Beats

Sound two tuning forks together that are almost, but not quite, the same pitch, and you do not hear two separate notes — you hear one note whose loudness throbs, rising and falling like a slow pulse. That throb is called a beat, and it is one of the simplest, most useful pieces of physics a musician ever leans on: a piano tuner, a violinist checking a string against a reference note, even a pilot balancing engine RPMs, all listen for the same rise-and-fall and adjust until it slows to nothing. Beats are a direct, audible consequence of superposition. Two waves of slightly different frequency drift in and out of step, alternately reinforcing and cancelling, and the ear hears that alternation as loudness pulsing in time rather than pitch changing in space. The ISC syllabus keeps this qualitative — no trigonometric derivation of the beat waveform is expected — but a clean wave-picture reasoning plus the counting rule for beat frequency is exactly what exam questions test, and that is what this article builds carefully.

The core insight

Two nearly-equal frequencies superpose to drift in and out of phase, and the ear hears that alternation as beats, with beat frequency simply the magnitude of the frequency difference, fbeat=∣f1−f2∣f_{beat}=|f_1-f_2|.

The core idea

Superposition of two close frequencies. When two sound waves of frequencies f1f_1 and f2f_2, with f1≈f2f_1 \approx f_2, travel through the same region of air, the principle of superposition says the net displacement at any instant is just the sum of the two individual displacements. Because the frequencies are close but not identical, the two waves cannot stay locked in step forever — one is completing cycles very slightly faster than the other, so the phase relationship between them keeps drifting.

In phase, out of phase. At some instants the two waves arrive crest-to-crest (and trough-to-trough): they are in phase, and their amplitudes add directly, producing a moment of louder sound. A little later the faster wave has pulled slightly ahead, until its crest lines up with the slower wave's trough: the waves are out of phase, and they largely cancel, producing a moment of near-silence. The waves then drift back into phase, and the cycle of loud-soft-loud repeats. One full loud-soft-loud cycle is one beat.

Counting the beats. Because f1f_1 is completing cycles slightly faster than f2f_2 (say f1>f2f_1>f_2), it gains exactly one whole extra cycle over f2f_2 every 1f1−f2\dfrac{1}{f_1-f_2} seconds. Each time it gains a full cycle, the two waves realign in phase, giving one loud pulse. So the number of beats heard per second is:

fbeat=∣f1−f2∣f_{beat} = |f_1 - f_2|

where: fbeatf_{beat} is the number of beats heard per second (Hz), and f1,f2f_1, f_2 are the frequencies of the two interfering sources (Hz). The formula only needs the two frequencies — no amplitude, no phase, no distance — which is exactly why it is such a fast, practical tool for tuning.

Why beats fade with a bigger gap. Beats are only cleanly perceptible when f1f_1 and f2f_2 are close — typically the difference stays below about 1010 to 15 Hz15\ \text{Hz}. The ear's loudness response is too sluggish to track pulses that come and go faster than roughly this rate; beyond it, instead of a throb, you simply hear two distinct notes played together.

Figure
Two nearby waveforms drawn together showing crests aligning at the start producing a loud beat, drifting out of alignment in the middle producing near silence, and realigning at the end producing the next loud beat
Two close frequencies drifting in and out of phase wave 1 (blue, f1) and wave 2 (red, f2 slightly different) slowly drift apart in phase resultant: loud where waves reinforce, quiet where they nearly cancel — the throb of a beat loud quiet loud loud time between successive loud moments = 1 / (f1 minus f2)
Derivation AWhy the beat frequency equals the difference of the two frequencies

Starting from the fact that two sources of frequency f1f_1 and f2f_2 (with f1>f2f_1>f_2) both begin a cycle together at t=0t=0, show step by step that the next instant at which they return to being exactly in phase — and hence the next loud beat — occurs after a time 1f1−f2\dfrac{1}{f_1-f_2}, which directly gives the counting rule fbeat=∣f1−f2∣f_{beat}=|f_1-f_2| used throughout this topic.

Step 1. At t=0t=0 both waves start a new cycle together, so they are in phase and the resultant is at its loudest.

Step 2. In any time tt, the first wave completes f1tf_1 t cycles and the second completes f2tf_2 t cycles. Since f1>f2f_1>f_2, the first wave is always slightly ahead; the number of extra cycles it has completed over the second wave by time tt is (f1−f2)t(f_1-f_2)t.

Step 3. The two waves return to being exactly in phase again precisely when the first wave has gained one whole extra cycle over the second — a lead of any non-integer number of cycles leaves them out of step. So the condition for the next in-phase (loud) moment is (f1−f2)t=1(f_1-f_2)t = 1.

