2 Two traveling sinusoidal waves are described by the wave functions
4 y_1 = (5.00\U{m})\cdot\sin[\pi(4.00x-1200t)]
8 y_2 = (5.00\U{m})\cdot\sin[\pi(4.00x-1200t-0.250)]
10 where $x$, $y_1$, and $y_2$ are in meters and $t$ is in seconds.
11 \Part{a} What is the amplitude of the resultant wave?
12 \Part{b} What is the frequency of the resultant wave?
13 \end{problem*} % problem 14.5
19 = 5.00\U{m}\cdot\p\{{\sin[\pi(4.00x-1200t)]+\sin[\pi(4.00x-1200t-0.250)}\}
20 = 5.00\U{m}\cdot\p[{\sin(\theta)+\sin(\theta-\pi/4)}] \;,
22 where $\theta\equiv\pi(4.00x-1200t)$. So $y_2$ trails $y_1$ by
23 $\pi/4=45\dg$. In terms of the reference circle, that looks like
30 pair q = rotate(-45)*p;
33 draw(scale(u)*unitcircle, dotted);
35 Angle a = Angle(p, (0,0), q, radius=6mm, "$\pi/4$");
37 draw(p--(p+q), blue+dashed);
38 draw(q--(p+q), red+dashed);
41 draw((0,0)--(p+q), green);
47 The amplitude of $y$ is then given by vector addition
49 A = 2\cdot A\cos(\phi/2)
50 = 2\cdot 5.00\U{m}\cdot\cos(\pi/8) = \ans{9.24\U{m}} \;,
52 where $\phi=\pi/4$ is the phase difference between $y_1$ and $y_2$.
55 Both $y_1$ and $y_2$ rotate around the reference circle with a frequency of
57 f = \frac{\omega}{2\pi} = \frac{1200\pi\U{rad/s}}{2\pi\U{rad/cycle}}
60 so $y$ must also rotate around the reference circle at $\ans{600\U{Hz}}$.