Beats — derivation, applications: Visual walkthrough

Step 1 — One wave at a point: what a single squiggle means

WHAT: We write the first wave as $y_{1} = a cos (2 π f_{1} t)$

Let's earn every symbol on that line:

$y_{1}$ — the displacement: how far the air is pushed from rest, up or down.
$a$ — the amplitude: the biggest push, the height of the wiggle's crest.
$t$ — time, ticking along the bottom.
$f_{1}$ — the frequency: how many complete up-down wiggles happen each second (unit: hertz, Hz).
$cos$ — a machine that turns a steadily-growing number into a smooth back-and-forth between $+ 1$ and $- 1$ . We use $cos$ (not a straight line) precisely because the air really does return to where it started and repeat — cosine is the simplest "returns and repeats" curve.
$2 π f_{1} t$ — the angle fed to cosine. Why the $2 π$ ? Because cosine completes one full trip every time its angle grows by $2 π$ . Multiplying by $f_{1} t$ means: after time $t = \frac{1}{f _{1}}$ (one period), the angle is exactly $2 π$ → exactly one full wiggle. The $2 π$ is a units-converter from "wiggles" to "radians the cosine wants."

WHY: We track a fixed point over time (not a snapshot in space) because a beat is something you hear at your ear — one location, changing loudness as the seconds pass.

PICTURE:

Figure — Beats — derivation, applications

Step 2 — The second wave: almost the same, but drifting

Everything is identical to $y_{1}$ except $f_{1} \to f_{2}$ , where $f_{2}$ is just barely different from $f_{1}$ . Same amplitude $a$ , same start.

WHY equal amplitude? So that when the two waves eventually oppose each other perfectly, they can cancel to total silence — the cleanest possible effect. Unequal amplitudes work too; they just never fully cancel (a leftover hum), which muddies the picture.

WHY same start? So at $t = 0$ both crests coincide — they begin perfectly in step, and we can watch them drift apart. (What if they don't start in step? We handle that in Step 8 — spoiler: the beat rate doesn't change, the whole envelope just slides sideways.)

PICTURE: Draw $y_{1}$ and $y_{2}$ on the same board. At the far left they march together; as time passes, the slightly-faster one nudges ahead until, later, its crest sits over the other's trough.

Step 3 — Superpose: let the ripples add

WHAT: $y = y_{1} + y_{2} = a cos (2 π f_{1} t) + a cos (2 π f_{2} t)$

Read it literally: at each instant, take the height of the blue curve, add the height of the pink curve, plant a dot at that total. Trace all the dots → the black resultant.

WHY: Because that black curve is the actual motion of the air at your ear — the thing that will drive your eardrum. Everything about beats lives inside its shape.

PICTURE: Where both crests point up → tall sum (loud). Where crest meets trough → they cancel to near-zero (silent). The black curve visibly bulges and pinches.

Step 4 — The trick that untangles the sum

We have a sum of two cosines. It's hard to see the throbbing in a sum. We want a product — one factor for the fast wiggle, one for the slow swelling. A trig identity converts exactly that:

WHY this tool and not another? We deliberately pick the sum→product identity (not, say, expanding or a Taylor series) because it turns "two things added" into "two things multiplied," and multiplication is exactly the language of a loud tone (one factor) whose volume is dialed up and down (the other factor). That volume-knob structure is a beat.

Here $C = 2 π f_{1} t$ and $D = 2 π f_{2} t$ . Substitute: $y = 2 a slow: half the difference cos (2 π \cdot \frac{f _{1} - f _{2}}{2} \cdot t) fast: the average cos (2 π \cdot \frac{f _{1} + f _{2}}{2} \cdot t)$

Note the dots: the division by $2$ applies only to the frequency $(f_{1} - f_{2})$ or $(f_{1} + f_{2})$ — it is $(\frac{f _{1} - f _{2}}{2})$ that gets multiplied by $2 π$ and by $t$ , never $t /2$ .

