Normal Modes of a String

A string can only sing at certain notes: its harmonics.

13 minUpdated 21 October 2026normal modeharmonicsovertonesfundamental frequencyresonance

Lesson at a glanceShow the key points

Waves reflected back and forth along a string reinforce the source only if L = nλ/2: the string resonates.
Fixed at both ends: νₙ = (n/2L)√(F/μ), n = 1, 2, 3, … (15.22, 15.26); fundamental ν₀ = (1/2L)√(F/μ).
All harmonics (integer multiples of ν₀) are allowed; the nth overtone is the (n + 1)th harmonic. This is why strings make musical sounds.
Normal mode n: y = 2A sin(nπx/L) cos ωt (15.27), with n − 1 nodes between the ends.
A plucked string vibrates in a mixture of normal modes.
Fixed at one end, free at the other: ν = (2n + 1)v/4L (15.28); fundamental v/4L and only odd harmonics.

A guitar string can’t vibrate at just any frequency. Its length, tension and weight pick out a ladder of special frequencies, and those are the notes it plays.

How resonance builds up

Tie one end of a string of length L to a wall and the other to a tuning fork vibrating gently along the string. The fork sends sine waves toward the wall; they reflect (inverted) at the fixed end and travel back; the heavy, barely moving fork acts like another fixed end and reflects them again (inverted again). After two reflections a wave has the same shape as the fork’s new output but has travelled an extra 2L.

If 2L is a whole number of wavelengths (L = nλ/2), the old wave and the new one are in step: they add to amplitude 2A, then 3A, and so on. The string soaks up energy from the fork until losses (air drag, stiffness) balance the input. The steady state is a standing wave with large antinodes: the string is in resonance with the fork. If the condition fails, successive reflections arrive out of step, the build-up doesn’t happen, and the string barely moves.

A string fixed at both ends

Analytically, waves travelling each way add to a standing wave. Both ends must be nodes: y = 0 at x = 0 fixes the phase, giving $y = 2A\sin kx\cos\omega t$ (15.24), and y = 0 at x = L needs $\sin kL = 0$ , so $kL = n\pi$ :

L = \frac{n\lambda}{2} \quad (15.25), \qquad n = 1, 2, 3, \dots

Natural frequencies of a string fixed at both ends(15.22, 15.26)

\nu_n = \frac{n}{2L}\sqrt{\frac F\mu}, \qquad \nu_1 = \frac1{2L}\sqrt{\frac F\mu}\ \text{(fundamental)}

These are the string’s natural (normal, resonant) frequencies. Frequencies above the fundamental are overtones; integer multiples of a frequency are its harmonics. For a string fixed at both ends, every harmonic is an allowed overtone: the first overtone is the 2nd harmonic, the second overtone the 3rd harmonic, and so on. That neat harmonic ladder is why strings make such musical sounds, from the violin to the sitar.

Normal modes

Vibrating at its nth natural frequency, the string is in its nth normal mode:

Normal mode n(15.27)

y = 2A\sin\frac{n\pi x}{L}\cos\omega_n t

Fundamental (n = 1): nodes only at the ends, one antinode in the middle.
First overtone (n = 2): a node at the middle, antinodes at L/4 and 3L/4.
Second overtone (n = 3): nodes at L/3 and 2L/3, antinodes at L/6, L/2, 5L/6.
In general the nth overtone has n nodes between the ends and n + 1 antinodes.

Interactive

The notes a string can play

A string fixed at both ends can vibrate steadily only in patterns that fit a whole number of half-wavelengths: ν = n·v/2L, every harmonic. With one end free the free end must be an antinode, so only odd harmonics occur: ν = (2n − 1)v/4L. A real pluck excites a mixture of modes.

v = \sqrt{F/\mu}

132m/s

harmonic 1102Hz

fundamental102Hz

Ends

Show

Mode number

n

Length

L

65cm

Tension

F

70N

Mass per unit length

\mu

4.0g/m

Worked example

A guitar string

A guitar string has a vibrating length of 65 cm, mass per unit length 4.0 g/m and tension 70 N. Find its fundamental frequency and its third harmonic.

A string fixed at one end only

Fix x = 0 but let x = L move freely sideways (in practice, tie it to a very light thread). Then x = L must be an antinode: $\sin kL = \pm1$ , so $kL = (n + \tfrac12)\pi$ and

String with one free end(15.28)

\nu = \Big(n + \frac12\Big)\frac{v}{2L} = \frac{(2n + 1)v}{4L}, \qquad n = 0, 1, 2, \dots

fundamental v/4L; only odd harmonics (3ν₀, 5ν₀, …) occur

Worked example

One end free

Waves travel at 40 m/s on a 1.0 m string fixed at one end and free at the other. List its three lowest natural frequencies.

What to remember

1Repeated reflections reinforce the driver only when L = nλ/2: resonance.
2Fixed both ends: $\nu_n = (n/2L)\sqrt{F/\mu}$ ; all harmonics occur.
3Normal mode n: y = 2A sin(nπx/L) cos ωt, with n − 1 internal nodes.
4A pluck excites a mixture of modes.
5One end free: $\nu = (2n + 1)v/4L$ ; odd harmonics only.

