Beats

Sit two flutes side by side and have them play what is meant to be the same note, one a hair out of tune, and you will hear something neither flute is playing: a slow, throbbing wah — wah — wah, the loudness swelling and fading a few times a second. Nothing is wrong with your ears. You are hearing beats — the direct, audible signature of two waves adding up, sometimes reinforcing, sometimes cancelling, in a rhythm set by exactly how far apart their frequencies are. Beats are the humble everyday face of superposition, and they are also a precision instrument. A piano tuner does not own a frequency meter; a tuner listens for the beats between a reference fork and a string and tightens the string until the throbbing slows, then stops. When the beats vanish, the two frequencies are equal to a fraction of a hertz — an accuracy no eye reading a dial could match. The same trick tunes a guitar, calibrates an oscillator, and once let Joseph Sauveur measure the pitch of an organ pipe in 1701 with nothing but his ears and a clock. The physics is one trigonometric identity. Add two tones of nearly equal frequency and the sum-to-product formula splits the result cleanly into a fast wave — the average pitch you actually hear — riding inside a slow envelope — the swell and fade. The rate at which the loudness pulses, the beat frequency, turns out to be simply the difference of the two frequencies, $f_{\text{beat}} = |f_1 - f_2|$ . Not half the difference, not twice it — the difference, and this article's main job is to show you exactly why the factor is one, because that single factor is where marks are won and lost.

Key takeaways

Beats are the slow rise and fall in loudness you hear when two tones of nearly equal frequency sound together.
By the sum-to-product identity, $y = 2A\cos\!\big(2\pi\tfrac{f_1-f_2}{2}t\big)\cos\!\big(2\pi\tfrac{f_1+f_2}{2}t\big)$ : a fast carrier at the average frequency, inside a slow envelope.
The loudness peaks twice per envelope cycle, so the beat frequency is the full difference: $f_{\text{beat}} = |f_1 - f_2|$ , not half of it.
The beat period is $T_{\text{beat}} = 1/|f_1 - f_2|$ — the time from one swell to the next.
Beats are the standard way to tune: adjust one source until the beating slows to zero, meaning the frequencies match.
Beats are heard only for close frequencies (a few hertz apart); far apart, the ear hears two separate tones instead.
Key idea: two nearly-equal tones add to a single tone at their average frequency whose loudness pulses at the difference frequency $|f_1 - f_2|$ — so counting beats per second directly measures how far apart the two pitches are.

The core insight

Superposing two equal-amplitude tones of nearby frequency gives a wave at the average frequency $(f_1+f_2)/2$ multiplied by a slow envelope; because loudness peaks at both bulges of that envelope, the beats occur at the difference frequency $f_{\text{beat}} = |f_1 - f_2|$ , which is why nulling the beats tunes two sources into unison.

Where this lives today

Beats are still the fastest tuning method in the world. A piano technician tunes the whole instrument by counting beats between strings and against a fork; a guitarist matches two strings by driving their beats to zero at the fifth fret. Beyond music, the same "difference frequency" is the engine of heterodyne electronics: mix an incoming radio signal with a local oscillator and the beat between them — the difference frequency — is what a radio actually amplifies and decodes, the principle inside every superheterodyne receiver since the 1910s. A laser interferometer measures a moving mirror by the beats between the outgoing and returning beams; a Doppler radar gun reads your car's speed from the beat between the emitted and reflected microwaves; and physicists compare atomic clocks by beating one against another and counting. Every one of these is equation (3): two waves, one difference, a rhythm you can measure. The next article in this chapter turns the tuning logic into a full problem-solving archetype — how to pin down an unknown frequency, sign and all, from the beats it makes.

