Pipes and Air Columns

Standing waves can form not only on strings but also in columns of air. When sound waves travel through a pipe, they can reflect from the ends of the pipe and interfere with incoming waves. At particular frequencies, this interference creates stable standing-wave patterns.

The possible patterns depend strongly on the boundary conditions at the ends of the pipe. A pipe may be:

• open at both ends
• closed at one end and open at the other

These two systems support different wavelengths and different sets of harmonics.

This physics is central to instruments such as flutes, recorders, clarinets, saxophones, trumpets, trombones, and organ pipes.

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Sound Waves in Air

Sound in air is a longitudinal wave.

Air particles oscillate back and forth approximately parallel to the direction in which the wave travels.

As they move, they create alternating regions of:

• compression
• rarefaction

The disturbance transfers energy through the air without the air itself traveling with the wave over large distances.

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Standing Waves in an Air Column

A sound wave traveling along a pipe reflects when it reaches an end or boundary.

The incoming and reflected waves overlap.

At certain frequencies, their interference creates a stable standing-wave pattern.

These frequencies are the natural frequencies of the air column.

The air column then resonates strongly at these frequencies.

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Displacement Nodes and Antinodes

Standing waves in air can be represented in terms of air-particle displacement.

A displacement node is a position where air particles have minimum displacement.

A displacement antinode is a position where air particles have maximum displacement.

For the simplified pipe models used here:

closed end → displacement node

open end → displacement antinode

These boundary conditions determine which standing-wave patterns are possible.

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Why a Closed End Is a Displacement Node

At a closed end, the air cannot move through the solid boundary.

Therefore, the air particles immediately next to the closed end have very little longitudinal displacement.

The closed end is therefore modeled as a:

displacement node

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Why an Open End Is a Displacement Antinode

At an open end, the air is free to move.

The air particles can therefore undergo relatively large displacement.

The open end is modeled as a:

displacement antinode

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Pressure Nodes and Antinodes

Sound can also be represented using pressure variations rather than particle displacement.

The pattern is reversed.

At an open end:

displacement antinode → pressure node

At a closed end:

displacement node → pressure antinode

Therefore:

This distinction is important when interpreting standing-wave diagrams.

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Open Pipes

An open pipe is open at both ends.

For particle displacement:

both ends must be antinodes

The standing-wave pattern must therefore fit between two displacement antinodes.

The simplest possible pattern is the fundamental mode.

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Fundamental Mode of an Open Pipe

In the fundamental mode:

• there is an antinode at each end
• there is one node in the middle
• half a wavelength fits inside the pipe

Therefore:

L = λ₁/2

so:

λ₁ = 2L

where L is the length of the air column.

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Fundamental Frequency of an Open Pipe

For any wave:

v = fλ

For an open pipe:

λ₁ = 2L

Therefore:

f₁ = v/(2L)

where:

• f₁ = fundamental frequency
• v = speed of sound
• L = pipe length

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Example: Fundamental of an Open Pipe

An open pipe has length:

L = 0.85 m

Take the speed of sound as:

v = 340 m/s

Fundamental wavelength:

λ₁ = 2(0.85)

λ₁ = 1.70 m

Frequency:

f₁ = 340/1.70

f₁ = 200 Hz

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Harmonics in an Open Pipe

An ideal open pipe supports all integer harmonics:

1st, 2nd, 3rd, 4th, 5th, ...

The general wavelength relationship is:

λₙ = 2L/n

The frequency relationship is:

fₙ = nv/(2L)

or:

fₙ = nf₁

where n = 1, 2, 3, 4, ...

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Open-Pipe Harmonic Relationships

As harmonic number increases:

• frequency increases
• wavelength decreases

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Second Harmonic in an Open Pipe

For the second harmonic:

L = λ₂

Therefore:

λ₂ = L

and:

f₂ = 2f₁

The displacement pattern contains:

antinode → node → antinode → node → antinode

across the pipe.

