $\phi$, we have
$\Delta \phi = \vert(20x - 30t) - (25x - 40t)\vert = \pm n\pi$
$\vert-5x + 10t\vert = \pm n\pi$
Plugging in when t = 2,
$\vert-5x + 20\vert = \pm n\pi$
We want a value close to the origin, so lets limit ourselves to the case x < 4, then the absolute value comes out of the equation, and we have
$-5x + 20 = \pm n\pi$
The value of $\pm n\pi$ that will minimize the left hand side of our equation is when n = 5
$x = 4 \pm \pi$
x = 7.14 or x = .858
We conclude the value closest to the origin, or
x = .858 cm