Check yourself

A string's fundamental is 200 Hz. Its second overtone is…

How many nodes (including the ends) does the 4th harmonic of a string fixed at both ends have?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchStanding waves on stringsKhan Academy WatchSinging plates: standing waves on Chladni platesPhysics Girl PlayNormal ModesPhET · CU Boulder PlayFourier: Making WavesPhET · CU Boulder ReadStanding waves and resonanceOpenStax University Physics §16.6 ReadHarmonics and patternsThe Physics Classroom ReadVibrating stringHyperPhysics ReadString vibrationWikipedia ReadNormal modeWikipedia

PreviousStanding Waves Next Laws of Vibrating Strings & the Sonometer

Normal Modes of a String

A string can only sing at certain notes: its harmonics.

13 minUpdated 21 October 2026normal modeharmonicsovertonesfundamental frequencyresonance

Lesson at a glanceShow the key points

Waves reflected back and forth along a string reinforce the source only if L = nλ/2: the string resonates.
Fixed at both ends: νₙ = (n/2L)√(F/μ), n = 1, 2, 3, … (15.22, 15.26); fundamental ν₀ = (1/2L)√(F/μ).
All harmonics (integer multiples of ν₀) are allowed; the nth overtone is the (n + 1)th harmonic. This is why strings make musical sounds.
Normal mode n: y = 2A sin(nπx/L) cos ωt (15.27), with n − 1 nodes between the ends.
A plucked string vibrates in a mixture of normal modes.
Fixed at one end, free at the other: ν = (2n + 1)v/4L (15.28); fundamental v/4L and only odd harmonics.

A guitar string can’t vibrate at just any frequency. Its length, tension and weight pick out a ladder of special frequencies, and those are the notes it plays.

How resonance builds up

A string fixed at both ends

L = \frac{n\lambda}{2} \quad (15.25), \qquad n = 1, 2, 3, \dots

Natural frequencies of a string fixed at both ends(15.22, 15.26)

\nu_n = \frac{n}{2L}\sqrt{\frac F\mu}, \qquad \nu_1 = \frac1{2L}\sqrt{\frac F\mu}\ \text{(fundamental)}

Normal modes

Vibrating at its nth natural frequency, the string is in its nth normal mode:

Normal mode n(15.27)

y = 2A\sin\frac{n\pi x}{L}\cos\omega_n t

Fundamental (n = 1): nodes only at the ends, one antinode in the middle.
First overtone (n = 2): a node at the middle, antinodes at L/4 and 3L/4.
Second overtone (n = 3): nodes at L/3 and 2L/3, antinodes at L/6, L/2, 5L/6.
In general the nth overtone has n nodes between the ends and n + 1 antinodes.

Interactive

The notes a string can play

v = \sqrt{F/\mu}

132m/s

harmonic 1102Hz

fundamental102Hz

Ends

Show

Mode number

n

Length

L

65cm

Tension

F

70N

Mass per unit length

\mu

4.0g/m

Worked example

A guitar string

A guitar string has a vibrating length of 65 cm, mass per unit length 4.0 g/m and tension 70 N. Find its fundamental frequency and its third harmonic.

A string fixed at one end only

Fix x = 0 but let x = L move freely sideways (in practice, tie it to a very light thread). Then x = L must be an antinode: $\sin kL = \pm1$ , so $kL = (n + \tfrac12)\pi$ and

String with one free end(15.28)

\nu = \Big(n + \frac12\Big)\frac{v}{2L} = \frac{(2n + 1)v}{4L}, \qquad n = 0, 1, 2, \dots

fundamental v/4L; only odd harmonics (3ν₀, 5ν₀, …) occur

Worked example

One end free

Waves travel at 40 m/s on a 1.0 m string fixed at one end and free at the other. List its three lowest natural frequencies.

What to remember

1Repeated reflections reinforce the driver only when L = nλ/2: resonance.
2Fixed both ends: $\nu_n = (n/2L)\sqrt{F/\mu}$ ; all harmonics occur.
3Normal mode n: y = 2A sin(nπx/L) cos ωt, with n − 1 internal nodes.
4A pluck excites a mixture of modes.
5One end free: $\nu = (2n + 1)v/4L$ ; odd harmonics only.

Check yourself

A string's fundamental is 200 Hz. Its second overtone is…

How many nodes (including the ends) does the 4th harmonic of a string fixed at both ends have?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousStanding Waves Next Laws of Vibrating Strings & the Sonometer

How resonance builds up#

A string fixed at both ends#

Normal modes

A string fixed at one end only#

Check yourself

Keep exploring

How resonance builds up#

A string fixed at both ends#

Normal modes

A string fixed at one end only#

Check yourself

Keep exploring

How resonance builds up

A string fixed at both ends

A string fixed at one end only

How resonance builds up

A string fixed at both ends

A string fixed at one end only