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:
\[ L = n\left(\frac{\lambda_n}{2}\right), \qquad n = 1, 2, 3, \dots \]
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.
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.
\[ L = n\left(\dfrac{\lambda_n}{2}\right), \qquad n = 1, 2, 3, \dots \]
Why. the condition for an open pipe: an antinode at each end, with a whole number of half-wavelengths between them.
\[ \lambda_n = \dfrac{2L}{n} \]
Why. solve step 1 for the wavelength of mode \(n\).
\[ f_n = \dfrac{v}{\lambda_n} \]
Why. for any wave, speed = frequency × wavelength. The speed of sound is the same for every mode, since it depends only on the air.
\[ f_n = n\left(\dfrac{v}{2L}\right) \]
Why. put step 2 into step 3.
\[ f_n = nf_1, \qquad f_1 = \dfrac{v}{2L} \]
Read it. the lowest frequency, with \(n = 1\), is the fundamental \(f_1\). Every other mode is a whole-number multiple of it.
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.
For a pipe open at both ends, which harmonics are present?
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 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.
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?
For an open pipe in its fundamental, \(L = \lambda_1/2\). So \(\lambda_1 = 2L = 2 \times 0.65 = 1.30\,\text{m}\).
Then \(f_1 = v/\lambda_1 = 343/1.30 \approx 264\,\text{Hz}\).
Answer: \(264\,\text{Hz}\). The example takes \(343\,\text{m/s}\), the speed of sound in air at about \(20\,°\text{C}\). This is the first tone that Listen plays in Fig 2.1.
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.
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?
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 \(20\,°\text{C}\).
A pipe \(0.60\,\text{m}\) long is open at both ends. Find the wavelength of its third harmonic, in m.
Why does a pipe open at both ends produce every harmonic, odd and even?
Both ends of an open pipe must be displacement antinodes. Adjacent antinodes are half a wavelength apart, so any whole number of half-wavelengths fits between the ends, \(L = n\lambda/2\) for every \(n\). Each fit gives a frequency \(f_n = nv/2L\), so every whole-number multiple of the fundamental is allowed.
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.
- Writing \(L = n\lambda\) for an open pipe.Adjacent antinodes are half a wavelength apart, so an open pipe holds half-wavelengths: \(L = n(\lambda/2)\).
- Using \(f_1 = v/L\) for an open pipe.The fundamental wavelength is \(2L\), so \(f_1 = v/2L\).
- 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.