The physical picture

Take two pure tones of equal amplitude $A$ , in phase at $t = 0$ , with frequencies $f_1$ and $f_2$ close together. Model each as a cosine oscillation of the air at your ear:

y_1 = A\cos(2\pi f_1 t), \qquad y_2 = A\cos(2\pi f_2 t). \qquad (1)

The ear feels the sum $y = y_1 + y_2$ , because pressure disturbances simply add — that is superposition. The sum of two cosines does not look like anything special until you apply the identity $\cos C + \cos D = 2\cos\!\big(\tfrac{C+D}{2}\big)\cos\!\big(\tfrac{C-D}{2}\big)$ , which turns a sum of two waves into a product of two:

y = 2A\,\cos\!\Big(2\pi\,\frac{f_1 - f_2}{2}\,t\Big)\,\cos\!\Big(2\pi\,\frac{f_1 + f_2}{2}\,t\Big) \qquad (2)

where: the second factor, $\cos\big(2\pi\tfrac{f_1+f_2}{2}t\big)$ , oscillates fast at the average frequency $f_{\text{avg}} = (f_1+f_2)/2$ — this is the pitch you hear; and the first factor, $2A\cos\big(2\pi\tfrac{f_1-f_2}{2}t\big)$ , oscillates slowly because $f_1 - f_2$ is tiny — this is a slowly changing amplitude, the envelope, that swells the loud fast tone up and down. The whole trick of beats is in that split: a single audible note whose volume is being turned up and down by a slow hand.

Figure

Two nearly equal tones add to a fast carrier at the average frequency wrapped inside a slow envelope, and the loudness swells to a maximum twice for each full cycle of the envelope

Reading off the beat frequency

Now the crucial step, and the one every exam tests. The envelope is $2A\cos\big(2\pi\tfrac{f_1-f_2}{2}t\big)$ , an oscillation at frequency $(f_1-f_2)/2$ . But your ear does not hear the sign of the amplitude — a loud burst with amplitude $+2A$ sounds exactly as loud as one with $-2A$ . Loudness follows the magnitude of the envelope, $\big|2A\cos(\dots)\big|$ , and a magnitude of a cosine peaks twice in every full cosine cycle — once when the cosine is $+1$ , once when it is $-1$ . So the loudness swells to a maximum at twice the envelope's frequency:

f_{\text{beat}} = 2 \times \frac{f_1 - f_2}{2} = |f_1 - f_2| \qquad (3)

where: the factor of $2$ from the two bulges exactly cancels the $\tfrac12$ in the envelope frequency, leaving the plain difference. This is the headline result: the number of beats you hear per second is the difference of the two frequencies, and the time between swells is its reciprocal,

T_{\text{beat}} = \frac{1}{|f_1 - f_2|}. \qquad (4)

Put numbers to it. Sound a $256\ \mathrm{Hz}$ fork (middle C) with a $260\ \mathrm{Hz}$ fork. The carrier sits at the average $f_{\text{avg}} = (256+260)/2 = 258\ \mathrm{Hz}$ — a single pitch between the two. The envelope oscillates at $(260-256)/2 = 2\ \mathrm{Hz}$ , but the loudness pulses at twice that: $f_{\text{beat}} = |256 - 260| = 4\ \mathrm{Hz}$ , four swells a second, a beat period of $T_{\text{beat}} = 1/4 = 0.25\ \mathrm{s}$ . If instead you strike a $512\ \mathrm{Hz}$ fork against a $508\ \mathrm{Hz}$ string, the same rule gives $|512 - 508| = 4\ \mathrm{Hz}$ — four beats a second again, even though the pitches are an octave higher, because beats care only about the difference.

Tuning: nulling the beats

Equation (3) is the tuner's whole method. The closer the two frequencies, the smaller $|f_1 - f_2|$ , and the slower the beats. Drive $f_2$ toward $f_1$ and the throbbing stretches out — from a rapid flutter, to a lazy pulse, to a single slow swell, to a dead-steady tone the instant the two match. The beats do not just get quieter; their rate falls to zero.

