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Standing Sound Waves · Lesson 2 of 5

Pipes Open at Both Ends

14 min read3 ideas5 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 3
An open pipe holds whole half wavelengths

A pipe open at both ends

A flute is a good example of a pipe open at both ends. The player blows across a hole near one end, and that end is open to the air. The far end is open too. Such a pipe is an open pipe.

A pipe open at both ends must have a displacement antinode at each end, as Lesson 1 showed. The simplest standing wave that meets this condition fits exactly half a wavelength into the length of the pipe, \(L\). That wave has an antinode at each end and a single node in the middle.

Other waves are allowed too, but each must also have antinodes at both ends. Adjacent antinodes are half a wavelength apart, as on a string. So the pipe must contain a whole number of half-wavelengths. This is the condition for an open pipe:

\[ L = n\left(\frac{\lambda_n}{2}\right), \qquad n = 1, 2, 3, \dots \]

Which waves fit a pipe open at both endsA horizontal pipe 0.65 metre long, open at both ends. Inside it, a pair of curves shows the displacement envelope of a standing wave whose wavelength is set by a slider. The envelope always starts with an antinode, marked A, at the open left end. A dot rides on the envelope at the right end. The wave fits only when the right end is an antinode too: when the length holds a whole number of half-wavelengths, at 1.30, 0.65 and 0.43 metre. Then an A appears at the right end. A readout gives the length in half-wavelengths and says whether the wave fits. A L = 1.30 half-wavelengths: no match
SimA pipe \(0.65\,\text{m}\) long, open at both ends. The slider sets the wavelength. The envelope always starts with an antinode at the left end. It fits the pipe only when the right end is an antinode too.

Start the slider in the figure above at \(1.00\,\text{m}\). The envelope begins with an antinode at the left end, but it meets the right end partway down a slope. The pipe then holds \(1.30\) half-wavelengths, and no standing wave can form. Slide to \(1.30\,\text{m}\), \(0.65\,\text{m}\) and \(0.43\,\text{m}\). At each of these the pipe holds \(1\), \(2\) or \(3\) half-wavelengths, and the right end becomes an antinode.

Every harmonic sounds in an open pipe

The allowed frequencies

Each allowed wavelength gives an allowed frequency through \(f_n = v/\lambda_n\). Here \(v\) is the speed of sound in the air of the pipe. The derivation below puts the two together.

Deriving \(f_n = n\left(\dfrac{v}{2L}\right) = nf_1\) step 1 of 5
1

\[ L = n\left(\dfrac{\lambda_n}{2}\right), \qquad n = 1, 2, 3, \dots \]

Why. the condition for an open pipe: an antinode at each end, with a whole number of half-wavelengths between them.

RememberThe harmonics of a pipe open at both ends
\[ \begin{aligned} L &= n\left(\dfrac{\lambda_n}{2}\right) \\ f_n &= n\left(\dfrac{v}{2L}\right) = nf_1, \quad n = 1, 2, 3, \dots \end{aligned} \]
Holds whena column of air of length \(L\), open at both ends, with sound travelling at speed \(v\) in it.
Symbols\(L\) the length of the pipe; \(\lambda_n\) the wavelength of mode \(n\); \(f_n\) its frequency; \(f_1 = v/2L\) the fundamental frequency; \(v\) the speed of sound; \(n\) the harmonic number
Units\(L\) and \(\lambda_n\) in m, \(f\) in Hz, \(v\) in m/s

The fundamental frequency is \(f_1 = v/2L\). Critically, an open pipe produces all whole-number multiples of the fundamental frequency. This complete set of harmonics, \(f_1, 2f_1, 3f_1, \dots\), is the harmonic series of the pipe. It is the same series that a string fixed at both ends has.

A real note from an instrument is a mixture. The fundamental sets the pitch, and many harmonics sound quietly with it. The mixture gives the note its quality, or timbre. The complete set of harmonics of an open pipe gives instruments like flutes and recorders their bright, rich timbre.

CheckpointUnanswered

For a pipe open at both ends, which harmonics are present?

The modes of an open pipe

The first three modes

Fig 2.1 shows the first three modes of vibration of a pipe open at both ends. Choose each chip in turn and press play. The pipe is \(0.65\,\text{m}\) long, the length of a flute.