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Third Harmonic in an Open Pipe

For the third harmonic:

L = 3λ₃/2

Therefore:

λ₃ = 2L/3

and:

f₃ = 3f₁

More nodes and antinodes appear as the harmonic number increases.

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Worked Example: Open-Pipe Harmonics

An open pipe has:

L = 1.00 m

and:

v = 340 m/s

Fundamental:

f₁ = 340/[2(1.00)]

f₁ = 170 Hz

Second harmonic:

f₂ = 340 Hz

Third harmonic:

f₃ = 510 Hz

Fourth harmonic:

f₄ = 680 Hz

The corresponding third-harmonic wavelength is:

λ₃ = 2(1.00)/3

λ₃ ≈ 0.667 m

Check:

v = fλ

v ≈ 510(0.667)

v ≈ 340 m/s

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Closed Pipes

In introductory standing-wave problems, a closed pipe usually means a pipe:

• closed at one end
• open at the other

The closed end must be a displacement node.

The open end must be a displacement antinode.

These different boundary conditions produce a different set of allowed standing waves.

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Fundamental Mode of a Closed Pipe

The shortest standing-wave pattern that can connect:

node → antinode

contains one quarter of a wavelength.

Therefore:

L = λ₁/4

so:

λ₁ = 4L

This is different from an open pipe, where:

λ₁ = 2L.

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Fundamental Frequency of a Closed Pipe

Using:

v = fλ

and:

λ₁ = 4L

we obtain:

f₁ = v/(4L)

Therefore, for pipes of the same length and with the same sound speed:

a closed pipe has half the fundamental frequency of an open pipe

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Example: Closed-Pipe Fundamental

A closed pipe has:

L = 0.50 m

Take:

v = 340 m/s

Fundamental wavelength:

λ₁ = 4(0.50)

λ₁ = 2.00 m

Frequency:

f₁ = 340/2.00

f₁ = 170 Hz

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Harmonics in a Closed Pipe

An ideal closed pipe supports only the odd harmonics:

1st, 3rd, 5th, 7th, ...

The allowed wavelengths are:

λₙ = 4L/n

where:

n = 1, 3, 5, 7, ...

The allowed frequencies are:

fₙ = nv/(4L)

or:

fₙ = nf₁

for odd n only.

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Closed-Pipe Harmonic Relationships

The ideal closed pipe does not support:

• second harmonic
• fourth harmonic
• sixth harmonic

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Why Even Harmonics Are Missing

A closed pipe must always satisfy both boundary conditions:

closed end = displacement node

and:

open end = displacement antinode

A second-harmonic pattern would not satisfy these two different conditions at the ends.

Only an odd number of quarter-wavelength sections can fit while maintaining a node at one end and an antinode at the other.

Therefore:

L = λ/4, 3λ/4, 5λ/4, 7λ/4, ...

This produces only the odd harmonics.

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Modes vs Harmonic Numbers

This distinction is especially important for closed pipes.

The first allowed mode is:

1st harmonic

The second allowed mode is:

3rd harmonic

The third allowed mode is:

5th harmonic

The fourth allowed mode is:

7th harmonic

Therefore, the phrase second mode does not mean second harmonic for a closed pipe.

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Worked Example: Closed-Pipe Harmonics

A closed pipe has:

L = 0.80 m

and:

v = 320 m/s

Fundamental frequency:

f₁ = 320/[4(0.80)]

f₁ = 320/3.20

f₁ = 100 Hz

Next allowed harmonic:

f₃ = 3(100)

f₃ = 300 Hz

Next:

f₅ = 5(100)

f₅ = 500 Hz

Next:

f₇ = 7(100)

f₇ = 700 Hz

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Finding Wavelength in a Closed Pipe

Using the same pipe:

L = 0.80 m

For the third harmonic:

λ₃ = 4L/3

λ₃ = 4(0.80)/3

λ₃ = 3.20/3

λ₃ ≈ 1.07 m

Check:

v = fλ

v ≈ 300(1.07)

v ≈ 320 m/s

The small difference results from rounding.