Figure

As two tones are tuned closer the beats slow down, from five swells a second to one, to a steady silence when the frequencies match

This is why "tune until the beats disappear" works to a fraction of a hertz. When two piano strings are set to $440$ and $442\ \mathrm{Hz}$ , you hear $|440 - 442| = 2\ \mathrm{Hz}$ — two lazy swells a second. Ease one string a whisker sharper toward $441\ \mathrm{Hz}$ and the beats slow to $1\ \mathrm{Hz}$ ; nudge it home to $440$ and the beating stops dead. A human ear can reliably resolve a beat as slow as one every few seconds, which corresponds to a mismatch well under half a hertz — far finer than reading any dial. Concert A is standardised at $440\ \mathrm{Hz}$ precisely so that this nulling works between instruments across an orchestra.

Derivation AThe beat formula, and why the factor is one not a half

The goal is to derive equation (3), $f_{\text{beat}} = |f_1 - f_2|$ , from the superposition of two tones, and in particular to nail down why the beat frequency is the full difference and not half of it — because "half the difference" is the single most common wrong answer, and it comes from stopping the derivation one step too early. The strategy is to collapse the sum of two cosines into a product, identify the slow factor as an envelope, and then reason carefully about what the ear actually counts.

Step 1. Superpose. Two equal-amplitude tones in phase at $t = 0$ : $y = A\cos(2\pi f_1 t) + A\cos(2\pi f_2 t)$ . The ear responds to this sum because pressure disturbances add linearly.

Step 2. Apply sum-to-product, $\cos C + \cos D = 2\cos\big(\tfrac{C+D}{2}\big)\cos\big(\tfrac{C-D}{2}\big)$ , with $C = 2\pi f_1 t$ and $D = 2\pi f_2 t$ :

y = 2A\cos\!\Big(2\pi\,\frac{f_1-f_2}{2}\,t\Big)\cos\!\Big(2\pi\,\frac{f_1+f_2}{2}\,t\Big).

The second factor is fast (the average pitch); the first is slow (the envelope) because $f_1 - f_2$ is small.

Step 3. Identify what the envelope's frequency is. The slow factor $\cos\big(2\pi\tfrac{f_1-f_2}{2}t\big)$ completes one full oscillation at frequency $(f_1-f_2)/2$ . If loudness followed the envelope's value, this would be the beat rate — and this is exactly where the "half the difference" error is born.

Step 4. Count what the ear actually hears. Loudness depends on the amplitude's magnitude, $\big|2A\cos(2\pi\tfrac{f_1-f_2}{2}t)\big|$ , because a burst of amplitude $-2A$ is just as loud as one of $+2A$ . The magnitude of a cosine reaches its maximum twice per cosine cycle (at the $+1$ and the $-1$ peaks) and drops to zero twice (as the cosine crosses zero). So the loudness swells $2 \times (f_1-f_2)/2 = |f_1 - f_2|$ times a second. The factor of two from the two bulges cancels the one-half in the envelope frequency, giving $f_{\text{beat}} = |f_1 - f_2|$ exactly. Verified: for 256 and 260 Hz, carrier 258 Hz, envelope 2 Hz, beat 4 Hz, period 0.25 s; and 512 with 508, 440 with 437, 440 with 442 give 4, 3, 2 Hz — recomputed in verify_hs2-beats.py.

Go deeper · Derivation B · 1 minTuning to unison: the beats vanish as the difference goes to zero

Derivation BTuning to unison: the beats vanish as the difference goes to zero

Here we make the tuning method quantitative: we show that as one source is driven toward the other, the beat frequency falls smoothly to zero, and that "no beats" is a genuine measurement of "equal frequencies" and not merely "quiet". The goal is to justify why a tuner trusts the null, using the locked piano numbers. The strategy is simply to track $f_{\text{beat}} = |f_1 - f_2|$ as $f_2$ approaches $f_1$ .