The first three modes of a pipe open at both endsA flute modelled as a pipe 0.65 metre long, open at both ends, filled with columns of air particles. Three chips, n = 1, 2 and 3, choose a mode. A dashed envelope inside the pipe shows the displacement amplitude: it is widest at both open ends and at every antinode, marked A above the pipe, and pinched to nothing at every node, marked N below it. Mode 1 has one node, mode 2 has two and mode 3 has three. When played, the columns swing along the pipe; mode n oscillates n times as fast as mode 1. A readout gives the wavelength, 2L over n, and the frequency with the speed of sound 343 metres per second: 1.30 metres and 264 hertz, 0.65 metre and 528 hertz, 0.43 metre and 792 hertz. A Listen chip plays the frequency of the chosen mode. A A N n = 1: λ = 2L = 1.30 m, f = 264 Hz
Fig 2.1The first three modes of a pipe open at both ends, \(0.65\,\text{m}\) long, with the speed of sound \(343\,\text{m/s}\). Both open ends are always antinodes (A), and mode \(n\) has \(n\) nodes (N) inside. Listen plays the chosen mode.

The fundamental mode, \(n = 1\). This is the simplest standing wave. It has an antinode at each end and one node in the middle. Each open end is a displacement antinode (A), and the centre is a node (N). One half-wavelength fits in the pipe, so \(L = \lambda_1/2\) and \(f_1 = v/2L\).

The second harmonic, \(n = 2\). There are antinodes at both ends, with two nodes in between. A third antinode sits at the centre. One full wavelength fits in the pipe, so \(L = \lambda_2\). The frequency of this mode is \(f_2 = 2f_1\).

The third harmonic, \(n = 3\). Both open ends are antinodes again, and there are three nodes inside. Three half-wavelengths fit in the pipe, so \(L = 3\lambda_3/2\). The frequency of this mode is \(f_3 = 3f_1\).

Press play on each chip. Mode \(n\) oscillates \(n\) times as fast as mode 1. The readout gives \(264\), \(528\) and \(792\,\text{Hz}\), in the ratio \(1 : 2 : 3\). Now press Listen on each chip. The second harmonic sounds like the first one, one octave higher, as on a string.

ExampleThe fundamental of a flute

A flute can be modelled as a tube open at both ends, with an effective length of \(0.65\,\text{m}\). The speed of sound in air is \(v = 343\,\text{m/s}\). What is the fundamental frequency produced when it is played?

Try it on paper first. The solution is one tap away when you are ready.

A pipe closed at one end behaves differently, and Lesson 3 studies it. Its closed end is a node and its open end an antinode. The shortest wave that fits has a node at one end and the next antinode at the other. From a node to the next antinode is a quarter-wavelength, not a half. The next two checks compare the two kinds of pipe.

CheckpointUnanswered

Two pipes of equal length are made. One is open at both ends, and the other is closed at one end. How do their fundamental frequencies compare?

CheckpointUnanswered

A pipe is \(0.5\,\text{m}\) long and open at both ends. The speed of sound is \(340\,\text{m/s}\). What is the fundamental frequency?

This check uses \(340\,\text{m/s}\), a round value for the speed of sound that many problems take. The flute example used \(343\,\text{m/s}\). The two differ by less than \(1\,\%\), about the change from \(15\,°\text{C}\) to \(20\,°\text{C}\).

Checkpoint · numericUnanswered

A pipe \(0.60\,\text{m}\) long is open at both ends. Find the wavelength of its third harmonic, in m.

m± 2 %
Checkpoint · in your own wordsUnanswered

Why does a pipe open at both ends produce every harmonic, odd and even?

In the world
Finger holes on a bansuri

A bansuri is a bamboo flute open at both ends, with six finger holes along its side. With every hole covered, the whole bamboo tube sounds, and the note is lowest. Lift the finger nearest the far end, and the air meets the atmosphere at that hole. The hole acts as a new open end, so the sounding column is shorter. A shorter column holds shorter half-wavelengths, so the note rises. Lifting the fingers one by one walks the note up the scale.

Exam traps
  1. Writing \(L = n\lambda\) for an open pipe.
    Adjacent antinodes are half a wavelength apart, so an open pipe holds half-wavelengths: \(L = n(\lambda/2)\).
  2. Using \(f_1 = v/L\) for an open pipe.
    The fundamental wavelength is \(2L\), so \(f_1 = v/2L\).
  3. Expecting an open pipe, like a string with a free end, to lack its even harmonics.
    Both ends of an open pipe are antinodes. Every whole number of half-wavelengths fits, so every harmonic is present.