In Physics 2 you saw that overlapping waves add: crest on crest builds up, crest on trough cancels. With wave functions and the sum formulas of trigonometry we can say exactly how much. The key quantity is the phase difference $\phi$ between the two waves.
Key idea
Two waves of equal amplitude $A$ and frequency, differing in phase by $\phi$, add to a wave of amplitude $$A_{\text{res}} = 2A\left|\cos\frac{\phi}{2}\right|.$$ A path difference $\Delta r$ produces the phase difference $$\phi = \frac{2\pi\,\Delta r}{\lambda}.$$ Constructive: $\phi = 0, 2\pi, 4\pi, \ldots$ ($\Delta r = m\lambda$). Destructive: $\phi = \pi, 3\pi, \ldots$ ($\Delta r = (m + \tfrac{1}{2})\lambda$).
Why it works
Use $\sin a + \sin b = 2\sin\frac{a + b}{2}\cos\frac{a - b}{2}$ on $y_1 = A\sin(kx - \omega t)$ and $y_2 = A\sin(kx - \omega t + \phi)$: $$y_1 + y_2 = \underbrace{2A\cos\frac{\phi}{2}}_{\text{amplitude}}\ \sin\!\left(kx - \omega t + \frac{\phi}{2}\right).$$ The sum is the same kind of wave, with an amplitude that depends only on $\phi$. And why $\phi = 2\pi\Delta r/\lambda$? The phase of a wave changes by $2\pi$ for every wavelength traveled, so an extra distance $\Delta r$ adds $\Delta r/\lambda$ cycles, which is $2\pi\Delta r/\lambda$ radians.
Worked examples
Worked example: Amplitude from a phase
Two waves on a string, each of amplitude 4.0 mm, differ in phase by $\pi/3$. Find the amplitude of their sum.
- $A_{\text{res}} = 2A\cos\frac{\phi}{2} = 2(4.0\ \text{mm})\cos\frac{\pi}{6}$.
- $= 8.0 \times 0.866 = 6.9$ mm.
Answer: $6.9$ mm
Worked example: Two speakers
Two speakers driven in phase play 680 Hz ($v = 343$ m/s). You stand 3.00 m from one and 3.38 m from the other. How loud is it compared with one speaker alone?
- $\lambda = v/f = 343/680 = 0.504$ m.
- $\Delta r = 0.38$ m, so $\phi = 2\pi\frac{0.38}{0.504} = 4.73$ rad (0.754 of a cycle).
- Amplitude: $2A|\cos(4.73/2)| = 2A \times 0.716 = 1.43A$.
- Intensity goes as amplitude squared: $(1.43)^2 = 2.05$ times one speaker's.
Answer: About 2.0 times the intensity of one speaker
Worked example: Speakers out of phase
Two speakers are wired so that they push opposite ways (a built-in phase difference of $\pi$). A listener is 0.50 m farther from one than the other. What is the lowest frequency that is loudest there? Use $v = 343$ m/s.
- Total phase difference: $\phi = \frac{2\pi\Delta r}{\lambda} + \pi$.
- Loudest needs $\phi = 2\pi$ (the smallest positive multiple), so $\frac{2\pi\Delta r}{\lambda} = \pi$, which means $\Delta r = \lambda/2$.
- $\lambda = 1.00$ m, so $f = v/\lambda = 343$ Hz.
Answer: $343$ Hz
Common mistake
Calling a half-wavelength path difference constructive. A path difference is a distance; the phase difference is $2\pi$ times the number of wavelengths in it. $\Delta r = \lambda/2$ gives $\phi = \pi$: crest meets trough, the most destructive case. Constructive needs whole wavelengths, $\Delta r = 0, \lambda, 2\lambda, \ldots$. Check by drawing two sine waves shifted by half a wavelength: they are mirror images.
Common mistake
Forgetting a phase difference built into the sources. Speakers wired backward, or a reflection that flips a wave, add $\pi$ before any path difference. Add the source phase to $2\pi\Delta r/\lambda$.