Step 4. Solving gives t=1f1−f2t = \dfrac{1}{f_1-f_2}: this is the time between one loud moment and the next, i.e. the period of one beat. Since frequency is the reciprocal of period, the number of beats per second is f1−f2f_1-f_2. Writing the result so it holds regardless of which frequency is larger gives the general rule fbeat=∣f1−f2∣f_{beat}=|f_1-f_2|, confirming the counting argument used in the core idea above.

Worked example

Two tuning forks, A and B, are sounded together and produce 6 beats in 2 seconds. Fork A has a known frequency of 256 Hz256\ \text{Hz}. What are the possible frequencies of fork B, and how would you find which one is correct?

Step 1. Find the beat frequency from the count given: fbeat=6 beats2 s=3 Hzf_{beat} = \dfrac{6\ \text{beats}}{2\ \text{s}} = 3\ \text{Hz}.

Step 2. Apply the rule fbeat=∣fA−fB∣f_{beat}=|f_A-f_B|, so ∣256−fB∣=3|256 - f_B| = 3, giving two possibilities: fB=256+3=259 Hzf_B = 256+3 = 259\ \text{Hz} or fB=256−3=253 Hzf_B = 256-3 = 253\ \text{Hz}.

Step 3. The beat count alone cannot distinguish between these, since it only measures the magnitude of the difference, not its sign. To resolve the ambiguity, a small piece of wax is stuck onto fork B, which lowers its frequency slightly. If the beat rate then increases, fork B's original frequency was the lower one, 253 Hz253\ \text{Hz} (waxing pushed it further from A); if the beat rate decreases, fork B's original frequency was the higher one, 259 Hz259\ \text{Hz} (waxing pulled it closer to A).

So fork B's frequency is either 259 Hz\mathbf{259\ Hz} or 253 Hz\mathbf{253\ Hz}, resolved by the waxing test.

Competitive stretch

A note of 440 Hz440\ \text{Hz} sounded with a second note gives 5 beats per second; loading the second source (which lowers its frequency) drops the beat rate to 3 beats per second. Initially ∣440−f2∣=5|440-f_2|=5, so f2=435 Hzf_2=435\ \text{Hz} or 445 Hz445\ \text{Hz}. Lowering f2f_2 moves it closer to 440 Hz440\ \text{Hz} only if f2f_2 started above it, and the beat rate did decrease — so f2=445 Hzf_2 = 445\ \text{Hz} is the answer; had f2f_2 been 435 Hz435\ \text{Hz}, lowering it further would have increased the beat rate, contradicting the observation.

Common traps

  • Thinking beats need the frequencies to be far apart — it is the opposite: beats are heard clearly only for small differences (roughly under 1010-15 Hz15\ \text{Hz}); a larger gap sounds like two separate notes, not a throb.
  • Assuming beats mean energy is destroyed during the quiet moments — no energy is lost; superposition only redistributes the same total energy in time, with near-zero displacement at some instants and large displacement at others.
  • Forgetting the sign ambiguity: a measured beat count gives only ∣f1−f2∣|f_1-f_2|, never which source is higher — a physical test such as waxing or loading one source, and watching whether the beat rate rises or falls, is needed to break the tie.
  • Mixing up beats with stationary (standing) waves — beats are the interference of two close but different frequencies in time, heard as pulsing loudness at one place, while stationary waves are the interference of identical frequencies travelling in opposite directions, producing fixed nodes and antinodes in space.

Ask the tutor: Want to see the in-phase/out-of-phase waveform sketch worked through frame by frame, or try another waxing-experiment problem with different numbers?

Why do we hear loudness change instead of pitch change during beats

Beats arise because the resultant amplitude of the superposed waves rises and falls as the two waves drift in and out of phase, and amplitude is what the ear perceives as loudness. The underlying pitch we hear is close to the average of the two frequencies, essentially unchanged; only the amplitude envelope pulses, which is why beats are heard as a throbbing loudness, not a wavering pitch.

Can beats occur with light waves the same way as with sound

The same superposition mathematics applies to any wave, including light, and beating between two close optical frequencies does happen (it is used in some laser and spectroscopy techniques). At the school level the topic is restricted to audible sound beats between two close acoustic frequencies, since that is what the ear can directly perceive as a throb and what the syllabus's qualitative treatment covers.

Is there a limit to how many beats per second the ear can actually detect as a throb

Yes — beyond roughly 1010 to 1515 beats per second the ear's loudness response is too sluggish to resolve the individual pulses, and the sensation smoothly turns into hearing two separate steady notes rather than one pulsing note. This is why tuning by ear works best when the two sources are already close, and the beat rate is deliberately reduced further as tuning proceeds.

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Where this leads