Harmonics on a string — Physics — Mathemagicians Library

Before this

This leads to

Standing waves in pipes
Resonance
Damped and forced oscillations · coming soon

Why a guitar string plays one note, and not any note

Pluck the thinnest string of a guitar. In standard tuning it sounds a note called E, at 329.63 Hz. Pluck it again, harder or softer. The note is louder or quieter, but it is the same note. The string seems to know which frequency it is allowed to have.

Now press the string down exactly halfway along its length and pluck again. The note jumps up by one octave, to about 659 Hz: double the frequency. Halve the length, double the frequency. Why such a clean rule?

The last lecture showed how a standing wave forms on a string: two waves travel in opposite directions and add up to a pattern of nodes and antinodes. It ended with a promise: "The next lecture works out exactly which wavelengths fit." That is the whole of this lecture. The answer comes from one simple fact about the ends of the string, and it explains guitar notes, octaves and even why music sounds musical.

The ends cannot move

A guitar or violin string is held tightly at both ends: at one end by the bridge, at the other by the nut or by a finger. Those two points cannot move. So whatever standing wave forms on the string, it must have a node at each end.

Physicists call a rule like this a boundary condition: a condition set by the edges of the system. It sounds like a small detail. It is not. It decides everything else in this lecture.

Recall two facts from the last lecture. Neighbouring nodes are always half a wavelength apart, $\frac{λ}{2}$ . And between two neighbouring nodes there is one loop. So if there is a node at each end, the string must hold a whole number of loops. One loop, two loops, three loops: yes. One and a half loops: impossible, because that would put an antinode at a fixed end.

Stop and think

A student says: "The string can vibrate at any frequency. If I shake one end at 250 Hz, the string vibrates at 250 Hz." Is the student right?

Show the answer

Partly. If you shake the string at 250 Hz, it does move at 250 Hz. But the movement is small and messy, and it does not grow into a steady standing wave unless a whole number of half-wavelengths fits between the ends. When you pluck a string and let it go, nothing forces it. It is then free to vibrate only in the patterns that fit, and those are the frequencies you hear. What happens when you force a string at one of its special frequencies is the subject of the resonance lecture.

Which wavelengths fit

Let the string have length $L$ . Count the loops.

One loop. The whole string is half a wavelength: $L = \frac{λ _{1}}{2}$ , so $λ_{1} = 2 L$ . This is the longest wavelength that fits. It is called the fundamental, or the first harmonic.

Two loops. Two half-wavelengths fit, so $L = λ_{2}$ . There is a new node in the middle of the string.

Three loops. $L = 3 \times \frac{λ _{3}}{2}$ , so $λ_{3} = \frac{2 L}{3}$ .

The pattern is clear. With $n$ loops, $n$ half-wavelengths fit along the string:

L = n \times \frac{λ _{n}}{2} so λ_{n} = \frac{2 L}{n}

The first four harmonics of a string of length

L

fixed at both ends. The solid blue line is the string at one extreme, and the dashed line half a period later. Every harmonic has a node (dark dot) at each end. The

n

th harmonic has

n

loops,

n + 1

nodes and a frequency

n

times the fundamental.

From wavelengths to frequencies

Now use the wave equation, $v = f λ$ , so $f = \frac{v}{λ}$ . Here $v$ is the speed of the travelling waves on the string, the waves that add up to make the standing wave.

This is the key step, so read it carefully. The speed $v$ does not depend on how many loops there are. It is set by the string itself: how tight it is, and how heavy each metre of it is. So every harmonic on one string has the same $v$ . Only the wavelength changes from one harmonic to the next, and so the frequency changes too:

f_{n} = \frac{v}{λ _{n}} = \frac{v}{2 L / n} = \frac{n v}{2 L}

For $n = 1$ this gives the fundamental frequency, $f_{1} = \frac{v}{2 L}$ . And every other harmonic is a whole-number multiple of it:

Harmonics on a string fixed at both ends

$λ_{n} = \frac{2 L}{n}$ and $f_{n} = \frac{n v}{2 L} = n f_{1}$ , for $n = 1, 2, 3, \dots$

The $n$ th harmonic has $n$ loops. Every whole number $n$ is allowed, so the frequencies are $f_{1}, 2 f_{1}, 3 f_{1}, 4 f_{1}, \dots$ , evenly spaced.

That is it. That is the whole idea. Two nodes at the ends force a whole number of half-wavelengths. A whole number of half-wavelengths forces a whole-number list of frequencies.