Term by term:

$2 a$ — the maximum possible amplitude: when both waves crest together, heights $a + a = 2 a$ .
$(\frac{f _{1} - f _{2}}{2})$ — half the difference of the frequencies. Since $f_{1} \approx f_{2}$ , this is tiny → a very slow cosine.
$(\frac{f _{1} + f _{2}}{2})$ — the average frequency. Since $f_{1} \approx f_{2}$ , this is basically the original pitch → a fast cosine.

PICTURE: Overlay the fast average-frequency wiggle and, on top of it, the slow half-difference curve hugging its peaks like an outline.

Step 5 — Name the two factors: pitch vs. envelope

So the full motion reads as "a steady pitch, turned up and down slowly by $A (t)$ ." That is exactly the same math as Amplitude Modulation in radio.

PICTURE: The envelope drawn as a smooth dashed boundary; the fast tone shivers inside it, filling it completely at the fat parts and squeezed to nothing at the pinches.

Step 6 — The factor of 2: why loudness peaks twice per envelope cycle

This is the crux — the one step most people get wrong.

WHAT we ask: How often does the sound get loud? That count-per-second is what we finally name the beat frequency, $f_{b e a t}$ .

WHY it's subtle: Loudness does not care whether the envelope is positive or negative. A big upward bulge is loud; a big downward bulge is equally loud — the air is swinging hard either way. So loudness follows the size of the envelope, i.e. $∣ A (t) ∣$ .

The consequence: A plain cosine dips below zero once per cycle. But $∣ cos θ ∣$ folds those negative dips back up into positive humps — so it has two humps for every one cycle of $cos θ$ .

$c o s envelope cycles at \frac{f _{1} - f _{2}}{2} take ∣ \cdot ∣ ∣ c o s ∣ loudness peaks at 2 \times \frac{f _{1} - f _{2}}{2}$

$f_{b e a t} = 2 \times \frac{f _{1} - f _{2}}{2} = ∣ f_{1} - f_{2} ∣$

Term by term: the $2$ comes from " $∣ cos ∣$ peaks twice"; it exactly cancels the $\frac{1}{2}$ hiding inside the envelope frequency, leaving the clean difference. We take $∣ f_{1} - f_{2} ∣$ (absolute value) because "3 beats" and " $- 3$ beats" mean the same thing — a count of loud moments can't be negative.

PICTURE: Envelope $cos$ curve on top; below it, its fold-up $∣ cos ∣$ — clearly showing two loud bumps in the span of one envelope wiggle, with the beat period marked.

Step 7 — Edge cases: pushing the formula to its limits

Every honest derivation must survive its extremes. Three to check:

(a) $f_{1} = f_{2}$ (identical forks). Then $∣ f_{1} - f_{2} ∣ = 0$ : zero beats. The envelope $A (t) = 2 a cos (0) = 2 a$ is a flat constant — no swelling at all, just a steady doubled-loudness tone. This is the tuning target: musicians tune until the throb slows to nothing.

(b) $f_{1} - f_{2}$ growing large — where "beats" stop being beats. As the gap widens, the envelope wiggles ever faster. The transition your ear experiences is gradual, not a sharp cliff, and passes through three regimes:

Up to a few Hz — you clearly count individual swells: waah… waah. These are true beats.
Roughly 10–20 Hz and up — the swells come too fast to count individually; the ear stops hearing separate loud-quiet events and instead perceives a harsh, buzzy quality musicians call roughness.
Beyond the ear's critical bandwidth (the frequency spacing your inner ear can no longer blend — very roughly 10–15% of the tone's frequency, so ~30–100 Hz for mid-range pitches) — the two tones separate into two distinct smooth pitches and the roughness fades.

So "beats need $∣ f_{1} - f_{2} ∣ ≲ 10$ Hz" is a rule of thumb for the countable regime, not a hard physical constant — it marks about where your ear can no longer track the swells one by one. The exact numbers depend on the pitch and the listener; the physics of the envelope (Step 6) is unchanged throughout — only your perception of it shifts.

(c) Unequal amplitudes $a_{1} \neq = a_{2}$ . The pinch no longer reaches zero — there's a leftover quiet hum instead of total silence. Same rhythm, shallower dips. The beat frequency is unchanged; only the depth of the fade shrinks.