Step 1. Fix the reference at $f_1 = 440\ \mathrm{Hz}$ (concert A) and start the second string sharp at $f_2 = 442\ \mathrm{Hz}$ . The beat frequency is $|440 - 442| = 2\ \mathrm{Hz}$ : two swells a second, easily counted.

Step 2. Ease the string down to $f_2 = 441\ \mathrm{Hz}$ . Now $f_{\text{beat}} = |440 - 441| = 1\ \mathrm{Hz}$ — one slow swell a second. The pitch you hear (the carrier at $440.5\ \mathrm{Hz}$ ) has barely moved, but the rate of throbbing has halved.

Step 3. Bring it home to $f_2 = 440\ \mathrm{Hz}$ . Then $f_{\text{beat}} = |440 - 440| = 0$ : the envelope no longer oscillates, the loudness is dead steady, and the two sources are in unison to within the fraction of a hertz your ear can resolve.

Step 4. Note the safeguard the null gives you. Because the beat rate falls to zero continuously, a slow residual beat tells you both that a mismatch remains and roughly how big it is — one beat per two seconds means the strings differ by half a hertz. That graded feedback is why ears out-tune dials. Verified: the tuning sequence 440 with 442, 441, 440 gives beat rates 2, 1, 0 Hz, and 440 with 445 gives 5 Hz — recomputed in verify_hs2-beats.py.

Limiting cases

case	result	meaning
$f_1 = f_2$	$f_{\text{beat}} = 0$	in tune: steady loudness, no throbbing — the tuner's target
$f_1 - f_2 = 1\ \mathrm{Hz}$	one swell per second	the slow beat a careful tuner listens for near the null
$f_1 - f_2$ a few Hz	clear, countable beats	the working range: distinct swell and fade
$f_1 - f_2 \gtrsim 15\ \mathrm{Hz}$	beats blur into roughness	too fast to count; the ear starts to hear two tones
$f_1 - f_2$ large	two separate pitches	no beats at all — the tones are simply resolved
amplitude at a swell	up to $2A$ ; at a null, $0$	the resultant swings between $2A$ and zero, redistributing energy in time

Common trapThe beat frequency is the difference, not half the difference

The envelope in equation (2) oscillates at $(f_1 - f_2)/2$ , and it is fatally tempting to call that the beat frequency. It is not. Your ear counts loudness maxima, and the loudness (the magnitude of the envelope) peaks twice for each envelope cycle, so the beats come at $|f_1 - f_2|$ — twice the envelope frequency. Whenever you see the number $(f_1 - f_2)/2$ , that is the envelope, an intermediate quantity, never the answer to "how many beats per second". A second, related trap: the resultant amplitude swings all the way from $2A$ down to $0$ , so the intensity (which goes as amplitude squared) swings between $4I$ and $0$ , four times the intensity of one tone at the loud moments and dead silence at the nulls. No energy is created; it is merely gathered into the swells and drained from the silences.

Worked examples

Board level — beats between two forks (CBSE and ICSE pattern)

Two tuning forks of frequencies $256\ \mathrm{Hz}$ and $260\ \mathrm{Hz}$ are sounded together. Find the frequency you hear, the beat frequency, and the beat period.

Step 1. The pitch heard is the average (carrier) frequency: $f_{\text{avg}} = (256 + 260)/2 = \boxed{258\ \mathrm{Hz}}$ . The two forks fuse into one tone at $258\ \mathrm{Hz}$ .

Step 2. The beat frequency is the difference (equation 3): $f_{\text{beat}} = |256 - 260| = \boxed{4\ \mathrm{Hz}}$ — four loudness swells every second.

Step 3. The beat period is its reciprocal (equation 4): $T_{\text{beat}} = 1/4 = \boxed{0.25\ \mathrm{s}}$ — a quarter of a second between swells.

