Harmonics on strings

Pluck a guitar string and it does not vibrate in just one simple shape. It vibrates simultaneously as a superposition of a whole ladder of allowed modes — the fundamental plus a series of higher-frequency partners called harmonics. Because both ends of the string are pinned down, only wave patterns that put a node exactly at each end can survive; everything else cancels itself out through interference. That single boundary condition is the entire story: it forces the allowed wavelengths into the neat sequence $\lambda_n = 2l/n$ , and with them the frequencies into $f_n = nf_0$ , integer multiples of the fundamental. This article builds on the fundamental-mode result of the stretched string and extends it to the full harmonic series: how to count nodes and antinodes for any mode, how harmonic numbering differs from overtone numbering, why a string sounds different from an air column, and how to solve for harmonic frequencies and wavelengths numerically.

The core insight

A string fixed at both ends supports only stationary-wave patterns with a node at each end, so its allowed wavelengths are $\lambda_n = 2l/n$ and its allowed frequencies form the complete integer series $f_n = nf_0$ , called the harmonics.

The core idea

Boundary condition sets the allowed shapes. A string of length $l$ , fixed rigidly at both ends, cannot support just any stationary wave — only patterns that place a node at $x=0$ and a node at $x=l$ survive repeated reflection without destructive interference wiping them out. Fitting a whole number of half-wavelengths between the ends,

l = n\frac{\lambda_n}{2} \implies \lambda_n = \frac{2l}{n}, \qquad n = 1, 2, 3, \dots

where: $l$ is the fixed length of the string, $\lambda_n$ is the wavelength of the $n$ -th allowed mode, and $n$ is the mode number (also the number of loops, or antinodes, in that mode). Each integer $n$ picks out one distinct standing-wave shape.

Turning wavelength into frequency. The wave speed on a stretched string depends only on its tension and mass per unit length, $v = \sqrt{T/m}$ , and this speed is the same for every mode on a given string. Using $v = f\lambda$ ,

f_n = \frac{v}{\lambda_n} = \frac{nv}{2l} = n\left(\frac{1}{2l}\sqrt{\frac{T}{m}}\right) = nf_0

where: $T$ is the tension, $m$ is the mass per unit length, $f_0 = \dfrac{1}{2l}\sqrt{T/m}$ is the fundamental frequency ( $n=1$ ), and $f_n$ is the frequency of the $n$ -th mode. Every harmonic frequency is simply an integer multiple of the same $f_0$ .

Mode shapes, nodes and antinodes. The figure below shows the first three modes. The $n$ -th mode has $n$ antinodes (loops) and $n+1$ nodes in total — the two fixed ends plus $(n-1)$ internal nodes.

Figure

Three stationary wave patterns on a fixed string showing one loop for the first harmonic two loops for the second harmonic and three loops for the third harmonic with nodes marked at every crossing point

Harmonics vs. overtones. The harmonics are the full frequency ladder $f_0, 2f_0, 3f_0, 4f_0, \dots$ , with the fundamental as the first harmonic. The overtones are counted only from the modes above the fundamental: $2f_0$ is the first overtone, $3f_0$ the second overtone, and so on. A stretched string produces all integer harmonics with none missing — its $n$ -th overtone is always the same as its $(n+1)$ -th harmonic. This is a special property; some other vibrating systems (an air column closed at one end, treated separately) skip every even harmonic.

Why tension and mass per unit length still matter. Although the harmonic series is a simple integer sequence, its actual pitch depends on the same physical quantities as the fundamental: $f_n = n\left(\dfrac{1}{2l}\sqrt{T/m}\right)$ . Raising the tension scales every harmonic up together (all $\propto \sqrt{T}$ ); a thicker or denser string scales every harmonic down together (all $\propto 1/\sqrt{m}$ ). The whole harmonic ladder shifts as one unit — it never changes shape, only pitch. This is also why a plucked or bowed string, which excites several modes at once, produces a rich blend of harmonics superposed on the fundamental, giving each instrument its characteristic timbre.

Derivation ADeriving the harmonic frequency series

f_n = nf_0

The goal is to show, starting only from the requirement that a string fixed at both ends must have a node at $x=0$ and a node at $x=l$ , that the allowed frequencies form the exact integer sequence $f_n = nf_0$ — and to make explicit why no in-between frequency, such as $1.5f_0$ , can ever survive on such a string.

Step 1. State the boundary condition. A stationary wave on the string is built from two oppositely travelling waves reflecting at the fixed ends. For the pattern to persist without cancelling itself, both ends must be points of zero displacement at all times — nodes. Any stationary-wave shape that does not place a node exactly at $x=0$ and $x=l$ destructively interferes with its own reflection and dies out.

Step 2. Count half-wavelengths that fit. Between two consecutive nodes of a stationary wave lies exactly half a wavelength. If the string supports $n$ loops (antinodes) between the two end nodes, then it must contain exactly $n$ half-wavelengths along its length $l$ , giving $l = n(\lambda_n/2)$ , so $\lambda_n = 2l/n$ for $n = 1, 2, 3, \dots$ . Only these discrete wavelengths satisfy the boundary condition; no continuous range of wavelengths is allowed.

Step 3. Convert wavelength to frequency using the wave speed. The transverse wave speed on the string, $v = \sqrt{T/m}$ , is fixed once the tension $T$ and mass per unit length $m$ are fixed — it does not depend on which mode is excited. Using $v = f\lambda$ for each mode, $f_n = v/\lambda_n = v/(2l/n) = nv/(2l)$ .

