# 5.8 Vibrating air column: standing waves in a cylindrical tube open at both ends

The frequencies at which standing waves can be set up in an air column enclosed by a cylindrical tube that is open at both ends can be determined quite easily. Because both ends are open to the atmosphere, the pressure at these positions always remains at atmospheric pressure. In other words, there is no fluctuation in the pressure at the open ends so they must be pressure nodes (think ‘no-deviation’ in pressure). So, as we saw with the string fixed at both ends, the length *L* of the air column must be equal to an exact number of half wavelengths:

where *n* is an integer number (i.e. 1, 2, 3, 4, …) and *λ* is the wavelength measured in metres. Following the same arguments as we did for the string, the resonance frequencies of the air column are:

where in this case *v* is the speed of sound in air, which is approximately 340 m/s.

The normal modes of vibration plotted in terms of *pressure* are shown in Figure 14. For each mode of vibration, I have superimposed a graphical representation of the pressure over a diagram of the air column.

The solid curve shows the pressure at one extreme of the cycle and the dashed curve shows the pressure one half-cycle later. The separation between the curved lines at any position along the tube represents the pressure amplitude at that position.

I've presented these standing-wave patterns in terms of pressure variations in the air column because this is the analysis that will be most useful when we come to describe individual woodwind and brass instruments. However, had we been trying to create an exact analogy with the standing waves on a string we should really have looked at the displacement of the air molecules from the mean position. Thinking in these terms, at the open end of an air column, the air molecules are free to move in and out of the tube. Therefore for an air column open at both ends, the ends must be displacement antinodes. Adjacent antinodes are separated by half a wavelength so the length of the air column must be an integer number of half wavelengths. Displacement nodes are then found at positions halfway between adjacent displacement antinodes.

The normal modes of vibration plotted in terms of displacement are shown in Figure 15. In this case, for each mode of vibration, I have superimposed a graphical representation of the displacement over a diagram of the air column. The solid curve shows the displacement of the air molecules at one extreme of the cycle and the dashed curve shows the displacement one half-cycle later. The separation between the curved lines at any position along the tube represents the displacement amplitude at that position.

You may well have already realised that pressure nodes correspond to displacement antinodes and that pressure antinodes correspond to displacement nodes. If not, compare the positions of the nodes and antinodes in Figures 14 and 15 to satisfy yourself that this is the case.

## Activity 17

Run the Flash animation below. This animation demonstrates standing waves in a cylindrical tube open at both ends. It shows graphical representations of both the pressure variation along the tube and the displacement of the air molecules from their mean position. This should help to clarify the relationship between these two quantities. It also shows the motion of the air molecules.

Use the controls below (Play/Pause, Step, Reset) to run flash animation. You can also select the mode.

Just as was the case for the string fixed at both ends, the frequencies at which standing waves are set up in a cylindrical air column with two open ends form a harmonic series. The first resonance frequency (the fundamental frequency of the air column) is *v*/2*L*. If this frequency is denoted *f*_{1}, then the frequencies at which the second, third, fourth and later modes of vibration occur are 2*f*_{1}, 3*f*_{1}, 4*f*_{1} …, *nf*_{1}.

## Activity 18

Calculate the first three resonance frequencies of a 1.5 m long cylindrical tube open at each end. Assume the speed of sound is 340 m/s.

### Answer

If the cylinder is open at both ends, then the formula that gives the resonance frequencies is:

where *L* is the length of the cylinder and *v* is the speed of sound.

If *L* = 1.5 m and *v* = 340 m/s, the first three resonance frequencies are determined as follows:

First resonance frequency *f*_{1} = *v*/2*L* = 340/(2 × 1.5) = 113.3 Hz.

Second resonance frequency *f*_{2} = 2*v*/2*L* = (2 × 340)/(2 × 1.5) = 226.7 Hz.

Third resonance frequency *f*_{3} = 3*v*/2*L* = (3 × 340)/(2 × 1.5) = 340 Hz.