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Beats and the Doppler Effect · Lesson 2 of 5

Hearing Beats and Tuning

13 min read3 ideas6 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 3
Hearing beats
Maths you need

Experiencing beats

The best way to understand beats is to hear them. We start with a reference tone of \(440\,\text{Hz}\), the note A. Choose a chip in Fig 2.1 and press Listen. The drawn curve follows the chip, and so does the sound.

Four sounds to hear: one tone, beats at 2 and 4 per second, and unisonA graph of the resultant displacement s against time t from 0 to 1 second. Four chips choose a sound: 440 hertz alone; 440 and 442 hertz; 440 and 444 hertz; and unison, 440 and 440 hertz at amplitude 0.7 each. The drawn curve follows the chip. For 440 alone its outline is flat. For 440 with 442 the dashed envelope pinches to zero twice in the second, and for 440 with 444 four times; a dimension line marks one beat between two silences. For unison the outline is flat again, at 1.4. The fast oscillation inside is drawn slowed, 12 cycles per second instead of 440. A readout gives the two frequencies, the beats per second and the pitch heard. A Listen chip plays the same tones for six seconds; nothing plays until it is clicked.Ots 1 s one beat 440 Hz alone: a steady tone
Fig 2.1Four sounds. Choose a chip, then press Listen to hear it for six seconds; nothing plays until you press Listen. The drawing spans one second. The dashed envelope is drawn at its true rate. The fast oscillation inside it is drawn slowed, \(12\) cycles instead of \(440\).

Reference tone (440 Hz). The first chip plays \(440\,\text{Hz}\) alone. Its loudness is steady, and its envelope in Fig 2.1 is flat.

Slow beats (2 Hz). Now add a second tone at \(442\,\text{Hz}\), so \(\Delta f = 2\,\text{Hz}\). Listen for the two slow pulses per second. In the drawing, the envelope pinches to zero twice in the second.

Faster beats (4 Hz). If we raise the second frequency to \(444\,\text{Hz}\), then \(\Delta f = 4\,\text{Hz}\). The pulsing becomes noticeably faster, four times a second. The pitch you hear rises a little too, to the average, \(442\,\text{Hz}\).

Perfect unison (0 Hz beats). When an instrument is perfectly in tune, the two frequencies match, and \(\Delta f = 0\). The beat frequency becomes zero, and the pulsing disappears. What remains is a steady, combined tone. The last chip plays two tones of \(440\,\text{Hz}\), each at amplitude \(0.7\). Their envelope is flat at \(0.7 + 0.7 = 1.4\).

CheckpointUnanswered

In the unison chip, two \(440\,\text{Hz}\) tones of amplitude \(0.7\) each sound together, in phase. What does the resultant look like?

Tuning by beats

Tuning musical instruments

This effect is fundamental to tuning instruments. A musician plays a reference note and the same note on the instrument, both at once. If beats are heard, the instrument is out of unison with the reference. The musician adjusts the tuning until the beats become slower. When the beats finally disappear, \(f_{\text{beat}} = 0\), and the two frequencies are perfectly matched.

In the world
Tuning a sitar against a tanpura

Before a concert, a sitar player sounds a string together with the steady drone of the tanpura. If the two notes differ by \(3\,\text{Hz}\), the sound throbs three times a second. The player turns the peg slowly. If the throbbing slows, the string is moving towards the drone; if it quickens, the peg is turned the wrong way. The player stops when the throbbing can no longer be heard.

CheckpointUnanswered

Two tuning forks produce \(4\) beats per second when sounded together. You increase the frequency of one fork slightly, and the beat frequency decreases to \(2\) beats per second. Which fork's frequency did you increase?

CheckpointUnanswered

A tuning fork of frequency \(384\,\text{Hz}\) gives \(3\) beats per second when sounded with another tuning fork. What are the possible frequencies of the second fork?

Waxing and filing a fork

One beat rate, two possible frequencies

The last check shows a limit of beats. A beat rate gives the size of the mismatch, \(|f_1 - f_2|\), but not its sign. An unknown fork can lie above the known one or below it by the same amount. To decide, we change the unknown fork a little in a known direction and listen again.

