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বাং

Why does a guitar string sing one note, and why does a flute play an octave higher when you overblow?

A guitar string, a flute and the air in a bottle all do the same trick: they trap a wave between two ends, and only certain wavelengths survive. This page lets you pick the harmonic, change the length, tension and temperature, and watch the nodes and antinodes appear while the frequency readout follows the textbook formula.

The wave right nowLargest displacementNode (no displacement)Antinode (largest displacement)
Speed

Controls

Scene

1
0.65 m
70 N
5.0 g/m

Readings

Frequency f
91.0Hz
Wavelength λ
1.30m
Harmonic number n
1
Fundamental f₁
91.0Hz
Wave speed
118.3m/s
Nodes
2
Antinodes
1

How to use this simulation

  1. Start in the String scene with the defaults: 0.65 m long, 70 N of tension and 5.0 g/m. The wave speed reads 118.3 m/s and the fundamental is 91.0 Hz.
  2. Raise the harmonic number n. The string breaks up into n loops. Count the green antinode circles and compare with the readout; the red nodes are the points that never move.
  3. Change the tension and watch f₁ move with the square root of T. Quadruple the tension and the pitch doubles.
  4. Switch to the pipe open at both ends. Every harmonic is allowed, and both ends are antinodes. Then switch to the pipe closed at one end and notice that only odd harmonics appear.
  5. In either pipe, raise the temperature. The speed of sound rises, so every frequency rises with it, while the wavelengths stay the same.
  6. Look at the blue dots in the tube. They are air particles. Near a node they hardly move; near an antinode they swing back and forth. That is what a sound wave in a pipe really is.

A wave that stays in one place

Shake one end of a long rope and a wave runs along it. Tie the other end to a post and the wave bounces back. Now the incoming and the returning waves overlap, and at just the right shaking rate something odd happens: the rope seems to stop travelling and just swings up and down in fixed loops. That is a standing wave.

Nothing actually stops. Two travelling waves of the same size run in opposite directions, and they add up so that some points always cancel (the nodes) and others always swing the most (the antinodes). The pattern stays put even though the waves behind it are moving.

This matters in music and in exams. Every string and every pipe instrument works by selecting standing waves, and the questions that follow are short: which frequencies are allowed, how many nodes, and how does the pitch change if something is changed.

Nodes, antinodes and how they sit

A node is a point that never moves. An antinode is a point that moves the most. Between two neighbouring nodes the medium swings as one piece, and it swings opposite to the piece next door.

The distance between two neighbouring nodes is half a wavelength. The distance between a node and the next antinode is a quarter of a wavelength. These two facts let you read the wavelength straight from a drawing. The simulation draws a bracket for one such segment, labelled with its length.

The rule that decides which patterns are possible is about the ends. A fixed end of a string must be a node, because it cannot move. An open end of a pipe must be an antinode, because the air is free to move. A closed end of a pipe must be a node, because the wall stops the air.

End of the mediumWhat it forcesWhy
Fixed end of a stringNodeThe string cannot move there
Open end of a pipeAntinodeThe air is free to swing
Closed end of a pipeNodeThe wall stops the air

A string fixed at both ends

Both ends are nodes, so the length of the string must hold a whole number of half wavelengths: L = n λ / 2. That gives λ = 2L / n for n = 1, 2, 3, and so on.

The frequency is wave speed divided by wavelength, so f = n v / 2L. The lowest, n = 1, is the fundamental, or first harmonic. The others are exact whole-number multiples of it: second, third, fourth harmonic. For a string, harmonics and overtones differ only in counting. The second harmonic is the first overtone.

The wave speed on a string is v = √(T / μ), where T is the tension and μ the mass per metre. A tighter string carries waves faster; a heavier string carries them more slowly. With the defaults, v = √(70 ÷ 5.0 g/m) = 118.3 m/s, and the fundamental is 91.0 Hz because the wavelength is twice the 0.65 m length, 1.30 m.

The table lists the first six harmonics for that string. Notice how the number of nodes is always n + 1, counting the two ends, and the number of antinodes is n.

