In the physics practical of Class 11, a student strikes a tuning fork on a rubber pad. She holds it over a tall glass tube of water. The water level is lowered slowly. For a long time the fork sounds faint. Then, at one level, the tube booms out loudly with the note of the fork. Lower the water further, and the boom fades, and later returns.
The open end, looked at closely
Lessons 1 to 3 treated the open end of a pipe as the exact place of the antinode. Real air does not stop at the mouth of the pipe. The air just outside the mouth is pushed and pulled along with the air inside. So the region of free, constant pressure begins a little way beyond the end.
The displacement antinode, which is the pressure node, therefore lies a small distance \(e\) outside the open end. This distance is the end correction. For a pipe of inner radius \(r\), it is close to \(e = 0.6r\). The standing wave then behaves as if the pipe were longer by \(e\) at every open end.
In Fig 4.1 the envelope does not reach its widest at the mouth. It keeps widening for another \(0.8\,\text{cm}\), and the antinode sits in the open air. The quarter-wavelength of the fundamental therefore spans \(L + e\). The length that sets the resonances is this effective length. A closed pipe has one open end, and an open pipe has two.
The other harmonics follow in the same way. A closed pipe has \(f_n = nv/4(L + e)\) for odd \(n\). An open pipe has \(f_n = nv/2(L + 2e)\) for every \(n\).
The closed organ pipe of Lesson 3 is \(0.40\,\text{m}\) long. Its inner radius is \(2.0\,\text{cm}\). With \(v = 340\,\text{m/s}\), find its fundamental frequency, allowing for the end correction.
The end correction is \(e = 0.6r = 0.6 \times 2.0 = 1.2\,\text{cm}\). The pipe has one open end, so its effective length is \(L + e = 0.40 + 0.012 = 0.412\,\text{m}\).
Then \(f_1 = v/4(L + e) = 340/(4 \times 0.412) = 340/1.648 \approx 206\,\text{Hz}\).
Without the correction, Lesson 3 found \(212.5\,\text{Hz}\). The real pipe sounds about \(6\,\text{Hz}\) lower, because it behaves as a longer pipe.
Using the measured length alone always overestimates the frequency of a pipe. The effective length is longer, so the true fundamental is lower. A wider pipe needs a larger correction, since \(e\) grows with \(r\).
An open pipe \(82.0\,\text{cm}\) long has an inner radius of \(2.5\,\text{cm}\). The speed of sound is \(340\,\text{m/s}\). Allowing for the end correction, find its fundamental frequency, in Hz.
The resonance tube
A resonance tube is a long glass tube standing upright, with water in its lower part. The water level can be raised or lowered, usually by moving a connected reservoir. The air column above the water is closed at the water surface, because the air cannot move into the water. It is open at the top.
A tuning fork of known frequency \(f\) is struck and held over the mouth of the tube. The fork drives the air column at the frequency \(f\). The column answers loudly only when \(f\) is one of its own resonant frequencies. The lengths of the air column at which this happens are the resonance lengths.
The air column is a closed pipe of effective length \(L + e\). Start with the water near the top and lower it slowly. The first loud resonance comes when \(L_1 + e = \lambda/4\), the fundamental. Lower the water further. The next resonance comes when \(L_2 + e = 3\lambda/4\), the third harmonic of the longer column.
Start the slider in the figure above at \(5\,\text{cm}\) and lengthen the column slowly. The air barely stirs, and the fork sounds faint. At \(15.8\,\text{cm}\) the loudness jumps to a sharp peak. The envelope then fills the column, with a node at the water surface. The next peak comes at \(49.0\,\text{cm}\), with one more node inside, and a third at \(82.2\,\text{cm}\). Press Listen and move the slider through a peak to hear the boom.