PICTURE: Three mini-panels side by side: flat (equal freq), a fast blur (large gap), and shallow dips (unequal amplitude).

Step 8 — Edge case: what if the waves DON'T start in step?

We assumed both waves begin with their crests together at $t = 0$ . Reality rarely obliges — strike two forks a heartbeat apart and one already leads the other. So give each wave its own starting phase:

$ϕ_{1}, ϕ_{2}$ — the initial phase: how far along its wiggle each wave already is at $t = 0$ . A phase of $0$ means "starting at a crest"; a phase of $π$ means "starting at a trough."

Run the same sum-to-product identity (now $C = 2 π f_{1} t + ϕ_{1}$ , $D = 2 π f_{2} t + ϕ_{2}$ ): $y = 2 a cos (2 π \cdot \frac{f _{1} - f _{2}}{2} \cdot t + \frac{ϕ _{1} - ϕ _{2}}{2}) cos (2 π \cdot \frac{f _{1} + f _{2}}{2} \cdot t + \frac{ϕ _{1} + ϕ _{2}}{2})$

Read the new bits: the extra $\frac{ϕ _{1} - ϕ _{2}}{2}$ inside the slow factor is just a constant added to the angle — it does not depend on $t$ . Adding a constant to a cosine's angle only slides the whole curve left or right in time; it never changes how fast the curve wiggles.

WHY it matters: the beat frequency is set by how fast the envelope oscillates, and that speed is still $\frac{f _{1} - f _{2}}{2}$ → folded → $∣ f_{1} - f_{2} ∣$ . Unchanged. The only visible effect is that the first loud moment happens at a slightly different instant. In the clapping picture: if the two friends begin a little out of sync, the "loud-quiet-loud" pattern is identical — it just starts partway through.

PICTURE: The same beat pattern from Step 5, drawn twice — once with $ϕ_{1} - ϕ_{2} = 0$ and once with a nonzero offset — showing the envelope bodily shifted sideways but with identical spacing between loud moments.

The one-picture summary

Everything at once: two near-twin ripples (blue, pink) → their black sum → the dashed envelope tracing its swell → the fold-up $∣ envelope ∣$ marking the loud moments, spaced one beat-period $1/∣ f_{1} - f_{2} ∣$ apart.

Recall Feynman retelling — the whole walkthrough in plain words

Two air-pushes arrive at your ear, wiggling at almost the same speed. Watch one point of air: each wave alone is a smooth cosine wiggle (Step 1–2). The air can only be in one spot, so its real motion is the two wiggles added (Step 3) — and where crests agree it's tall, where crest meets trough it flattens.

That "sum of two wiggles" is clumsy, so we use a trig identity to rewrite it as a product (Step 4): a fast wiggle at the average speed — that's the pitch you hear — times a slow wiggle at half the difference — that's a volume knob turning up and down (Step 5).

Here's the trick everyone trips on (Step 6): your ear hears a downward bulge just as loud as an upward one, so it folds the slow curve's negatives up into positives — giving two loud moments for each slow cycle. That doubling cancels the "half," and out pops the clean answer: loud moments per second $= ∣ f_{1} - f_{2} ∣$ .

Check the corners (Steps 7–8): identical forks → no throb (tune to this!); as the forks get more different the swells speed up until you can't count them — first a buzzy roughness, then two separate notes; mismatched loudness → the dips don't quite hit silence; and if the forks start out of step, the whole throb just slides sideways in time — same rhythm. That's the entire story, and it's the same math radios use to pull a station out of the air.

Connections

Superposition Principle — Step 3 is superposition, nothing more.
Interference of Waves — beats are interference laid out in time.
Standing Waves — same sum-to-product algebra, but for oppositely-travelling waves.
Amplitude Modulation — the envelope of Step 5 is identical to a radio carrier's modulation.
Doppler Effect — supplies the frequency difference that radar reads out as a beat.
Simple Harmonic Motion — each single wave (Step 1) is SHM at one point.