Some books call the harmonics above the fundamental "overtones". Be careful, because the counting is different: the first overtone is the second harmonic. In this course we count harmonics.

Explore the patterns below. Change the harmonic number, and watch how the wavelength and frequency follow.

Explore — the harmonics of one string

Change the harmonic number n first. Count the loops and nodes, and read the wavelength and frequency. Then change the length or the wave speed, and see which quantities change.

Harmonic number n = 1

Length L = 0.65 m

Wave speed v = 420 m/s

The first four harmonics of a string 0.65 m long, with a wave speed of 420 m/s.

$n$	Loops	Nodes	$λ_{n} = 2 L / n$ in m	$f_{n}$ in Hz
1	1	2	1.30	≈ 323
2	2	3	0.65	≈ 646
3	3	4	≈ 0.433	≈ 969
4	4	5	0.325	≈ 1292

Doubling the length halves every frequency. Doubling the wave speed doubles every frequency. The wavelengths depend only on the length.

What should you have seen? Each step up in $n$ adds one loop and one node. The wavelength shrinks to $\frac{1}{n}$ of the fundamental's, and the frequency grows to $n$ times the fundamental. The higher harmonics also vibrate faster on the screen. A longer string lowers every frequency. A faster wave speed raises every frequency. But neither of them changes the shape of the pattern: the wavelengths depend only on the length.

What a musician changes

The formula $f_{1} = \frac{v}{2 L}$ tells a musician exactly how to change a note. The note you hear as "the pitch" is set mainly by the fundamental frequency.

Change the length. Pressing a string against a fret makes the vibrating part shorter. A smaller $L$ gives a higher $f_{1}$ . Pressing at exactly half the length halves $L$ , and so doubles the frequency: one octave higher, as in the hook.
Change the tension. Turning a tuning peg tightens the string. A tighter string carries waves faster, so $v$ rises and $f_{1}$ rises. This is how a player tunes a string without changing its length.
Change the string. A heavier string, with more mass in each metre, carries waves more slowly. That is why the low strings of a guitar are thick: a small $v$ gives a low $f_{1}$ without needing a very long string.

At Pre-University you will meet the formula that links the wave speed to the tension and the mass per metre. For now, the directions are enough: tighter means faster, heavier means slower.

A real string never vibrates in only one harmonic. A pluck or a bow starts many harmonics at once, each with its own amplitude, and you hear them all together as one note. The mix of harmonics is called the timbre, or tone quality. A guitar and a violin playing the same note have the same fundamental, but a different mix of harmonics. That is how you can tell them apart with your eyes closed.

Working with harmonics

Worked Example 1 — The first three harmonics (invented)

A guitar string 65 cm long carries waves at 420 m/s. Find the frequencies of its first three harmonics.

Convert first: $L = 0.65 m$ . Then $f_{1} = \frac{v}{2 L} = \frac{420}{2 \times 0.65} = \frac{420}{1.3} \approx 323 Hz$ .

The others are whole-number multiples: $f_{2} = 2 f_{1} \approx 646 Hz$ and $f_{3} = 3 f_{1} \approx 969 Hz$ .

Check with $f_{3}$ : $λ_{3} = \frac{2 \times 0.65}{3} \approx 0.433 m$ , and $f_{3} = \frac{420}{0.433} \approx 969 Hz$ . ✓

Worked Example 2 — Reading a pattern

A string 80 cm long vibrates with 4 loops. The wave speed is 200 m/s. Which harmonic is this? Find its wavelength, its frequency and the fundamental frequency.

Four loops means four half-wavelengths: the 4th harmonic. So $0.80 = 4 \times \frac{λ}{2}$ , which gives $λ = 0.40 m$ .

$f_{4} = \frac{v}{λ} = \frac{200}{0.40} = 500 Hz$ . The fundamental is $f_{1} = \frac{500}{4} = 125 Hz$ .

Check: $f_{1} = \frac{v}{2 L} = \frac{200}{1.6} = 125 Hz$ . ✓

Worked Example 3 — Two routes to one answer

A string 1.2 m long vibrates in its third harmonic. The wave speed is 360 m/s. Find the frequency.

Route 1, through the wavelength. $λ_{3} = \frac{2 L}{3} = \frac{2.4}{3} = 0.80 m$ , so $f_{3} = \frac{360}{0.80} = 450 Hz$ .