Step 4. Guard against the classic slip: the envelope frequency here is $(260-256)/2 = 2\ \mathrm{Hz}$ , but that is not what you are asked for. The beats — the swells you actually count — come at $4\ \mathrm{Hz}$ , twice the envelope, because loudness peaks at both bulges.

Board level — an octave up, same beats (JEE-Main and NEET pattern)

A $512\ \mathrm{Hz}$ tuning fork is sounded with a violin string tuned to $508\ \mathrm{Hz}$ . How many beats per second are heard? Compare with the $256$ -and- $260$ case above.

Step 1. Apply equation (3) directly: $f_{\text{beat}} = |512 - 508| = \boxed{4\ \mathrm{Hz}}$ — four beats a second.

Step 2. Compare. This pair is a full octave higher than the $256$ -and- $260$ forks, yet the beat rate is identical at $4\ \mathrm{Hz}$ . Beats depend only on the difference of the frequencies, not on how high the pitches sit — a $4\ \mathrm{Hz}$ gap beats four times a second whether the tones are near $258$ or near $510\ \mathrm{Hz}$ .

Step 3. The carrier does differ: here $f_{\text{avg}} = (512+508)/2 = 510\ \mathrm{Hz}$ , so you hear a $510\ \mathrm{Hz}$ tone throbbing four times a second, an octave above the earlier $258\ \mathrm{Hz}$ . Difference sets the beat; average sets the pitch.

Competitive level — finding an unknown frequency by beats (JEE and NEET pattern)

A string of unknown frequency is sounded with a $440\ \mathrm{Hz}$ fork and gives $3$ beats per second. The string is then tightened slightly and the beat rate falls to $1$ per second. What was the string's original frequency? Take concert A as exactly $440\ \mathrm{Hz}$ .

Step 1. Three beats a second means the string differs from the fork by $3\ \mathrm{Hz}$ : it is either $440 + 3 = 443\ \mathrm{Hz}$ or $440 - 3 = 437\ \mathrm{Hz}$ . Beats alone cannot yet tell which — this ambiguity is the heart of the beat-problem archetype later in this chapter.

Step 2. Use the change. Tightening a string raises its frequency (recall $f \propto \sqrt{T}$ ). The beat rate fell from $3$ to $1$ , meaning the string moved closer to $440\ \mathrm{Hz}$ as it rose.

Step 3. Deduce the side. A string that rises and gets closer to $440$ must have started below $440$ . So the original frequency was $\boxed{437\ \mathrm{Hz}}$ — raising it toward $440$ shrinks the gap from $3\ \mathrm{Hz}$ to $1\ \mathrm{Hz}$ , exactly as observed. Had it started at $443$ , tightening would have widened the gap and increased the beats.

Step 4. Check for consistency: at $1$ beat per second the string is now at $440 - 1 = 439\ \mathrm{Hz}$ , having risen $2\ \mathrm{Hz}$ from $437$ under the extra tension — a sensible small change. The logic "which way did the beats move when I changed the source" is the master tool for resolving the $\pm$ ambiguity. Timing note: about 90 seconds. The two-answer trap ( $437$ or $443$ ) is deliberate — never stop at $f_{\text{fork}} \pm n$ . The tie-breaker is always the direction the beat count moves when you deliberately change one source, and knowing that tightening raises pitch.

Exam traps

Beat frequency is $|f_1 - f_2|$ , full stop. Not $(f_1 - f_2)/2$ , not $(f_1 + f_2)/2$ . The half-difference is the envelope; the sum-over-two is the carrier (the pitch). The beats you count per second are the plain difference. Distractor lists on JEE and NEET always seed the half-value — it is there to catch a derivation stopped early.

Two answers from a beat count. "The fork and the wire beat $n$ times a second" means the wire is at $f_{\text{fork}} + n$ or $f_{\text{fork}} - n$ . You cannot choose between them from the beat count alone — you need extra information (loading, filing, or tightening the source and watching which way the beats move). Writing a single answer where two exist loses the mark.

