Beats — derivation, applications: Worked examples

This page is the drill-ground for Beats. The parent note gave you the one formula that runs everything:

The trap that hides everywhere: $∣ f_{1} - f_{2} ∣$ throws away the sign. It tells you how far apart two frequencies are, but not which one is bigger. Half of the examples below exist purely to recover that lost sign.

The scenario matrix

Below is every distinct kind of question this topic can ask. Each later example is tagged with the cell (C1, C2, …) it demolishes.

Cell	Case class	What makes it different	Example
C1	Plain difference	Both frequencies known → just subtract	Ex 1
C2	Sign-hidden (load with wax)	$f_{?}$ could be above or below; mass added ⇒ $f$ drops	Ex 2
C3	Sign-hidden (file/tension up)	Same, but the fix raises $f$ instead	Ex 3
C4	Degenerate: zero beats	$f_{1} = f_{2}$ → beats vanish (the tuning goal)	Ex 4
C5	Limiting: difference too big	$	f_1-f_2
C6	Time/period reading	Convert beat rate ↔ time between silences	Ex 6
C7	Real-world word problem (Doppler radar)	Beat = transmitted vs reflected frequency	Ex 7
C8	Exam twist: two beats give two clues	Combine two measurements to pin an exact value	Ex 8

Example 1 — Plain difference (cell C1)

Example 2 — Wax lowers the fork (cell C2)

Worked example The classic sign resolver

An unknown fork gives 6 beats/s against a $480$ Hz standard. A tiny blob of wax is stuck on the unknown fork; now the beats rise to 8/s. Find the unknown frequency.

Forecast: $∣ f_{?} - 480∣ = 6$ means $f_{?}$ is either $474$ or $486$ . Which one, and why?

List the two candidates: $f_{?} = 480 - 6 = 474$ Hz or $f_{?} = 480 + 6 = 486$ Hz. Why this step? $∣ f_{1} - f_{2} ∣$ is symmetric, so both are algebraically valid; the experiment must break the tie.
Wax adds mass ⇒ frequency drops. From Simple Harmonic Motion, $f = \frac{1}{2 π} k / m$ , so bigger $m$ gives smaller $f$ . The unknown's frequency decreases after waxing. Why this step? We need a directional nudge to expose which side of $480$ we sit on.
Trace both candidates under a downward nudge:
- If $f_{?} = 486$ (above 480): dropping it moves it toward 480 ⇒ gap shrinks ⇒ beats decrease.
- If $f_{?} = 474$ (below 480): dropping it moves it away from 480 ⇒ gap grows ⇒ beats increase.
Match the observation: beats went $6 \to 8$ (increased) ⇒ we are in the " $474$ " case.

Answer: $f_{?} = 474$ Hz.

Verify: Suppose waxing lowered $474$ by $2$ Hz to $472$ : new gap $= ∣472 - 480∣ = 8$ Hz. ✓ Exactly the observed $8$ beats/s.

Example 3 — Filing raises the fork (cell C3)

Worked example Same trick, opposite direction

An unknown fork gives 4 beats/s with a $256$ Hz standard. The unknown fork is filed (metal removed from the prongs), and the beats drop to 2/s. Find the unknown.

Forecast: Candidates are $252$ or $260$ . Filing removes mass — which way does $f$ move, and which candidate survives?

Two candidates: $f_{?} = 256 \pm 4 = 252$ or $260$ Hz. Why? Same symmetric ambiguity as before.
Filing removes mass ⇒ frequency rises. Smaller $m$ in $f = \frac{1}{2 π} k / m$ ⇒ larger $f$ . The unknown's frequency increases. Why? This is the exact opposite nudge to waxing — the sign of the argument flips.
Trace both under an upward nudge:
- If $f_{?} = 252$ (below 256): raising it moves toward 256 ⇒ beats decrease.
- If $f_{?} = 260$ (above 256): raising it moves away from 256 ⇒ beats increase.
Match: beats went $4 \to 2$ (decreased) ⇒ the " $252$ " case.

Answer: $f_{?} = 252$ Hz.

Verify: If filing raised $252$ by $2$ Hz to $254$ : gap $= ∣254 - 256∣ = 2$ Hz. ✓ Matches observed 2 beats/s.

Example 4 — Degenerate case: zero beats (cell C4)

Worked example Tuning to silence

A guitarist tunes a string against a $440$ Hz reference. She hears $3$ beats/s. She tightens the string slightly and now hears 0 beats/s. What is the string's frequency, and what was it before?

Forecast: Zero beats means what physically? And did tightening raise or lower the pitch?

Zero beats ⇒ exact match. $f_{beat} = ∣ f_{string} - 440∣ = 0 \Rightarrow f_{string} = 440$ Hz. Why this step? This is the degenerate limit $f_{1} = f_{2}$ : the slow envelope $cos (2 π \frac{f _{1} - f _{2}}{2} t)$ becomes $cos (0) = 1$ , a constant — the loudness never dips. Silence-of-beats is the goal of tuning.
Tightening raises frequency (higher tension ⇒ higher pitch on a string). Since tightening removed the 3-beat gap, the string was below $440$ before. Why? Raising a below-reference string closes the gap to zero.
Original frequency: $440 - 3 = 437$ Hz.

Answer: Now $440$ Hz; was $437$ Hz.

Verify: $∣437 - 440∣ = 3$ Hz ✓ (matches the pre-tuning beats), and $∣440 - 440∣ = 0$ ✓ (the silent match).

Figure — Beats — derivation, applications

Look at the figure: the amber envelope shrinks as the two frequencies converge, and at the top (equal frequencies) it becomes a flat ceiling — no throb at all. That flat line is what "zero beats" looks like.

Example 5 — Limiting case: gap too large (cell C5)

Worked example When there are no beats to hear

A $200$ Hz fork and a $260$ Hz fork are sounded together. What is the "beat frequency," and what does a listener actually perceive?

Forecast: Plug into the formula — but then ask: can an ear hear that as a throb?

Formula still gives a number: $∣200 - 260∣ = 60$ Hz. Why this step? The algebra never fails — the product-of-cosines is always valid.
But the ear cannot resolve it. A beat is heard only when the envelope is slow — roughly $∣ f_{1} - f_{2} ∣ ≲ 10$ Hz. At $60$ Hz the "swells" arrive faster than the ear can separate; the two tones are simply heard as two distinct notes (a musical interval, roughly a minor third here). Why this step? The parent note's condition $∣ f_{1} - f_{2} ∣$ small isn't optional decoration — it's what makes the phenomenon audible.

Answer: Mathematically $60$ Hz; perceptually no beats — you hear two separate pitches.

Verify: $60$ Hz $> 10$ Hz threshold ⇒ envelope too fast for the ear ⇒ not perceived as beating. ✓ Consistent with the definition's small-difference requirement.

Example 6 — Reading time between silences (cell C6)

Worked example From beat rate to a stopwatch

Two pipes at $340$ Hz and $348$ Hz sound together. (a) Beat frequency? (b) How much time between two successive moments of silence? (c) How many silences in $5$ seconds?

Forecast: Silences are the complete-destructive dips. How often do they come?

Beat frequency: $f_{beat} = ∣340 - 348∣ = 8$ Hz. Why? Difference of frequencies, as always.
Beat period = time for one loud–soft–loud cycle: $T_{beat} = \frac{1}{f _{beat}} = \frac{1}{8} = 0.125$ s. Why this step? Frequency and period are reciprocals: $T = 1/ f$ . One silence occurs per beat cycle, so silences are spaced by $T_{beat}$ .
Count in 5 s: $5 \times 8 = 40$ silences. Why? $8$ per second for $5$ seconds.

Answers: (a) $8$ Hz, (b) $0.125$ s, (c) $40$ silences.

Verify: $0.125 s \times 8 beats/s = 1$ s per second of beats — the reciprocal is self-consistent, and $40 = 8 \times 5$ . ✓ Units: $[s] \times [Hz] = [s] \times [1/ s] =$ dimensionless count. ✓

Example 7 — Real-world word problem: Doppler radar (cell C7)

Worked example Speed gun as a beat detector

A radar transmits $f_{0} = 10.000 000$ GHz at a car. The reflected wave returns Doppler-shifted to $10.000 002$ GHz. The receiver mixes transmitted and reflected signals and reads the beat. What beat frequency appears, and what does it encode?

Forecast: The two frequencies differ by an almost invisible amount. What tiny number pops out of the mixer?

Beat = difference of the two frequencies: $f_{beat} = ∣10.000 002 - 10.000 000∣ GHz = 0.000 002 GHz = 2000 Hz .$ Why this step? Exactly the beats mechanism — mixing two near-equal frequencies leaves the slow difference as a countable tone. This is called heterodyning.
Interpret: the $2000$ Hz beat is proportional to the car's speed (via the Doppler shift formula). A gigahertz difference of $2$ kHz — one part in five million — is measured easily as an audible-range beat. Why this step? This is the whole point of beats as a tool: they magnify an unmeasurably-tiny frequency gap into an easy-to-count rhythm.

Answer: $f_{beat} = 2000$ Hz, which encodes the target's speed.

Verify: $10.000 002 - 10.000 000 = 0.000 002$ GHz. Convert: $0.000 002 \times 1 0^{9} Hz = 2000$ Hz. ✓ Units: GHz $= 1 0^{9}$ Hz, so the arithmetic and scaling both hold.

Example 8 — Exam twist: two clues nail one value (cell C8)

Worked example Combining two beat measurements

Fork X beats with a $500$ Hz standard at $5$ beats/s. Fork X also beats with a $508$ Hz standard at $3$ beats/s. Find $f_{X}$ exactly — no wax needed.

Forecast: Each clue alone gives two candidates. Can the overlap of the two lists give a unique answer?

Clue 1 candidates: $∣ f_{X} - 500∣ = 5 \Rightarrow f_{X} \in {495, 505}$ . Why? Symmetric difference from the first standard.
Clue 2 candidates: $∣ f_{X} - 508∣ = 3 \Rightarrow f_{X} \in {505, 511}$ . Why? Symmetric difference from the second standard.
Intersect the two lists: the only value in both is ${495, 505} \cap {505, 511} = {505}$ . Why this step? $f_{X}$ must simultaneously satisfy both experiments; only the common value can be the true frequency.

Answer: $f_{X} = 505$ Hz.

Verify: $∣505 - 500∣ = 5$ ✓ and $∣505 - 508∣ = 3$ ✓ — both measurements reproduced by the single value. ✓

Recall Which fix nudges which way?

Wax adds mass → frequency decreases. Filing removes mass (or higher tension) → frequency increases. Beats grow after the fix ::: your fork was on the side the fix pushes you away from. Beats shrink after the fix ::: your fork was on the side the fix pushes you toward.

Connections

Superposition Principle — every example rests on displacements adding.
Interference of Waves — beats are interference in time.
Doppler Effect — supplies the frequency shift read as a beat in Ex 7.
Simple Harmonic Motion — the $f = \frac{1}{2 π} k / m$ argument behind wax/file (Ex 2–3).
Standing Waves — same superposition algebra, opposite-travelling waves.
Amplitude Modulation — the envelope maths reused in radio and heterodyne radar.

Worked examples — Beats — derivation, applications