Route 2, through the fundamental. $f_{1} = \frac{360}{2.4} = 150 Hz$ , so $f_{3} = 3 \times 150 = 450 Hz$ .

Both routes agree. ✓ Route 2 is usually safer: find $f_{1}$ once, then multiply by $n$ . It keeps the 2 and the $n$ in separate steps, where they cannot be mixed up.

Worked Example 4 — Finding the wave speed from two patterns (invented)

In a laboratory, a string 1.20 m long is shaken by a vibration generator. Steady standing waves appear at 375 Hz and at 450 Hz, and at no frequency in between. Which harmonics are these, and what is the wave speed?

Neighbouring harmonics are $f_{1}$ apart, because $f_{n + 1} - f_{n} = (n + 1) f_{1} - n f_{1} = f_{1}$ . So $f_{1} = 450 - 375 = 75 Hz$ .

Then $375 = 5 \times 75$ and $450 = 6 \times 75$ : the 5th and 6th harmonics.

From $f_{1} = \frac{v}{2 L}$ : $v = 2 L f_{1} = 2 \times 1.20 \times 75 = 180 m/s$ .

Check: $f_{6} = \frac{6 \times 180}{2 \times 1.20} = 450 Hz$ . ✓

Counting loops correctly is where most marks are lost. The working below contains one wrong step. Find it.

Spot the error — which harmonic?

A student finds the frequency of a standing wave. One line is wrong. Tap the line you think contains the error, then read the explanation.

A student's working. The question: a string 1.5 m long, with a wave speed of 240 m/s, shows a standing wave with 6 nodes, counting the two ends. Find its frequency. The error is in line 2.

Line 1. The string shows 6 nodes, including the two ends.
Line 2. So there are 6 loops, and this is the 6th harmonic. This is the error. The loops sit between the nodes, so 6 nodes make 5 loops: the 5th harmonic.
Line 3. λ = 2L/n = 2 × 1.5 / 6 = 0.5 m.
Line 4. f = v/λ = 240 / 0.5 = 480 Hz.
Line 5. Check: f₁ = 240 / 3 = 80 Hz, and 6 × 80 = 480 Hz. The check uses the same wrong n, so it cannot catch the mistake.

The correct working: 5 loops, λ = 3/5 = 0.6 m, f = 240/0.6 = 400 Hz = 5 × 80 Hz.

What should you have learned? Nodes and loops are counted differently: a string with $n$ loops has $n + 1$ nodes. Sketch the pattern before you write $n$ .

Three classic mistakes

Trap 1 — Taking the fundamental wavelength as L

For the fundamental, the string holds one loop, which is half a wavelength. So $λ_{1} = 2 L$ , not $L$ . Using $λ_{1} = L$ doubles every frequency you calculate.

Trap 2 — Counting nodes instead of loops

The harmonic number is the number of loops. A string with 3 loops has 4 nodes, counting both ends. If you count nodes, subtract 1.

Trap 3 — Thinking higher harmonics travel faster

All harmonics on one string share the same wave speed $v$ , set by the tension and the string. A higher harmonic has a shorter wavelength and a higher frequency, and $f λ$ stays equal to $v$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A string 0.80 m long carries waves at 240 m/s. Find the wavelength and frequency of its first three harmonics.

Hint 1
Use $λ_{n} = \frac{2 L}{n}$ for the wavelengths.

Hint 2
Find $f_{1} = \frac{v}{2 L}$ first, then multiply by 2 and 3.

Full answer
$λ_{1} = 1.6 m$ , $λ_{2} = 0.80 m$ , $λ_{3} \approx 0.533 m$ . $f_{1} = \frac{240}{1.6} = 150 Hz$ , $f_{2} = 300 Hz$ , $f_{3} = 450 Hz$ . Check: $450 \times 0.533 \approx 240 m/s$ . ✓
A string vibrates with 3 loops at 270 Hz. (a) What is its fundamental frequency? (b) At what frequency would it show 5 loops? (c) How many nodes does the 5-loop pattern have, counting the ends?

Hint 1
3 loops means the 3rd harmonic, so $270 = 3 f_{1}$ .

Hint 2
The 5-loop pattern is the 5th harmonic, $5 f_{1}$ . There is one more node than loops.