λ = 2L / nn = 1, 2, 3, …

f = n v / 2LHarmonics of a fixed string

v = √(T / μ)Wave speed on the string

f = (n / 2L) √(T / μ)The two combined

nFrequency (Hz)Wavelength (m)Nodes (with ends)Antinodes
191.01.3021
2182.00.6532
3273.00.4343
4364.10.3354
5455.10.2665
6546.10.2276

Why pitch depends on length, tension and thickness

Three knobs set the note, and the formula shows each at a glance. Shorter string, higher pitch: f ∝ 1/L. Press a finger at the middle and the length halves, so the fundamental becomes 182.0 Hz, an octave above 91.0 Hz. Tighter string, higher pitch: f ∝ √T. A thicker string, with larger μ, plays lower: f ∝ 1/√μ.

The table changes only the tension. Notice the last column: the frequency relative to the 70 N value. To double the pitch you need to quadruple the tension, not double it. This is why a tuning peg turns so little between notes.

That is also why the bass strings of a guitar are thick and wound with metal: the extra mass per metre lowers the note without a big drop in tension.

Tension T (N)Wave speed (m/s)Fundamental (Hz)Relative to the default
3583.764.40.71
70118.391.01.00
140167.3128.71.41
280236.6182.02.00

A pipe open at both ends

In a flute or an open organ pipe, air moves freely at both ends, so both ends are antinodes. Between two antinodes there must be a whole number of half wavelengths, so the condition is the same as for a string: λ = 2L / n and f = n v / 2L.

All harmonics are present. The pipe of 0.50 m at 20 °C, where sound travels at 343.0 m/s, has a fundamental of 343.0 Hz and a second harmonic of 686.0 Hz, exactly one octave higher.

The table gives the first four. The number of antinodes is n + 1 and the number of nodes is n, the opposite of the string, because the ends are now antinodes.

λ = 2L / nOpen pipe, n = 1, 2, 3, …

f = n v / 2LAll harmonics are present

nFrequency (Hz)Wavelength (m)NodesAntinodes (with ends)
1343.01.0012
2686.00.5023
31,029.00.3334
41,372.00.2545

A pipe closed at one end

Close one end and the picture changes. The open end is an antinode and the closed end is a node, so the pipe holds an odd number of quarter wavelengths: L = n λ / 4 with n = 1, 3, 5, and so on. Hence λ = 4L / n and f = n v / 4L.

Only odd harmonics survive. The fundamental of the same 0.50 m pipe is 171.5 Hz, half of the open pipe’s 343.0 Hz: closing one end drops the note by an octave. The next resonance is not at twice the fundamental but at three times: 514.5 Hz.

This is why a clarinet, which behaves like a closed pipe, sounds hollow and low next to a flute of the same length. The missing even harmonics change the tone colour.

In the simulation, the slider for a closed pipe moves in steps k, with n = 2k − 1. Step 1 gives n = 1, step 2 gives n = 3, step 3 gives n = 5. Check that the count of nodes and antinodes is (n + 1) / 2 for each.

λ = 4L / nClosed pipe, n = 1, 3, 5, …

f = n v / 4LOnly odd harmonics

f(closed) = ½ f(open)Same length, fundamentals only

nFrequency (Hz)Wavelength (m)NodesAntinodes (with open end)
1171.52.0011
3514.50.6722
5857.50.4033
71,200.50.2944

Temperature changes the pitch of a pipe

For a string, the pitch follows tension. For air in a pipe, it follows the speed of sound, and that depends on temperature: v ≈ 331 + 0.6 T in metres per second, with T in degrees Celsius. The wavelengths are fixed by the pipe’s length; the frequency is v ÷ λ, so it rises when the air warms up.

The table uses the closed 0.50 m pipe at three temperatures. A cold concert hall pushes wind instruments flat, which is why players warm their instruments before tuning.

A string instrument goes the other way: a warm metal string expands and loses tension, so the pitch drops. Brass and strings can drift apart in a hot room.

Temperature (°C)Speed of sound (m/s)Closed pipe f₁ (Hz)
0331.0165.5
20343.0171.5
40355.0177.5

Measuring the speed of sound with a tube

The resonance-tube experiment is a favourite practical. A tuning fork of known frequency is held over a tube of water; the water level is lowered until the tube rings loudly. That is resonance, and the air column is a closed pipe, with the water as the closed end.

The first loud sound is at L₁ = λ/4, and the next at L₂ = 3λ/4. The difference of the two lengths is exactly half a wavelength, which removes the awkward end correction at the open mouth of the tube. So v = 2 f (L₂ − L₁).

With a 512 Hz fork at 20 °C, the first resonance comes at 0.167 m and the second at 0.502 m. The difference is 0.335 m, and 2 × 512 × 0.335 gives 343.0 m/s, the speed of sound.