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Comparing Open and Closed Pipes

The major differences can be summarized as follows:

These differences come directly from the different boundary conditions.

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Same Length, Different Fundamental Frequency

Suppose an open pipe and a closed pipe both have:

L = 1.00 m

and:

v = 340 m/s

Open pipe:

f₁ = 340/[2(1.00)]

f₁ = 170 Hz

Closed pipe:

f₁ = 340/[4(1.00)]

f₁ = 85 Hz

Therefore:

f_closed = 1/2 f_open

for equal lengths and sound speeds.

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Finding Pipe Length from Frequency

The equations can also be rearranged.

For an open pipe:

f₁ = v/(2L)

Therefore:

L = v/(2f₁)

For a closed pipe:

f₁ = v/(4L)

Therefore:

L = v/(4f₁)

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Example: Designing an Open Pipe

An open pipe must have a fundamental frequency of:

425 Hz

Take:

v = 340 m/s

Then:

L = 340/[2(425)]

L = 340/850

L = 0.40 m

An ideal open pipe approximately 0.40 m long would therefore have the required fundamental frequency.

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Example: Designing a Closed Pipe

A closed pipe must have a fundamental frequency of:

170 Hz

Take:

v = 340 m/s

Then:

L = 340/[4(170)]

L = 340/680

L = 0.50 m

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Resonance in an Air Column

An air column has natural frequencies determined by:

• its length
• its boundary conditions
• the speed of sound

If the air column is driven at one of these frequencies, a large-amplitude standing wave can form.

This is resonance.

At resonance, energy is transferred efficiently from the source to the vibrating air column.

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Resonance with a Tuning Fork

A common experiment uses a tuning fork and an adjustable air column.

The tuning fork produces sound at a fixed frequency.

As the air-column length changes, certain lengths satisfy the standing-wave condition.

At these lengths:

• a standing wave forms
• resonance occurs
• the sound becomes noticeably louder

The increase in volume provides evidence that the air column is resonating.

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Why Resonance Amplifies Sound

When the driving frequency matches or closely approaches a natural frequency, each cycle of the source transfers energy efficiently to the air column.

The oscillation builds.

This produces:

• larger pressure variations
• larger particle oscillations
• stronger sound

The amplitude does not increase forever because energy is continually lost through processes such as:

• sound radiation
• friction
• viscosity
• thermal effects

Eventually, energy input and energy loss balance.

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Wind Instruments

Wind instruments use resonating air columns to produce sound.

Examples include:

• flute
• recorder
• clarinet
• saxophone
• trumpet
• trombone
• organ pipes

The musician creates vibrations that excite particular natural frequencies of the air column.

The resulting resonances determine which frequencies are strongly produced.

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Changing Pitch by Changing Air-Column Length

For an open pipe:

f₁ = v/(2L)

For a closed pipe:

f₁ = v/(4L)

In both cases:

f ∝ 1/L

Therefore:

longer air column → lower frequency

shorter air column → higher frequency

This principle is used throughout wind instruments.

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Flutes and Recorders

Opening and closing holes changes the effective length of the resonating air column.

Opening a hole closer to the mouthpiece can shorten the effective air column.

A shorter effective length produces:

higher frequency

Closing holes can increase the effective length and produce:

lower frequency

Real instruments are more complicated than ideal pipes, but the standing-wave model explains the basic behaviour.

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Trombones

A trombone changes pitch by moving its slide.

Extending the slide increases the effective air-column length.

Therefore:

longer air column → lower frequency

Moving the slide inward shortens the air column and raises the frequency.

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Trumpets and Similar Instruments

Instruments such as trumpets use valves to change the effective length of the air path.

Pressing different valve combinations adds sections of tubing.

Increasing the air-column length generally lowers the resonant frequencies.

The player's lips provide the vibrating source that excites resonances in the instrument.