A tuning fork vibrates like a mass on a spring. Its prongs are the mass, and the bending of the metal is the spring. So the frequency of a fork depends on the mass of its prongs:

  • Waxing a prong adds mass. More mass lowers the frequency.
  • Filing the ends of the prongs removes mass. Less mass raises the frequency.
Deeper

Where the metal is filed matters. Filing near the ends of the prongs removes mass where it moves most, so the frequency rises. Filing near the base, where the prongs join, weakens the bend that acts as the spring, and the frequency falls instead. Problems mean the ends unless they say otherwise.

Waxing or filing a fork moves both candidate frequenciesA frequency axis from 246 to 266 hertz. Fork A is marked at 256 hertz. Fork B gives 4 beats per second with A, so B may be at 252 or at 260 hertz; small open dots mark these two starting values. Three chips: before, B loaded with wax, and B filed. Waxing moves both candidates 2 hertz lower, to 250 and 258 hertz; filing moves both 2 hertz higher, to 254 and 262 hertz. Dimension lines join A to each candidate and give the beats per second. Two readouts say what each possible starting value would give now. A Listen chip plays A with the fork of the worked example, which started at 260 hertz.246248250252254256258260262264266f (Hz) 4 beats/s 4 beats/s A B from 260 Hz: now 260 Hz, 4 beats/s B from 252 Hz: now 252 Hz, 4 beats/s
Fig 2.2Fork A is \(256\,\text{Hz}\). Fork B gives \(4\) beats per second with A, so B may be \(252\) or \(260\,\text{Hz}\). A small change moves both candidates the same way, \(2\,\text{Hz}\) down for wax and \(2\,\text{Hz}\) up for filing. One candidate moves towards A and the other away. Listen plays A with the B of the worked example below.

Fig 2.2 shows the rule at work. Waxing B moves both possible values down by \(2\,\text{Hz}\). The value \(260\,\text{Hz}\) moves to \(258\,\text{Hz}\), towards A, so its beats fall to \(2\) per second. The value \(252\,\text{Hz}\) moves to \(250\,\text{Hz}\), away from A, so its beats rise to \(6\) per second. So the change in the beat rate tells us which candidate was true.

ExampleWhich side of 256 Hz?

A tuning fork A of frequency \(256\,\text{Hz}\) gives \(4\) beats per second with a fork B of unknown frequency. When a little wax is put on the prongs of B, the beats fall to \(2\) per second. Find the frequency of B before it was waxed.

Reveal one step at a time and finish the rest yourself.
Common mistake

The rule assumes a small change. Suppose so much wax is added that \(260\,\text{Hz}\) falls to \(252\,\text{Hz}\). Then the beats are still \(4\) per second, because B has crossed over to the other side of A. A beat rate that does not change after waxing means exactly this. B was above A and dropped by twice the beat frequency.

CheckpointUnanswered

A fork of unknown frequency gives \(5\) beats per second with a \(512\,\text{Hz}\) fork. The ends of the prongs of the unknown fork are filed a little, and the beats become \(7\) per second. What was the unknown frequency before filing?

Checkpoint · numericUnanswered

Fork P has a frequency of \(300\,\text{Hz}\), and fork Q gives \(4\) beats per second with it. This time a little wax is put on P, the known fork, and the beats become \(6\) per second. What is the frequency of Q, in Hz?

Hz± 0 %
Checkpoint · in your own wordsUnanswered

A tuner hears \(3\) beats per second between a guitar string and a reference fork. Why can the beat rate alone not tell the tuner whether to tighten or loosen the string, and what should the tuner do?

In the world
Two engines in step

The two propellers of a twin-engine aircraft each hum at a frequency set by the speed of its engine. If the two speeds differ a little, passengers hear a slow, tiring throb, which is beats. Pilots, and in newer aircraft an automatic synchroniser, adjust one engine until the throb disappears. The engines then turn at the same rate, and the cabin is quieter.

Exam traps
  1. Giving only one value for an unknown frequency from a beat rate.
    \(|f_1 - f_2| = f_{\text{beat}}\) has two solutions, \(f_2 = f_1 \pm f_{\text{beat}}\). Keep both until more information decides.
  2. Thinking that wax raises the frequency of a fork.
    Wax adds mass, and more mass vibrates more slowly. Filing the ends of the prongs removes mass and raises the frequency.
  3. Forgetting which fork was changed.
    If the known fork is waxed or filed, move the known value, then compare it with both candidates for the unknown one.