Full answer
(a) $f_{1} = \frac{270}{3} = 90 Hz$ . (b) $f_{5} = 5 \times 90 = 450 Hz$ . (c) 5 loops have 6 nodes.
On a steel-string guitar, the vibrating length of each string is about 0.648 m. The thinnest string is tuned to 329.63 Hz. (a) Find the wave speed on this string. (b) The player presses the string exactly halfway along. What frequency sounds now? (c) A note a "perfect fifth" higher has 1.5 times the frequency. What vibrating length gives it?

Hint 1
(a) The note is the fundamental, so $v = 2 L f_{1}$ .

Hint 2
(b) and (c): the wave speed stays the same, so $f_{1}$ is proportional to $\frac{1}{L}$ .

Full answer
(a) $v = 2 \times 0.648 \times 329.63 \approx 427 m/s$ . (b) Half the length gives double the frequency: $2 \times 329.63 \approx 659 Hz$ , one octave higher. (c) 1.5 times the frequency needs $\frac{1}{1.5}$ of the length: $0.648 \div 1.5 = 0.432 m$ . Check: $\frac{427.2}{2 \times 0.432} \approx 494 Hz = 1.5 \times 329.63$ . ✓ On a real guitar, the 7th fret leaves almost exactly this length, about 0.4325 m.
Guitarists play "natural harmonics". They touch a string very lightly at one point, without pressing it down, and pluck. The light touch stops the string moving at that point, so it forces a node there, but the whole string still vibrates. (a) The player touches the low E string (82.41 Hz) at exactly one third of its length. Which harmonics can still sound? What is the lowest frequency you hear? (b) Explain why touching at the midpoint gives a note one octave above the open string. (c) Plucking a string exactly at its midpoint, without touching, produces no 2nd harmonic at all. Suggest why.

Hint 1
(a) Draw the first few harmonics. Which of them already have a node at one third of the length?

Hint 2
(c) Where is the midpoint in the 2nd harmonic's pattern: a node or an antinode? Can a pluck there start that pattern?

Full answer
(a) A harmonic survives only if it already has a node at $\frac{L}{3}$ . The nodes of the $n$ th harmonic are at $\frac{L}{n}, \frac{2 L}{n}, \dots$ , so $\frac{L}{3}$ is a node when $n$ is 3, 6, 9 and so on. The lowest is the 3rd harmonic: $3 \times 82.41 \approx 247 Hz$ . (b) At the midpoint, only harmonics with a node there survive: $n$ = 2, 4, 6 and so on. The lowest is $2 f_{1}$ , double the frequency: one octave higher. (c) In the 2nd harmonic, the midpoint is a node: it never moves. A pluck at the midpoint pulls the string sideways exactly there, which is the opposite of what the 2nd harmonic needs. So that pattern is not started, and neither are the 4th, 6th and other even harmonics, which also have a node at the middle. This is one reason why where you pluck changes the timbre.

The bigger idea

The link between strings and number is one of the oldest results in science. Ancient Greek writers credited the school of Pythagoras with discovering that pleasant musical intervals come from simple ratios: halve a string for the octave (2 : 1), take two thirds of it for the fifth (3 : 2). A famous story says Pythagoras found these ratios by listening to blacksmiths' hammers. That story cannot be true, because the pitch of a hammer does not follow its weight in that way. But early Pythagoreans such as Hippasus did study the ratios with care, and the string ratios are real.

This lecture shows why they work. The harmonics of one string are $f_{1}, 2 f_{1}, 3 f_{1}, \dots$ . Two notes an octave apart share many harmonics, because every harmonic of the higher note ( $2 f_{1}, 4 f_{1}, \dots$ ) is also a harmonic of the lower one. Notes a fifth apart share fewer, but still many. The more harmonics two notes share, the more smoothly they sound together. So a boundary condition, two fixed ends, sits underneath a great deal of music. The next lecture changes the boundary condition: a pipe with open or closed ends gives a different list of harmonics, and a different sound.

Key concepts in this lesson

A string fixed at both ends must have a node at each end, so it can hold only a whole number of loops.
The $n$ th harmonic has $n$ loops, each half a wavelength long, so $λ_{n} = \frac{2 L}{n}$ and the fundamental has $λ_{1} = 2 L$ .
All harmonics on one string share the same wave speed, so $f_{n} = \frac{n v}{2 L} = n f_{1}$ and the harmonics are evenly spaced in frequency.
A shorter, tighter or lighter string has a higher fundamental frequency, which is how musicians change and tune notes.
A plucked string vibrates in many harmonics at once, and the mix of harmonics, the timbre, is why different instruments sound different on the same note.