Harmonics and overtones
The chapter Standing Waves on a String named the modes in two ways. A harmonic is named by its multiple of the fundamental: \(f_n = nf_1\) is the \(n\)th harmonic. An overtone counts only the frequencies above the fundamental, in order. The first overtone is the next resonant frequency after \(f_1\).
For an open pipe every harmonic is present. So the overtones are \(2f_1, 3f_1, 4f_1, \dots\), and the \(n\)th harmonic is the \((n - 1)\)th overtone. This is the same rule as for a string fixed at both ends.
For a closed pipe the even harmonics are missing. So the first overtone is \(3f_1\), the third harmonic. The second overtone is \(5f_1\), the fifth harmonic. In general the \(m\)th overtone of a closed pipe is the \((2m + 1)\)th harmonic.
The two numbers are easy to mix up in a closed pipe. "The second overtone" means the third resonant frequency, \(5f_1\). It does not mean \(2f_1\) or \(3f_1\). Always list the frequencies in order first, then count.
To count the resonant frequencies below some limit, list the allowed \(n\) and stop at the limit. The example below does this for an open pipe and a closed pipe of the same length.
Two pipes are each \(0.85\,\text{m}\) long. One is open at both ends, and the other is closed at one end. Take \(v = 340\,\text{m/s}\). How many resonant frequencies of each pipe lie below \(1250\,\text{Hz}\), and how many of those are overtones?
Open pipe: \(f_1 = v/2L = 340/1.70 = 200\,\text{Hz}\). Its resonances are \(200, 400, 600, 800, 1000\) and \(1200\,\text{Hz}\); the next, \(1400\,\text{Hz}\), is above the limit. That is 6 resonant frequencies, of which 5 are overtones.
Closed pipe: \(f_1 = v/4L = 340/3.40 = 100\,\text{Hz}\). Its resonances are the odd multiples, \(100, 300, 500, 700, 900\) and \(1100\,\text{Hz}\); the next, \(1300\,\text{Hz}\), is above the limit. That is also 6, of which 5 are overtones.
The closed pipe starts an octave lower, but it skips every even multiple. Up to this limit the two pipes have the same number of resonances.
A pipe \(0.50\,\text{m}\) long is open at both ends, and \(v = 340\,\text{m/s}\). How many overtones does it have below \(1500\,\text{Hz}\)?
The second overtone of a pipe closed at one end is its
Two pipes in unison
Two pipes sound in unison when they have the same fundamental. Take an open pipe of length \(L_o\) and a closed pipe of length \(L_c\). Their fundamentals are equal when
\[ \frac{v}{2L_o} = \frac{v}{4L_c}, \qquad \text{so} \qquad L_c = \frac{L_o}{2}. \]
The closed pipe needs only half the length. Its quarter-wavelength equals the half-wavelength of the open pipe.
Look at the two ladders of Fig 5.1. They share the bottom rung, \(170\,\text{Hz}\). Above it the open pipe has a rung at every multiple, \(340\), \(510\), \(680\,\text{Hz}\) and so on. The closed pipe has rungs only at \(510\) and \(850\,\text{Hz}\), and dashed gaps where the even multiples would be.
Now press Listen on each chip. Each sound is the pipe's first four resonances played together, each softer than the one below it. The two sounds have the same pitch, \(170\,\text{Hz}\). Yet the open pipe sounds full and bright, and the closed pipe sounds hollow. This is the difference of timbre that Lesson 3 described, now heard directly.
An open pipe is \(0.60\,\text{m}\) long. What length of closed pipe has the same fundamental, in m?
Why does a closed pipe need only half the length of an open pipe to sound the same fundamental?
The same fundamental means the same wavelength, since the speed of sound is the same. An open pipe holds half a wavelength between its two antinodes. A closed pipe holds only a quarter-wavelength, from the node at the closed end to the antinode at the open end. A quarter-wavelength is half of a half-wavelength, so the closed pipe is half as long.
A pipe and a tuning fork
A pipe resonates with a tuning fork when the fork's frequency equals one of the pipe's allowed frequencies. To find the mode, divide the fork's frequency by the pipe's fundamental. The result must be a whole number, and odd for a closed pipe.
A tuning fork of \(510\,\text{Hz}\) is sounded near two pipes, with \(v = 340\,\text{m/s}\). One is a closed pipe \(0.50\,\text{m}\) long, and the other is an open pipe \(1.00\,\text{m}\) long. Does each pipe resonate, and in which mode?
Closed pipe: \(f_1 = v/4L = 340/2.0 = 170\,\text{Hz}\). Then \(510/170 = 3\), an odd number, so the pipe resonates in its third harmonic. That is its first overtone.
Open pipe: \(f_1 = v/2L = 340/2.0 = 170\,\text{Hz}\). Then \(510/170 = 3\) again, so the pipe resonates in its third harmonic. For an open pipe that is the second overtone.
The two pipes of Fig 5.1 answer the fork in the same harmonic, but it is a different overtone for each.
A tuning fork of \(340\,\text{Hz}\) should make a pipe resonate in its fundamental, with \(v = 340\,\text{m/s}\). Which pipe will do?
The pressure picture
Every mode has a second picture, drawn with pressure instead of displacement. Lesson 1 showed the rule at the ends. It holds all along the pipe, because \(\Delta P = -B\,\partial s/\partial x\). Where the displacement is largest, its slope is zero, so there is a pressure node. Where the displacement is always zero, its slope is largest, so there is a pressure antinode.
Step through the chips of Fig 5.2. In the open pipe the pressure envelope is pinched at both ends, which are pressure nodes. In its fundamental the pressure swings most at the centre, the displacement node. In the closed pipe the pressure envelope is widest at the wall. The two envelopes are a quarter-wavelength apart in every mode.
This picture tells an experimenter where to look. A microphone, which feels pressure, is loudest at a displacement node. A light powder, which is shaken by the moving air, dances most at a displacement antinode, as in Kundt's tube.
A pipe open at both ends sounds its second harmonic. Which statements are true?
The longest pipes of a large church organ are open at both ends and about \(5\,\text{m}\) long. Take \(v = 340\,\text{m/s}\). Estimate the fundamental frequency of such a pipe, in Hz.
An organ needs very low notes, and an open pipe for a low note must be very long. Organ builders therefore use stopped pipes, closed with a wooden plug at the top. A stopped pipe sounds the same fundamental as an open pipe twice its length. It saves height and metal. Its tone is soft and hollow, because its even harmonics are missing, and organists choose it for exactly that sound.
- Calling \(2f_1\) the first overtone of a closed pipe.A closed pipe has no \(2f_1\). Its first overtone is \(3f_1\), the third harmonic.
- Counting the fundamental among the overtones.Overtones are the resonances above the fundamental. A pipe with four resonances below a limit has three overtones there.
- Placing a pressure node at a closed end.A closed end is a displacement node, so it is a pressure antinode. Pressure nodes lie at the displacement antinodes.