A pipe closed at one end
A clarinet is a good example of a pipe closed at one end. The player's lips and the reed close the mouthpiece end, and the bell end is open. Such a pipe is a closed pipe. It has mixed boundary conditions: a displacement node at the closed end and a displacement antinode at the open end.
The simplest wave that meets this requirement has the pipe length equal to exactly one quarter of a wavelength. From a node to the next antinode is a quarter-wavelength, so the whole pipe spans just that. The next possible mode must add another node and antinode pair inside. That fits three quarters of a wavelength into the pipe, and so on.
Try the slider in the figure above. The wave fits at \(1.60\,\text{m}\), \(0.53\,\text{m}\) and \(0.32\,\text{m}\). The pipe then holds \(1\), \(3\) or \(5\) quarter-wavelengths, and the closed end is a node. Now set \(0.80\,\text{m}\). The pipe holds \(2\) quarter-wavelengths, which is half a wavelength. The closed end would then be an antinode, but a wall cannot let the air swing.
The derivation below turns this into the condition for a closed pipe.
\[ L = \dfrac{\lambda}{4} \]
Why. the simplest fit: from the node at the closed end to the antinode at the open end is a quarter-wavelength.
\[ L = \dfrac{\lambda}{4} + m\left(\dfrac{\lambda}{2}\right), \qquad m = 0, 1, 2, \dots \]
Why. each further mode adds a node and an antinode, which is half a wavelength. The ends keep their node and antinode.
\[ L = (2m + 1)\left(\dfrac{\lambda}{4}\right) \]
Why. write \(m(\lambda/2)\) as \(2m(\lambda/4)\) and collect the quarter-wavelengths.
\[ L = n\left(\dfrac{\lambda_n}{4}\right), \qquad n = 2m + 1 = 1, 3, 5, \dots \]
Read it. the number of quarter-wavelengths is always odd. Name it \(n\), and the wavelength of that mode \(\lambda_n\).
The allowed frequencies
Solving the condition for the frequency, with \(f_n = v/\lambda_n\) and \(\lambda_n = 4L/n\), gives
\[ f_n = n\left(\frac{v}{4L}\right) = nf_1, \qquad n = 1, 3, 5, \dots \]
The fundamental frequency is \(f_1 = v/4L\).
Notice the striking result. A pipe closed at one end can produce only the odd whole-number multiples of its fundamental frequency. The even harmonics, \(2f_1, 4f_1, \dots\), are missing. An even harmonic would need a node at the closed end and a node at the open end too. The open end cannot provide one.
The missing even harmonics change the sound. They give an instrument like the clarinet its characteristic hollow, or reedy, timbre. That timbre is distinctly different from the sound of an open-pipe instrument playing the same pitch. A flute and a clarinet can play the same note and still sound unlike each other.
The trombone is also closed at one end, by the player's lips, and it is sometimes named beside the clarinet. Yet it does not sound only odd harmonics. Its flared bell and its mouthpiece shift its resonances, until they form an almost complete harmonic series. The clarinet, whose bore is a nearly uniform cylinder, is the clean example of a closed pipe.
For a pipe closed at one end, which harmonics are present?
The first three allowed modes
Fig 3.1 shows the first three allowed modes of vibration of a pipe closed at one end. The pipe is an organ pipe \(0.40\,\text{m}\) long, open at the left and closed at the right. Choose each chip and press play.
The fundamental mode, \(n = 1\). There is a displacement node at the closed end and an antinode at the open end. One quarter-wavelength fits in the pipe, so \(L = \lambda_1/4\) and \(f_1 = v/4L\).
The third harmonic, \(n = 3\). The open end is an antinode and the closed end is a node. One more node and one more antinode fit between them. Three quarter-wavelengths fit in the pipe, so \(L = 3\lambda_3/4\) and \(f_3 = 3f_1\). Note that the even harmonics are absent: this mode comes straight after the fundamental.
The fifth harmonic, \(n = 5\). The open end is an antinode and the closed end is a node, with two more pairs between them. Five quarter-wavelengths fit in the pipe, so \(L = 5\lambda_5/4\) and \(f_5 = 5f_1\). The pattern continues with only the odd harmonics.