Beats need close frequencies. The whole picture assumes $f_1$ and $f_2$ are within a few hertz. If they differ by tens of hertz, the "swells" come faster than about $15$ a second, the ear stops counting them, and above that you simply hear two distinct tones. Do not apply the beat formula to two far-apart notes and report " $300$ beats per second" — that is not a perceptible beat.

Intensity swings, energy is conserved. At a swell the amplitude is $2A$ and the intensity $4I$ ; at a null both are zero. Students sometimes think energy is being created at the loud moments — it is not. The two waves redistribute a fixed total energy in time, piling it into the swells and emptying the silences.

Board scope. Beats are examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. For A-Level (CIE 9702) beats are thin in the current specification — check your syllabus version before relying on them. Beats are not an assessed objective on IB (2023) or AP Physics 2, and IGCSE 0625 has no beats at all; if you sit one of those, this topic is enrichment, not exam content.

Origins · 1 minLéon Brillouin and how this was discovered

Léon Brillouin

1889–1969 · Paris & New York

Léon Brillouin spent his life on the physics of wave groups — bundles of waves of nearly equal frequency travelling together — and a beat pattern is the simplest wave group there is. When two close tones superpose, the swelling envelope you hear is a group, and the speed at which such an envelope travels is the group velocity, a quantity Brillouin (with Sommerfeld, from 1914) pinned down with unusual care, showing how it differs from the speed of the individual crests and how signals really propagate through a dispersive medium. His name survives in the Brillouin zone of solid-state physics and in Brillouin scattering, in which light is deflected by sound waves in a material and its tiny frequency shift reads off the speed of sound directly. Brillouin's groups were of electrons and light in crystals; the two-tone acoustic beat in this article is the very same superposition, slowed down until a human ear can count it.

History

1636

— Marin Mersenne describes the slow throbbing heard when two nearly-tuned organ pipes sound together, the first written account of beats.

1701

— Joseph Sauveur turns beats into a measuring tool, counting the beat rate between two pipes to fix their absolute frequencies by ear and clock.

1802

— Thomas Young explains beats as the interference of two wave trains, folding them into his wave theory of sound and light.

1863

— Hermann von Helmholtz, in On the Sensations of Tone, links the roughness of slow beats to musical dissonance and the sensation of being out of tune.

1914

— Arnold Sommerfeld and Léon Brillouin analyse how wave groups and their envelopes travel, the framework in which a beat is the simplest group.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), the section on beats · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves", beats and their applications · Halliday, Resnick and Walker, "Waves — II", beats · Cambridge A-Level 9702 (superposition topic; confirm your syllabus version covers beats).

Provenance. Standard trigonometric result. Recompute: sum-to-product gives $f_{\text{avg}} = (f_1+f_2)/2$ , $f_{\text{beat}} = |f_1 - f_2|$ ; $256$ and $260$ : carrier $258$ , envelope $2\ \mathrm{Hz}$ , beat $4\ \mathrm{Hz}$ , period $0.25\ \mathrm{s}$ ; $512$ and $508$ : $4\ \mathrm{Hz}$ , carrier $510$ ; $440$ and $437$ : $3\ \mathrm{Hz}$ ; $440$ and $442$ : $2\ \mathrm{Hz}$ ; tuning sequence $440$ with $445, 441, 440$ : $5, 1, 0\ \mathrm{Hz}$ ; the unknown-string case resolves to $437\ \mathrm{Hz}$ ; loud-moment intensity $4I$ .

Prerequisites. Periodic Motion and the Definition of SHM — each tone is simple harmonic; beats are two SHMs superposed. Leads to. ARCHETYPE: Beat-Frequency and Tuning Problems — resolving the $\pm$ ambiguity with loading and filing, and chaining beats onto the string and pipe formulas.

Go deeper (UG). The full treatment of beats as a wave group, and the distinction between phase velocity and group velocity, is developed in Beats in the university mechanics track. Enrichment only — nothing here depends on it.