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Clarinet and Closed-Pipe Behaviour

A clarinet is often introduced as an approximate example of a closed-open air column because the reed end behaves approximately like a closed end.

This helps explain why its resonance pattern strongly emphasizes odd harmonics.

Real clarinets are more complicated because of:

• tone holes
• bell shape
• bore geometry
• reed behaviour

The ideal closed-pipe model nevertheless captures important features.

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Flute and Open-Pipe Behaviour

A flute is often modeled approximately as an open-open air column.

This means it can support a sequence resembling:

f₁, 2f₁, 3f₁, 4f₁, ...

Again, real instruments have corrections due to their geometry and openings.

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Organ Pipes

Organ pipes provide particularly clear examples of open and closed air columns.

An organ can contain both:

• open pipes
• stopped or closed pipes

A closed pipe can produce a lower fundamental frequency than an open pipe of the same length.

This allows organ builders to obtain low frequencies using shorter pipes.

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Temperature and the Speed of Sound

The speed of sound in air depends on temperature.

At approximately room temperature, it is often taken as:

340–343 m/s

As temperature increases, the speed of sound increases.

Since:

f = v/λ

a change in sound speed can change the resonant frequencies of an air column whose physical length remains fixed.

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End Correction

The simple pipe equations assume that the displacement antinode occurs exactly at an open end.

In reality, the oscillating air extends slightly beyond the physical opening.

Therefore, the effective length of the air column is slightly longer than its physical length.

This effect is called:

end correction

For many introductory calculations, end correction is ignored.

For precise acoustic measurements, it can become important.

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Strings and Air Columns

Strings and air columns both support standing waves, but their physical waves are different.

A string supports primarily:

transverse waves

Sound in an air column consists of:

longitudinal waves

Both systems can nevertheless be analyzed using:

• nodes
• antinodes
• wavelength
• frequency
• harmonics
• resonance

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Fixed String vs Open Pipe

A string fixed at both ends has:

nodes at both ends

An open pipe, using displacement representation, has:

antinodes at both ends

Despite this difference, both have the same fundamental wavelength relationship:

λ₁ = 2L

Both also support:

all integer harmonics

Therefore:

fₙ = nf₁

for both ideal systems.

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Fixed String vs Closed Pipe

A string fixed at both ends supports:

all integer harmonics

A closed-open pipe supports:

odd harmonics only

The difference comes from the boundary conditions.

String:

node → node

Closed pipe:

node → antinode

The node-antinode requirement changes the allowed wavelengths.

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Comparison of Standing-Wave Systems

This comparison shows why boundary conditions are so important in standing-wave systems.

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Similarities Between Strings and Air Columns

Both systems:

• form standing waves through interference
• have natural frequencies
• exhibit resonance
• contain nodes and antinodes
• support a fundamental mode
• can support higher harmonics
• obey v = fλ
• produce different frequencies when their effective length changes

These shared principles allow the same wave ideas to explain many different musical instruments.

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Important Differences

The physical oscillation is different.

For strings:

the material of the string moves transversely

For sound in pipes:

air particles oscillate longitudinally

Wave speed is also controlled differently.

On a string:

v = √(T/μ)

and depends on tension and linear mass density.

In air:

wave speed depends mainly on properties such as:

• temperature
• gas composition

The boundary conditions are also different.

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Complete Open-Pipe Problem

An open pipe is:

L = 0.60 m

Take:

v = 360 m/s

Fundamental:

f₁ = v/(2L)

f₁ = 360/[2(0.60)]

f₁ = 300 Hz

Third harmonic:

f₃ = 3(300)

f₃ = 900 Hz

Third-harmonic wavelength:

λ₃ = 2L/3

λ₃ = 1.20/3

λ₃ = 0.40 m

Check:

v = fλ

v = 900(0.40)

v = 360 m/s

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Complete Closed-Pipe Problem

A closed pipe is:

L = 0.75 m

Take:

v = 330 m/s

Fundamental:

f₁ = v/(4L)

f₁ = 330/[4(0.75)]

f₁ = 330/3.00

f₁ = 110 Hz

Third harmonic:

f₃ = 3(110)

f₃ = 330 Hz

Fifth harmonic:

f₅ = 5(110)

f₅ = 550 Hz

There is no second or fourth harmonic in the ideal model.

