3.2.4.3. Pipes

Table of Contents

Air Columns in Pipes

A pipe can support standing sound waves in the air inside it. In this context, the vibrating quantity is mainly the motion of the air and the associated pressure changes. The pipe acts like a resonator, which means only certain wave patterns fit well inside it. These allowed patterns produce the musical notes of wind instruments and many everyday acoustic effects.

To understand pipes, the most important idea is that the ends of the pipe impose boundary conditions. These conditions determine where nodes and antinodes occur, and therefore which wavelengths are allowed.

Open and Closed Pipe Boundaries

In an open end, the air can move more freely, so the displacement of air is largest there. This makes an open end a displacement antinode. At the same place, the pressure variation is very small, so it is a pressure node.

In a closed end, the air cannot move back and forth at the wall, so the displacement must be zero there. This makes a closed end a displacement node. At the same place, pressure variation is largest, so it is a pressure antinode.

This distinction is the key to all pipe resonances.

For air columns in pipes:
Open end $\rightarrow$ displacement antinode, pressure node
Closed end $\rightarrow$ displacement node, pressure antinode

Open Pipes

An open pipe has both ends open. Since both ends are displacement antinodes, only wave patterns with antinodes at both ends are allowed.

For the simplest standing wave, the pipe length contains half a wavelength:

$$
L = \frac{\lambda_1}{2}
$$

so the fundamental wavelength is

$$
\lambda_1 = 2L
$$

Using the wave relation $v = f\lambda$, the fundamental frequency is

$$
f_1 = \frac{v}{2L}
$$

Higher resonances occur when the pipe contains one full extra half wavelength each time:

$$
L = n\frac{\lambda_n}{2}, \quad n = 1,2,3,\dots
$$

So the allowed wavelengths are

$$
\lambda_n = \frac{2L}{n}
$$

and the allowed frequencies are

$$
f_n = \frac{nv}{2L}, \quad n = 1,2,3,\dots
$$

These frequencies are all integer multiples of the fundamental. Therefore, an open pipe supports all harmonics.

For an open pipe of length $L$:
$$
f_n = \frac{nv}{2L}, \quad n = 1,2,3,\dots
$$
All harmonics are present.

Rendering scientific sketch…

Standing wave patterns in an open pipe

Closed Pipes

A closed pipe has one end closed and one end open. The closed end must be a displacement node, and the open end must be a displacement antinode. This changes the allowed wave patterns.

For the simplest resonance, the pipe length contains one quarter of a wavelength:

$$
L = \frac{\lambda_1}{4}
$$

$$
\lambda_1 = 4L
$$

and the fundamental frequency is

$$
f_1 = \frac{v}{4L}
$$

The next possible pattern must still satisfy a node at one end and an antinode at the other. This happens when the length contains three quarters of a wavelength, then five quarters, and so on:

$$
L = \frac{(2n-1)\lambda_n}{4}, \quad n = 1,2,3,\dots
$$

Thus,

$$
\lambda_n = \frac{4L}{2n-1}
$$

and

$$
f_n = \frac{(2n-1)v}{4L}, \quad n = 1,2,3,\dots
$$

Only odd multiples of the fundamental appear. So a closed pipe supports only odd harmonics.

For a pipe closed at one end:
$$
f_n = \frac{(2n-1)v}{4L}, \quad n = 1,2,3,\dots
$$
Only odd harmonics are present, $f_1, 3f_1, 5f_1, \dots$

Rendering scientific sketch…

Standing wave patterns in a closed pipe

Allowed Resonances

The difference between open and closed pipes becomes very clear when the resonance frequencies are compared.

Pipe type	Boundary condition	Fundamental wavelength	Fundamental frequency	Harmonics
Open at both ends	antinode at each end	$\lambda_1 = 2L$	$f_1 = \dfrac{v}{2L}$	all
Closed at one end	node and antinode	$\lambda_1 = 4L$	$f_1 = \dfrac{v}{4L}$	odd only

An open pipe of a given length has a fundamental frequency twice that of a closed pipe of the same length.

For the same length $L$ and sound speed $v$:
Open pipe: $f_1 = \dfrac{v}{2L}$
Closed pipe: $f_1 = \dfrac{v}{4L}$
So the open pipe has a fundamental frequency that is twice as large.

Pressure Patterns in Pipes

Although pipe standing waves are often drawn using displacement, they can also be described using pressure. The pressure pattern is opposite to the displacement pattern at the ends.

End of pipe	Displacement	Pressure
Open end	antinode	node
Closed end	node	antinode

This is very useful in real acoustics, because sound is often detected as pressure variation. A microphone placed at different points in a pipe would detect stronger or weaker pressure oscillations depending on whether it is near a pressure antinode or pressure node.

Examples from Instruments

Many wind instruments behave approximately like pipes. A flute is roughly an open pipe, because both ends act approximately as open ends. A clarinet is closer to a closed pipe, because the reed end behaves approximately like a closed end. This is why clarinets emphasize odd harmonics more strongly than flutes.

These are approximations, because real instruments have mouthpieces, tone holes, changing diameters, and end corrections. Still, the simple pipe model captures the main resonance behavior.

End Correction

In a real pipe, the displacement antinode at an open end is not located exactly at the physical end of the pipe. It lies slightly outside. Because of this, the effective length of the pipe is a little longer than its actual length.

If the physical length is $L$, the effective length may be written as

$$
L_{\text{eff}} = L + \Delta L
$$

where $\Delta L$ is the end correction. For an open pipe with two open ends, both ends contribute. For a pipe closed at one end, only the open end contributes.

This correction matters when accurate frequencies are needed, especially in musical acoustics.

In real pipes, use the effective length rather than the physical length when high accuracy is needed:
$$
f \approx \frac{v}{2L_{\text{eff}}} \quad \text{or} \quad f \approx \frac{v}{4L_{\text{eff}}}
$$

Simple Calculation

Suppose an open pipe has length $L = 0.50 \, \text{m}$ and the speed of sound is $v = 340 \, \text{m/s}$. Its fundamental frequency is

$$
f_1 = \frac{v}{2L} = \frac{340}{2(0.50)} = 340 \, \text{Hz}
$$

The next harmonics are

$$
f_2 = 680 \, \text{Hz}, \quad f_3 = 1020 \, \text{Hz}
$$

Now consider a pipe of the same length closed at one end:

$$
f_1 = \frac{v}{4L} = \frac{340}{4(0.50)} = 170 \, \text{Hz}
$$

The next resonances are

$$
f_2 = 3f_1 = 510 \, \text{Hz}, \quad f_3 = 5f_1 = 850 \, \text{Hz}
$$

Notice that $340 \, \text{Hz}$ is allowed for the open pipe but not for the closed pipe as the second resonance. The closed pipe skips the even harmonics.

Main Ideas to Remember

Pipes support standing sound waves because only certain wavelengths satisfy the boundary conditions at the ends. Open ends are displacement antinodes, while closed ends are displacement nodes. An open pipe supports all harmonics with frequencies $f_n = \dfrac{nv}{2L}$. A pipe closed at one end supports only odd harmonics with frequencies $f_n = \dfrac{(2n-1)v}{4L}$. Real pipes often require an effective length that is slightly larger than the physical length.

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