Why do I hear only one pitch when two forks are sounded, not two?

Because the two frequencies are so close that your ear cannot separate them into distinct pitches — instead it fuses them into a single tone at their average frequency, $(f_1+f_2)/2$ , and reports the small difference as a change in loudness over time rather than as a second note. The maths says exactly this: equation (2) is one fast cosine (the average-frequency pitch you hear) multiplied by a slow one (the swell). Only when the frequencies are far enough apart — tens of hertz — does the ear resolve them as two separate pitches, and then the beats disappear because there is no longer a single fused tone to swell and fade.

Can I tell which of the two tones is higher just by listening to the beats?

No — and this is the single most important limitation of beats. A $440\ \mathrm{Hz}$ fork beating $3$ times a second with a string tells you the string is $3\ \mathrm{Hz}$ away, but the string could be at $437$ or $443\ \mathrm{Hz}$ ; the beat count is the same either way, because it depends on $|f_1 - f_2|$ , which does not know the sign. To break the tie you must change one source and watch the beats respond: tighten the string (which raises its pitch) and see whether the beats speed up (the string was above) or slow down (it was below). The next article, the beat-problem archetype, is built entirely around resolving this ambiguity with loading and filing.

What is the difference between the envelope frequency and the beat frequency?

The envelope frequency is $(f_1 - f_2)/2$ — the rate at which the slow cosine in equation (2) completes a full cycle, going positive, then negative, then positive again. The beat frequency is $|f_1 - f_2|$ — the rate at which you actually hear the loudness swell to a maximum. They differ by a factor of two because loudness follows the magnitude of the envelope, and a cosine reaches maximum magnitude twice per cycle (once at its positive peak, once at its negative peak). So there are two loud moments for every one envelope cycle. In an exam, the beat frequency is almost always what is wanted; the envelope frequency is an intermediate step, and quoting it as the answer is the classic error.

Does the amplitude of the two tones have to be equal for beats to happen?

Not for beats to occur, only for them to be complete. If the two amplitudes are equal, the nulls are total silence — the waves cancel perfectly at the low point of each swell. If the amplitudes differ, you still hear beats at the same rate $|f_1 - f_2|$ , but the loudness never falls all the way to zero, because the larger wave is never fully cancelled by the smaller. The rate is unchanged; only the depth of the throbbing shrinks. This is why a very quiet mistuned string still lets you hear a gentle beating against a loud reference, enough to tune by, even though the silence between swells is not perfect.

Why does a beat pattern count as a wave group?

Because a wave group is exactly what you get when you add waves of nearly equal frequency: they drift in and out of step, reinforcing in some regions and cancelling in others, producing a travelling packet of amplitude — the envelope. A beat is the one-dimensional, two-frequency version of that, slowed down to a few hertz so you can hear it. The envelope of a beat travels at the group velocity, which for sound in ordinary air is essentially the same as the speed of sound, but in a dispersive medium (where wave speed depends on frequency) the group can move at a noticeably different speed from the individual crests. That distinction — crest speed versus envelope speed — is Léon Brillouin's subject, and the humble acoustic beat is its most audible example.

Is 440 Hz special, or could an orchestra tune to anything?

There is nothing physically special about $440\ \mathrm{Hz}$ — it is a convention, standardised (as concert A) so that instruments built and tuned in different places can play together without beating against one another. Historically the reference pitch drifted from town to town and century to century (baroque ensembles often use $415\ \mathrm{Hz}$ , a semitone lower), and some modern orchestras tune slightly sharp, to $442$ or $443\ \mathrm{Hz}$ , for a brighter sound. What matters is that everyone agrees, because the beat method only tunes instruments to each other: it nulls the difference. Pick any common reference and the same physics — drive $|f_1 - f_2|$ to zero — brings the whole ensemble into unison.