Step 4. Identify the fundamental and factor it out. Setting $n=1$ gives $f_0 = v/(2l) = \dfrac{1}{2l}\sqrt{T/m}$ , the fundamental frequency. Substituting back, $f_n = n \cdot \dfrac{v}{2l} = nf_0$ . Because $n$ can only be a positive integer, the string's allowed frequencies are exactly $f_0, 2f_0, 3f_0, \dots$ — the complete harmonic series, with no fractional multiples possible.

Worked example

A string of length $0.75\ \text{m}$ is fixed at both ends and has a fundamental frequency of $220\ \text{Hz}$ . Find (a) the frequencies of the first three harmonics, (b) the wavelength of the third harmonic, and (c) which harmonic has a frequency closest to $1000\ \text{Hz}$ .

Step 1. List the first three harmonics using $f_n = nf_0$ with $f_0 = 220\ \text{Hz}$ : $f_1 = 220\ \text{Hz}$ (fundamental, first harmonic), $f_2 = 2 \times 220 = 440\ \text{Hz}$ (second harmonic), $f_3 = 3 \times 220 = 660\ \text{Hz}$ (third harmonic).

Step 2. Find the wavelength of the third harmonic using $\lambda_n = 2l/n$ with $l = 0.75\ \text{m}$ and $n = 3$ : $\lambda_3 = \dfrac{2 \times 0.75}{3} = \dfrac{1.5}{3} = 0.5\ \text{m}$ .

Step 3. Find which harmonic sits closest to $1000\ \text{Hz}$ by solving $f_n = nf_0$ for $n$ : $n = 1000/220 = 4.545$ . This is not an integer, so test the two neighbouring integers: $n=4$ gives $f_4 = 880\ \text{Hz}$ , and $n=5$ gives $f_5 = 1100\ \text{Hz}$ .

Step 4. Compare the two distances from $1000\ \text{Hz}$ : $|1000 - 880| = 120\ \text{Hz}$ versus $|1000 - 1100| = 100\ \text{Hz}$ . Since $100 < 120$ , the fifth harmonic is closer. The first three harmonics are $220\ \text{Hz}$ , $440\ \text{Hz}$ and $660\ \text{Hz}$ ; the third harmonic has wavelength $0.5\ \text{m}$ ; and the fifth harmonic, $1100\ \text{Hz}$ , lies closest to $1000\ \text{Hz}$ .

Competitive stretch

A sonometer wire vibrates in its third harmonic ( $n=3$ ) between fixed ends $0.9\ \text{m}$ apart, under tension $50\ \text{N}$ with mass per unit length $5\times10^{-4}\ \text{kg/m}$ . Step 1. Fundamental: $f_0 = \dfrac{1}{2l}\sqrt{T/m} = \dfrac{1}{1.8}\sqrt{\dfrac{50}{5\times10^{-4}}} = \dfrac{1}{1.8}\sqrt{1\times10^5} = \dfrac{316.2}{1.8} = 175.7\ \text{Hz}$ . Step 2. Actual vibrating frequency: $f_3 = 3f_0 = 3 \times 175.7 = 527.1\ \text{Hz}$ . Step 3. For $n=3$ , antinodes $= n = 3$ and nodes $= n+1 = 4$ (two at the fixed ends, two internal). The wire vibrates at $527.1\ \text{Hz}$ with 3 antinodes and 4 nodes.

Common traps

Assuming a fixed string can vibrate at a non-integer multiple of $f_0$ , such as $1.5f_0$ — the fixed-node boundary condition at both ends permits only whole-number multiples, $f_n = nf_0$ .
Confusing "harmonic number" with "overtone number" — the fundamental is the first harmonic but is not an overtone at all; the first overtone is the second harmonic. Check which convention a question uses.
Miscounting nodes and antinodes — the $n$ -th mode has $n$ antinodes but $(n+1)$ nodes, because the two fixed ends always count as nodes.
Assuming every vibrating system produces all integer harmonics like a string does — some air columns (closed pipes) only produce odd harmonics, so the "complete series" rule does not generalise automatically.

Ask the tutor: Still mixing up harmonic and overtone numbering, or want to work through a few more frequency and wavelength problems together?

Why can a string only vibrate at these specific frequencies and not anything in between?

Because both ends are physically pinned down, they must always be points of zero displacement — nodes. Only stationary-wave patterns that place a node exactly at each end survive being reflected back and forth without cancelling themselves through interference. That requirement only allows a whole number of half-wavelengths to fit along the string, which is a discrete condition, not a continuous one. So only the wavelengths $\lambda_n = 2l/n$ , and the frequencies $f_n = nf_0$ , are physically possible; everything in between destructively interferes and dies out almost instantly.

Is the second harmonic the same as the first overtone?

Yes, for a stretched string specifically. Harmonics are counted from the fundamental as the first harmonic, so the sequence is $f_0$ (first harmonic), $2f_0$ (second harmonic), $3f_0$ (third harmonic), and so on. Overtones are counted only from the first mode above the fundamental, so $2f_0$ is the first overtone, $3f_0$ is the second overtone. Because a string produces every integer harmonic with none skipped, its $n$ -th overtone always equals its $(n+1)$ -th harmonic — but this neat matching is specific to strings and does not automatically hold for every vibrating system.

Does a real plucked guitar string vibrate in only one harmonic at a time?

No. A real pluck or bow excites many modes simultaneously — the fundamental plus several harmonics superposed on top of it, each with its own amplitude depending on where and how the string was disturbed. The string's actual shape at any instant is the sum of all these modes together, not a single clean loop pattern. It is this particular mixture of harmonic strengths, not just the fundamental frequency, that gives an instrument its distinctive timbre, and why a violin and a guitar sound different even when playing the identical fundamental note.

Harmonics on strings

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads