In a college physics laboratory, a brass tube divides into two curved branches that meet again at an ear-piece. One branch can slide in and out, like the slide of a trombone. A tuning fork sounds at the open end. As a student pulls the slide out slowly, the tone at the ear-piece fades almost to nothing, then grows loud again, then fades once more.
One sound, two paths
The last two lessons used two speakers. Quincke's tube makes interference with only one source. The sound enters the tube, and at a junction it divides into two branches. The two parts travel along different paths and meet again at a second junction, which leads to the detector, an ear-piece or a microphone. The detector receives the sum of the two waves, and the sum depends on their path difference.
Quincke's tube is an apparatus in which sound from one source is divided between two tubes of different lengths and recombined at a detector. One tube is a U that slides like a trombone, so the path difference can be changed and measured.
Splitting one sound is the simplest way to get coherent sources. The two parts come from the same vibration, so they have the same frequency. Their phase difference at the detector is fixed by the paths alone, and it stays constant for as long as the paths do. Two separate forks of "the same" frequency would drift apart in phase, and the pattern would wander.
Sliding the U-tube
Now pull the sliding branch out by a distance \(x\). Each of its two straight legs becomes longer by \(x\). The sound goes out along one leg and back along the other, so its path grows by \(2x\), not by \(x\). If the paths already differed by \(\Delta L_0\) before the slide moved, the path difference becomes \[ \Delta L = \Delta L_0 + 2x. \]
Try it in the figure below. The source plays \(1700\,\text{Hz}\). With the speed of sound taken as \(340\,\text{m/s}\), as for every new number in the rest of this chapter, the wavelength is \(340/1700 = 0.200\,\text{m}\), or \(20\,\text{cm}\). The two branches are equal at \(x = 0\), so \(\Delta L_0 = 0\).
At \(x = 0\) the paths are equal, the waves meet in phase, and the reading is \(1.00\). At \(x = 5\,\text{cm}\) the path difference is \(10\,\text{cm}\), half a wavelength, and the reading falls to \(0.00\). At \(x = 10\,\text{cm}\) the path difference is a whole wavelength, and the reading is back to \(1.00\). At \(x = 15\,\text{cm}\) it is zero again.
The sliding U-tube of a Quincke's tube is pulled out by \(3\,\text{cm}\). By how much does the path difference change?
Minima half a wavelength apart
The conditions of the first lesson apply at the detector. The sound is loudest when \(\Delta L_0 + 2x = n\lambda\), and it is quietest when \(\Delta L_0 + 2x = (n + \tfrac12)\lambda\). Take two successive minima, at slides \(x_n\) and \(x_{n+1}\). Their path differences differ by exactly one wavelength: \[ 2x_{n+1} - 2x_n = \lambda. \] So successive minima are \(\lambda\) apart in path, and \(\lambda/2\) apart in slide. The unknown \(\Delta L_0\) cancels, so it never needs to be known.
This turns the tube into a measuring instrument. Read the slide at two successive minima, and double the gap to get the wavelength. With a tuning fork of known frequency, the speed of sound follows as \(v = f\lambda\). Minima are used rather than maxima because the ear, or a meter, finds the quietest point much more sharply than the loudest.
A tuning fork sounds at the inlet of a Quincke's tube. The detector falls to a minimum when the slide reads \(4.2\,\text{cm}\), and to the next minimum at \(21.2\,\text{cm}\). The speed of sound is \(340\,\text{m/s}\). Find the wavelength and the frequency of the fork.
Two successive minima: \(x_n = 4.2\,\text{cm}\) and \(x_{n+1} = 21.2\,\text{cm}\).
Successive minima are \(\lambda/2\) apart in slide: \(x_{n+1} - x_n = \lambda/2\), and \(f = v/\lambda\).
\(x_{n+1} - x_n = 21.2 - 4.2 = 17.0\,\text{cm}\), so \(\lambda = 34.0\,\text{cm} = 0.340\,\text{m}\). Then \(f = 340/0.340 = 1000\,\text{Hz}\).
The wavelength is \(34.0\,\text{cm}\) and the fork sounds \(1000\,\text{Hz}\). The first minimum came at \(4.2\,\text{cm}\), not at \(\lambda/4 = 8.5\,\text{cm}\). So the branches were not equal at zero slide, and the difference of the readings removed that error.
A \(1000\,\text{Hz}\) fork is used with a Quincke's tube. Two successive minima are found \(17.0\,\text{cm}\) apart on the slide. What is the speed of sound, in m/s?
A \(680\,\text{Hz}\) source is used with a Quincke's tube, and the speed of sound is \(340\,\text{m/s}\). How far must the slide move from one minimum to the next, in cm?
A real detector never reads exactly zero at a minimum. The sound in the longer branch loses a little more energy to the walls of the tube. So the two waves arrive with slightly unequal amplitudes. The minimum is deep and sharp, but not perfectly silent. This does not move the minimum, so the measurement of \(\lambda\) is unaffected.
At a minimum of a Quincke's tube the ear-piece still gives a faint tone. What is the most likely reason?
In 1833 the English astronomer John Herschel suggested that sound divided between two paths of different length could be made to cancel itself. In 1866 the German physicist Georg Quincke built such interference tubes and described them. With a sliding branch he could make a tone vanish and return at will, and he used the slide to measure wavelengths of sound. The device is still called the Herschel–Quincke tube.
A large fan in an air-conditioning duct often makes a steady hum at one frequency. Engineers can add a side branch to the duct, a pipe that leaves it and rejoins it a little further on. Let the side branch be longer than the main path by half a wavelength of the hum. The two parts of the hum then meet crest on trough and cancel. It is Quincke's tube, built into a building to silence one annoying note.
Now return to the brass tube in the laboratory. The fork's sound splits between the two branches and meets again at the ear-piece. Each pull of the slide adds twice its length to one path. Every half wavelength of slide, the waves meet crest on trough and the tone dies away. The distance between two such silences, doubled, is the wavelength.
- Taking the slide itself as the change in path difference.The slide lengthens both legs of the U, so the path changes by twice the slide.
- Taking the gap between successive minima as a whole wavelength of slide.Successive minima are one wavelength apart in path, which is half a wavelength of slide.
- Trying to find the path difference at zero slide before using the tube.Use the difference of two readings. The starting path difference cancels.