The readout of Fig 3.1 gives \(212.5\), \(637.5\) and \(1062.5\,\text{Hz}\), in the ratio \(1 : 3 : 5\). Press Listen on each chip. There is no tone between the first two at \(425\,\text{Hz}\), because the second harmonic does not exist in this pipe.
A closed organ pipe, closed at one end and open at the other, is \(0.40\,\text{m}\) long. What are its lowest three resonant frequencies if \(v = 340\,\text{m/s}\)?
For closed pipes, \(L = n\lambda_n/4\) with \(n = 1, 3, 5, \dots\)
Fundamental, \(n = 1\): \(\lambda_1 = 4L = 1.60\,\text{m}\), so \(f_1 = v/\lambda_1 = 340/1.60 = 212.5\,\text{Hz}\).
Next, \(n = 3\): \(\lambda_3 = 4L/3 = 0.533\,\text{m}\), so \(f_3 = 340/0.533 \approx 638\,\text{Hz}\). The exact value is \(3f_1 = 637.5\,\text{Hz}\).
Next, \(n = 5\): \(\lambda_5 = 4L/5 = 0.32\,\text{m}\), so \(f_5 = 340/0.32 = 1062.5 \approx 1063\,\text{Hz}\).
Answers: \(213\,\text{Hz}\), \(638\,\text{Hz}\) and \(1063\,\text{Hz}\), rounded from \(212.5\), \(637.5\) and \(1062.5\,\text{Hz}\). The example takes the round value \(340\,\text{m/s}\) for the speed of sound.
A clarinet player wants to produce a note at \(146.8\,\text{Hz}\), the note D\(_3\). What should be the length of the air column, a pipe closed at one end, to give this fundamental? Assume \(v = 343\,\text{m/s}\).
Closed pipe: \(f_1 = v/(4L)\), so \(L = v/(4f_1)\).
\(L = 343/(4 \times 146.8) = 343/587.2 \approx 0.584\,\text{m}\).
Answer: \(0.584\,\text{m}\). This example takes \(343\,\text{m/s}\), the speed of sound in air at about \(20\,°\text{C}\).
A pipe closed at one end has a length of \(0.25\,\text{m}\). The speed of sound is \(340\,\text{m/s}\). What is the fundamental frequency?
A pipe closed at one end has a fundamental of \(150\,\text{Hz}\). At which of these frequencies can it resonate?
The closed organ pipe of the example, \(0.40\,\text{m}\) long with \(v = 340\,\text{m/s}\), has modes at \(212.5\), \(637.5\) and \(1062.5\,\text{Hz}\). Find its next resonant frequency above \(1062.5\,\text{Hz}\), in Hz.
Why are the even harmonics missing in a pipe closed at one end?
The closed end must be a displacement node and the open end a displacement antinode. From a node to an antinode is an odd number of quarter-wavelengths, so \(L = n(\lambda/4)\) with \(n\) odd, and \(f_n = nv/4L\) with \(n\) odd. An even \(n\) would fit a whole number of half-wavelengths, which would put a node at the open end, and an open end cannot be a node.
Blow harder into a flute, and its note jumps up by one octave, from \(f_1\) to \(2f_1\). Press the register key of a clarinet, and its note jumps much further, to \(3f_1\). Musicians call that interval a twelfth. The clarinet has no \(2f_1\) to jump to, so it goes straight to its next mode, the third harmonic. The fingering of the whole instrument is built around this odd step.
Organ builders had long known that a pipe stopped at its top sounds about an octave below an open pipe of the same length. In a long memoir on organ pipes written in 1762, Daniel Bernoulli explained why. He treated the air in the pipe as a vibrating column with conditions at its two ends. He showed that the stopped pipe's resonances form only the odd multiples of its fundamental. His reasoning is the one this lesson follows.
- Using \(f_1 = v/2L\) for a pipe closed at one end.A closed pipe holds a quarter-wavelength in its fundamental, so \(f_1 = v/4L\).
- Taking the second harmonic of a closed pipe as the mode after the fundamental.A closed pipe has no even harmonics. The mode after \(f_1\) is \(3f_1\).
- Counting the harmonic number of a closed pipe as 1, 2, 3.The harmonic number \(n\) is the number of quarter-wavelengths in the pipe, and it is always odd: 1, 3, 5.