Solved problems

Five problems in the exam style, solved with the same functions as the animation.

Problem 1: harmonics of a string

A string 0.80 m long is under 100 N and has a mass of 4 g per metre. Find the fundamental frequency and the frequency of the third harmonic.

v = √(T / μ) = √(100 ÷ (4 × 10⁻³)) = 158.1 m/s. The fundamental is f₁ = v / 2L = 98.8 Hz. The third harmonic is three times that, 296.5 Hz, with a wavelength of 2L / 3 = 0.533 m.

Problem 2: halve the length

How does the fundamental of a string change if you press it at the middle, keeping the tension? Take the default string of 0.65 m with a fundamental of 91.0 Hz.

f ∝ 1/L, so halving L doubles f: 182.0 Hz, one octave up. This is exactly what a guitarist does when the finger presses on the twelfth fret.

Problem 3: an A pipe

A pipe must sound a fundamental of 440 Hz at 20 °C. How long must it be if it is closed at one end? And if it is open at both ends?

With v = 343.0 m/s: closed, L = v / 4f = 0.195 m. Open, L = v / 2f = 0.390 m. The open pipe has to be twice as long, which is why closed pipes save space in an organ.

Problem 4: resonance tube

A 512 Hz fork gives resonance at 0.167 m and again at 0.502 m in a tube. Find the speed of sound.

Two consecutive resonances differ by λ/2: 0.502 − 0.167 = 0.335 m, so λ = 2 × 0.335 m. v = f λ = 343.0 m/s.

Problem 5: compare open and closed

Write down the first three possible frequencies of a 0.6 m pipe when open at both ends and when closed at one end, at 20 °C.

Open: 285.8, 571.7 and 857.5 Hz. Closed: 142.9, 428.8 and 714.6 Hz. Call the open fundamental f. The open pipe gives f, 2f, 3f; the closed pipe gives f/2, 3f/2, 5f/2.

Mistakes that cost marks

Check this list before an exam. Each one has taken marks from someone.

  • Using L = n λ / 2 for a closed pipe. A closed pipe holds odd quarter wavelengths, L = n λ / 4.
  • Counting harmonics of a closed pipe as 1, 2, 3. They are 1, 3, 5, since the even ones are missing.
  • Mixing up the first overtone with the first harmonic. In an open pipe or a string the first overtone is the second harmonic; in a closed pipe it is the third.
  • Forgetting that the speed on a string depends on T and μ, not on the frequency. Changing the frequency changes the wavelength, not the speed.
  • Taking μ in grams per metre. The formula needs kilograms per metre.
  • Calling the ends of a string antinodes. A fixed end is a node.
  • Ignoring the end correction in the resonance tube. Subtracting two resonance lengths removes it.

Where else you meet standing waves

A microwave oven has standing waves too, which is why a plate must turn: the hot spots are antinodes of the electromagnetic field. Lasers use a standing wave of light between two mirrors. The electron in an atom is described by standing waves around the nucleus, which is where the idea of allowed energy levels comes from.

If you want to see what happens when two waves of nearly equal frequency meet and do not form a still pattern, try the wave-interference animation next. Standing waves are the special case where the two waves run in opposite directions.

Frequently asked questions

What is a standing wave?

A pattern formed when two identical waves travel in opposite directions and overlap. Some points, the nodes, never move, and the points halfway between them, the antinodes, swing the most.

What is the formula for the harmonics of a string fixed at both ends?

f = n v / 2L, where n = 1, 2, 3, … and v = √(T / μ). All whole-number multiples of the fundamental are allowed.

Which harmonics exist in an open pipe and in a closed pipe?

An open pipe has all harmonics, f = n v / 2L. A closed pipe has only the odd ones, f = n v / 4L with n = 1, 3, 5, … .

How far apart are a node and an antinode?

A quarter of a wavelength. Two neighbouring nodes, or two neighbouring antinodes, are half a wavelength apart.

Why does the pitch of a wind instrument rise when it warms up?

The speed of sound in air rises with temperature, about 0.6 m/s per degree. The pipe length fixes the wavelength, so the frequency, v divided by λ, rises.

How do you find the speed of sound with a resonance tube?

Find two successive resonance lengths for a fork of known frequency f. Their difference is λ/2, so v = 2 f (L₂ − L₁). This also removes the end correction.

Keep studying this topic

The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.

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