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Complete Comparison Problem

An open pipe and a closed pipe both have length:

0.50 m

Take:

v = 340 m/s

Open pipe:

f₁ = 340/[2(0.50)]

f₁ = 340 Hz

Closed pipe:

f₁ = 340/[4(0.50)]

f₁ = 170 Hz

Therefore:

the open pipe has twice the fundamental frequency of the closed pipe

when their lengths and sound speeds are equal.

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Identifying a Pipe from Its Harmonics

Suppose a pipe has resonant frequencies:

120 Hz, 360 Hz, 600 Hz, 840 Hz

These frequencies are:

1 × 120

3 × 120

5 × 120

7 × 120

Only odd harmonics are present.

This strongly suggests an ideal:

closed-open pipe

By contrast, frequencies such as:

120 Hz, 240 Hz, 360 Hz, 480 Hz

match the pattern expected from an ideal:

open-open pipe

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A Reliable Strategy for Pipe Problems

First identify the pipe.

Open-open pipe:

λₙ = 2L/n

fₙ = nv/(2L)

n = 1, 2, 3, 4, ...

Closed-open pipe:

λₙ = 4L/n

fₙ = nv/(4L)

n = 1, 3, 5, 7, ...

Then:

1. identify the harmonic
2. determine the correct wavelength relationship
3. use v = fλ if necessary
4. check whether the harmonic is allowed
5. include units
6. verify that the answer fits the boundary conditions

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Checking Answers Physically

After calculating, ask:

• Does a longer pipe give a lower fundamental frequency?
• Does a shorter pipe give a higher frequency?
• Does an open pipe include all integer harmonics?
• Does a closed pipe include only odd harmonics?
• Is the open-pipe fundamental wavelength 2L?
• Is the closed-pipe fundamental wavelength 4L?
• Does the standing-wave pattern satisfy the required nodes and antinodes?

These checks can reveal mistakes before the final answer is accepted.

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Common Mistakes

Treating every pipe as an open pipe

First identify the boundary conditions.

Calling both ends of an open pipe nodes

For displacement, open ends are antinodes.

Calling the open end of a closed pipe a node

For displacement, it is an antinode.

Confusing pressure and displacement diagrams

Pressure nodes correspond approximately to displacement antinodes, and vice versa.

Using λ = 2L for a closed-pipe fundamental

For a closed-open pipe:

λ₁ = 4L

Including even harmonics in an ideal closed pipe

Closed-open pipes support:

1, 3, 5, 7...

Calling the second allowed mode the second harmonic

For a closed pipe, the second allowed mode is the third harmonic.

Forgetting that open pipes support all integer harmonics

Ideal open-open pipes support:

1, 2, 3, 4...

Confusing string boundary conditions with open-pipe conditions

Fixed string ends are nodes.

Open pipe ends are displacement antinodes.

Ignoring the speed of sound

Use the value provided in the question rather than automatically assuming 340 m/s.

Ignoring end correction in precise experiments

The effective length of a real open pipe can be slightly greater than its physical length.

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Key Terms

Air column — volume of air within a pipe that can support standing sound waves.

Open pipe — pipe open at both ends.

Closed pipe — in this context, a pipe closed at one end and open at the other.

Standing wave — stationary interference pattern produced by waves traveling in opposite directions.

Displacement node — location where air-particle displacement is minimal.

Displacement antinode — location where air-particle displacement is maximum.

Pressure node — location where pressure variation is minimal.

Pressure antinode — location where pressure variation is maximum.

Boundary condition — physical requirement imposed at the end of a standing-wave system.