Two lengths, and the correction cancels
The end correction is hard to measure. It sits in the open air, where nothing marks it. The resonance tube avoids the difficulty by using two resonance lengths, as the derivation below shows.
\[ L_1 + e = \dfrac{\lambda}{4} \]
Why. the first resonance: the fundamental of the air column, whose effective length is \(L_1 + e\).
\[ L_2 + e = \dfrac{3\lambda}{4} \]
Why. the second resonance: three quarter-wavelengths in the longer column. The end correction is the same, since the tube and its mouth are the same.
\[ L_2 - L_1 = \dfrac{\lambda}{2} \]
Why. subtract step 1 from step 2. The unknown \(e\) cancels.
\[ v = f\lambda = 2f(L_2 - L_1) \]
Why. the speed of sound is frequency × wavelength, with \(\lambda = 2(L_2 - L_1)\) from step 3.
\[ e = \dfrac{L_2 - L_1}{2} - L_1 = \dfrac{L_2 - 3L_1}{2} \]
Read it. put \(\lambda/4 = (L_2 - L_1)/2\) from step 3 into step 1, and solve for \(e\).
A tuning fork of \(512\,\text{Hz}\) is held over a resonance tube. The first resonance is found at \(L_1 = 15.8\,\text{cm}\) and the second at \(L_2 = 49.0\,\text{cm}\). Find the speed of sound, the end correction and the inner radius of the tube.
The air column is closed at the water and open at the top. So \(L_1 + e = \lambda/4\) and \(L_2 + e = 3\lambda/4\).
Subtracting cancels \(e\): \(\lambda = 2(L_2 - L_1)\), and \(v = f\lambda\). Then \(e = (L_2 - 3L_1)/2\) and \(r = e/0.6\).
\(\lambda = 2(49.0 - 15.8) = 66.4\,\text{cm}\), so \(v = 512 \times 0.664 = 340\,\text{m/s}\). Next \(e = (49.0 - 47.4)/2 = 0.8\,\text{cm}\), and \(r = 0.8/0.6 \approx 1.3\,\text{cm}\).
The speed of sound is \(340\,\text{m/s}\), the end correction is \(0.8\,\text{cm}\), and the tube has an inner radius of about \(1.3\,\text{cm}\). These are the numbers of the tube in the figure above.
Compare this with a careless method. From the first resonance alone, \(v = 4fL_1\) gives \(4 \times 512 \times 0.158 \approx 324\,\text{m/s}\). That is about \(5\,\%\) too low, because it ignores the end correction. The difference \(L_2 - L_1\) needs no correction at all.
Now return to the booming tube of the practical. The boom is resonance: the fork's frequency matches a resonant frequency of the air column. The first boom marks \(L_1\) and the second marks \(L_2\). Half a wavelength separates them, whatever the end correction. So two readings of a metre scale and the number on the fork give the speed of sound.
A fork of \(425\,\text{Hz}\) gives resonances in a tube at \(L_1 = 19.0\,\text{cm}\) and \(L_2 = 59.0\,\text{cm}\). Find the speed of sound, in m/s.
For the same tube, with \(L_1 = 19.0\,\text{cm}\) and \(L_2 = 59.0\,\text{cm}\), find the end correction, in cm.
In the example, \(L_2 = 49.0\,\text{cm}\) is a little more than three times \(L_1 = 15.8\,\text{cm}\). Why is it not exactly three times?
Put the steps of the resonance-tube experiment in order.
A student finds the fundamental of a closed pipe as \(v/4L\), using its measured length \(L\). Compared with this value, the true fundamental is
In The Theory of Sound, first published in 1877, Lord Rayleigh studied the open end of a pipe with care. He argued that the air just beyond the mouth moves with the air inside, so the pipe acts as if it were longer. He set bounds on the extra length and compared them with experiments. The exact value for a thin pipe with a plain end waited until 1948. Harold Levine and Julian Schwinger then calculated it as \(0.6133\,r\), the \(0.6r\) of this lesson.
- Taking the second resonance length as \(2L_1\) or as \(3L_1\) exactly.The second resonance is the third harmonic, and the end correction adds \(2e\): \(L_2 = 3L_1 + 2e\).
- Adding only one end correction to an open pipe.An open pipe has two open ends, so its effective length is \(L + 2e\).
- Finding the speed of sound from \(L_1\) alone as \(4fL_1\).That ignores the end correction and gives too low a value. Use \(v = 2f(L_2 - L_1)\).