Play two notes that are almost the same pitch and you hear a single tone whose loudness throbs: wah, wah, wah. These are beats. Musicians use them to tune: as two strings come into tune, the beats slow down and finally stop. Beats are interference in time instead of in space.
Key idea
Two tones of frequencies $f_1$ and $f_2$ played together sound like one tone at the average frequency $\frac{f_1 + f_2}{2}$, swelling and fading at the beat frequency $$f_{\text{beat}} = |f_1 - f_2|.$$
Why it works
At one spot, the two sound waves are $\cos(2\pi f_1t)$ and $\cos(2\pi f_2t)$. The sum formula gives $$\cos(2\pi f_1t) + \cos(2\pi f_2t) = 2\cos\!\left(2\pi\,\frac{f_1 - f_2}{2}\,t\right)\cos\!\left(2\pi\,\frac{f_1 + f_2}{2}\,t\right).$$ The fast factor is the tone you hear. The slow factor is an envelope that sets the loudness. The envelope's frequency is $\frac{|f_1 - f_2|}{2}$, but loudness depends on the envelope's size, which peaks twice per cycle (at $+2$ and at $-2$). So the loudness throbs at twice that: $|f_1 - f_2|$. Said simply: the two waves drift in and out of step, and they get back in step once every $1/|f_1 - f_2|$ seconds.
Worked examples
Worked example: Count the beats
A tuning fork of 440 Hz and a piano string of 443 Hz sound together. What do you hear?
- Beat frequency: $|443 - 440| = 3$ Hz, so three swells of loudness per second.
- The tone itself is at $\frac{440 + 443}{2} = 441.5$ Hz.
Answer: A 441.5 Hz tone that throbs 3 times a second
Worked example: Tuning a guitar
A guitar string sounded with a 330 Hz fork gives 4 beats per second. Tightening the string slightly makes the beats faster. What was the string's frequency?
- The string is at $330 + 4 = 334$ Hz or $330 - 4 = 326$ Hz.
- Tightening raises a string's frequency ($f \propto \sqrt{F_T}$).
- If it were at 326 Hz, raising it would move it toward 330 Hz and slow the beats. The beats sped up, so it moved away from 330 Hz.
Answer: $334$ Hz (loosen it to tune)
Worked example: A small change in tension
Two identical strings both sound 220 Hz. The tension in one is raised by 2.0%. What beat frequency do you hear?
- For a string $f = \frac{1}{2L}\sqrt{F_T/\mu}$, so $f \propto \sqrt{F_T}$.
- New frequency: $220\sqrt{1.020} = 220 \times 1.00995 = 222.19$ Hz.
- Beats: $222.19 - 220 = 2.2$ Hz.
Answer: About $2.2$ beats per second
Common mistake
Taking the beat frequency as the average $\frac{f_1 + f_2}{2}$ or the sum $f_1 + f_2$. The average is the pitch you hear; the beats come from the difference. Another slip: using half the difference, the envelope's own frequency. Check with a case you can picture: tones at 100 Hz and 101 Hz get back in step once every second, so there is 1 beat per second, not 0.5.