Beats of two close frequencies

Problem

Superimpose two waves of $f_1 = 440\;\mathrm{Hz}$ and $f_2 = 444\;\mathrm{Hz}$. Calculate the perceived beat frequency and explain how this principle is used to tune instruments.

Solution

Setup. Adding two waves of nearby frequencies, the resulting amplitude is modulated by an envelope that periodically vanishes and builds up: these are the beats. The beat frequency (number of intensity maxima per second) is

$f_b = |f_1 - f_2|$

Substitution.

$f_b = |444 - 440| = 4\;\mathrm{Hz}$

$\ev{f_b = 4\;\mathrm{Hz}}$

The note heard is at the average frequency, $(440+444)/2 \approx 442\;\mathrm{Hz}$, whose intensity pulses 4 times per second.

Tuning instruments. Playing together the string to be tuned and a reference (tuning fork or another string), you hear the “wah-wah” of the beats. As the string’s frequency is brought closer to the reference, $f_b$ decreases; when the beats disappear ($f_b \to 0$) the two frequencies coincide and the instrument is in tune. It is a very sensitive method because the ear easily perceives even very slow beats.

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