Fundamental frequency — lowest natural frequency of a system.

Harmonic — allowed frequency related to the fundamental frequency.

Natural frequency — frequency at which a system naturally tends to oscillate.

Resonance — large-amplitude oscillation produced when a system is driven near a natural frequency.

Driving frequency — frequency supplied by an external vibrating source.

Effective length — length of the air column that behaves acoustically as though it were vibrating.

End correction — adjustment accounting for vibration extending slightly beyond an open end.

Longitudinal wave — wave in which particles oscillate parallel to the direction of wave travel.

Transverse wave — wave in which particles oscillate perpendicular to the direction of wave travel.

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Key Takeaways

• Standing sound waves can form when waves reflect and interfere inside air columns.
• Pipe boundary conditions determine the allowed standing-wave patterns.
• An open end is a displacement antinode and approximately a pressure node.
• A closed end is a displacement node and approximately a pressure antinode.
• An open-open pipe has displacement antinodes at both ends.
• The fundamental wavelength of an open pipe is λ₁ = 2L.
• The fundamental frequency of an open pipe is f₁ = v/(2L).
• Open pipes support all integer harmonics.
• For an open pipe, λₙ = 2L/n and fₙ = nf₁.
• A closed-open pipe has a displacement node at the closed end and an antinode at the open end.
• The fundamental wavelength of a closed pipe is λ₁ = 4L.
• The fundamental frequency of a closed pipe is f₁ = v/(4L).
• Ideal closed pipes support only odd harmonics.
• The second allowed mode of a closed pipe is the third harmonic.
• Resonance occurs when a driving frequency matches or approaches a natural frequency of the air column.
• Resonance produces a large-amplitude standing wave and amplified sound.
• Longer air columns generally produce lower frequencies.
• Shorter air columns generally produce higher frequencies.
• Wind instruments control pitch partly by changing their effective air-column length.
• Fixed strings and open pipes both have λ₁ = 2L and support all integer harmonics, despite having different boundary conditions.
• Closed pipes differ because their node-antinode boundary conditions allow only odd harmonics.
• Standing waves in strings are transverse, while sound waves in air columns are longitudinal.
• The same principles of interference, natural frequency, harmonics, and resonance connect strings, pipes, and musical instruments.

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Check Your Understanding

1. Explain how a standing wave forms inside an air column.
2. For particle displacement, identify whether an open end and a closed end are nodes or antinodes.
3. Explain why pressure nodes and displacement nodes occur at different locations in an air column.
4. An open pipe is 0.80 m long. Determine the wavelength of its fundamental mode.
5. An open pipe is 0.75 m long and the speed of sound is 330 m/s. Calculate its fundamental frequency.
6. For the pipe in Question 5, calculate the frequencies of the second, third, and fourth harmonics.
7. A closed pipe is 0.60 m long. Determine the wavelength of its fundamental mode.
8. A closed pipe is 0.50 m long and the speed of sound is 340 m/s. Calculate its fundamental frequency.
9. For the pipe in Question 8, calculate the frequencies of the next three allowed harmonics after the fundamental.
10. Explain why an ideal closed pipe supports only odd harmonics.
11. An open pipe and a closed pipe are each 1.00 m long. If the speed of sound is 340 m/s, calculate and compare their fundamental frequencies.
12. An open pipe has a fundamental frequency of 400 Hz. If the speed of sound is 320 m/s, determine the pipe length.
13. A pipe produces strong resonances at 150 Hz, 450 Hz, 750 Hz, and 1050 Hz. Determine whether the pipe behaves more like an ideal open pipe or closed pipe and explain your reasoning.
14. Explain how changing the effective length of an air column allows a flute, clarinet, or trombone to produce different pitches.
15. Compare a string fixed at both ends, an open-open pipe, and a closed-open pipe. Include their boundary conditions, fundamental wavelengths, allowed harmonics, type of wave, and one musical-instrument application of each.