Creating musical sounds
Creating musical sounds

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Creating musical sounds

5.12 Vibrating air column: pitches of notes produced by wind instruments

In a wind instrument, the air column is the primary vibrator. To excite the air column, a musician either blows across it (e.g. flute) or blows down it via a mouthpiece (e.g. trumpet) or reed (e.g. oboe). This supplies energy, which starts the air column vibrating. The air column isn't just forced to vibrate in one single mode; as with the string, it vibrates in a combination of several modes.

To a good approximation, the air column of a flute is cylindrical with two open ends and, as a result, its resonance frequencies form a complete harmonic series. The resonance frequencies of an oboe's air column also form a complete harmonic series but in this case it is because the air column is conical in shape. So, when these instruments are played, the sound radiated will be composed of any of the harmonics of the series.

The clarinet, on the other hand, has an air column that is cylindrical in shape but has one end effectively closed. Although the resonance frequencies are harmonically related, the even harmonics are missing. Therefore, when the clarinet is played, the radiated sound is predominantly composed of odd harmonics.

Just as with a stringed instrument, all of these harmonics will be fused into a single pitch sensation at the frequency of the fundamental. This will still be true in the case of notes produced by the clarinet where the even harmonics are missing – only the timbre of the notes will be affected, not their perceived pitch.

For all three configurations of air column we have looked at, the first resonance frequency is inversely proportional to the length of the air column. In other words, the longer the air column, the lower its fundamental frequency. One might expect, therefore, that altering the length of the air column of a wind instrument should lead to a change in the pitch of the note it produces.

Activity 21 – Optional

This optional activity requires the following additional materials:

Part 1: a descant recorder and three cardboard tubes that fit the end of the mouthpiece of the recorder as tightly as possible. The tubes should measure 100 mm, 200 mm and 300 mm in length.

Part 2: a microphone and sound editor software, such as Audition or Audacity, which is free to download from audacity.sourceforge.net.

Part 1

Part 1 requires a descant recorder and three cardboard tubes that fit the end of the mouthpiece of the recorder as tightly as possible. The tubes should measure 100 mm, 200 mm and 300 mm in length.

(a) Which tube would you expect to produce notes of the lowest pitch?

Remove the mouthpiece from the descant recorder and attach it to the 0.1 m tube to create a simple wind instrument. The window (the slit in the mouthpiece positioned above the U-shaped gouge) ensures that when this simple instrument is played there is a pressure node at the tip of the mouthpiece. Measure the distance in metres from the tip of the mouthpiece to the end of the 0.1 m tube. (Make sure your measurement is in metres – divide the result by 100 if you measure in centimetres, or 1000 if your measurement is in millimetres.)

(b) Treating the simple instrument as having a cylindrical air column open at each end, calculate the air column's first resonance frequency (its fundamental frequency). Assume that the speed of sound is 340 m/s but ignore end corrections.

(c) Replace the 0.1 m tube with the 0.2 m tube. Measure the distance from the tip of the mouthpiece to the end of this tube and perform a calculation to determine the fundamental frequency of the new air column.

(d) Finally, replace the 0.2 m tube with the 0.3 m tube. Again, measure the distance from the mouthpiece tip to the tube end and perform a calculation to determine the fundamental frequency of the new air column.

Answer

(a) You should have predicted that the longest tube will provide the lowest pitch (in other words, the 0.3 m tube).

(b) When the 0.1 m tube is attached to the mouthpiece, the overall length of the simple instrument is 0.2 m. The first resonance frequency (the fundamental frequency) is therefore f1 = v/2L = 340/(2 × 0.2) = 850 Hz. (You should replace the overall length in this equation with your own measurement.)

(c) When the 0.2 m tube is attached, the overall length of the instrument is 0.3 m. The fundamental frequency is thus f1 = v/2L = 340/(2 × 0.3) = 566.7 Hz. (You should replace the overall length in this equation with your own measurement.)

(d) When the 0.3 m tube is attached, the overall length of the instrument is 0.4 m. The fundamental frequency is thus f1 = v/2L = 340/(2 × 0.4) = 425 Hz. (You should replace the overall length in this equation with your own measurement.)

Part 2

Part 2 requires a microphone and sound editor software, again such as Audition or Audacity, which is free to download from audacity.sourceforge.net.

Make sure the microphone is connected to your computer. Install and run your sound-editor program, and open a new untitled window.

Reattach the recorder mouthpiece to the 0.1 m tube. Set your computer to record and play the simple instrument by blowing gently down the mouthpiece. Stop recording.

Select about 10 cycles of the waveform from the central steady-state part of your recording. Display a frequency spectrum of this section of the waveform and identify the frequency of the first peak in the spectrum. Compare this with the prediction for the frequency of the first mode of vibration of the air column that you made in Part 1(b).

Repeat the experiment for the 0.2 m and 0.3 m tubes. There is no need to save any of your results.

Discussion

Comment

You should find that, for each tube/recorder mouthpiece combination, the frequency of the first peak in the frequency spectrum is close to the prediction of the first resonance frequency that you made in Part 1.

Part 3

Suggest reasons why the air column fundamental frequency predictions that you calculated in Part 1 may not exactly match the frequencies of the first peaks in the spectra that you measured in Part 2.

Answer

The theoretical predictions of the air-column fundamental frequencies that you calculated in Part 1 may be slightly inaccurate because:

  • The end correction at the open end has not been taken into account, so the effective length of the tube is not the length you measured (i.e. not the length from the end of the tube to the tip of the recorder mouthpiece). Note that there is no end correction to be taken into account at the mouthpiece end, as the length measurement has been taken from the tip of the recorder where there is a pressure node.

  • The bore of a tube/recorder combination does not act as a purely cylindrical air column.

  • The speed of sound is not exactly 340 m/s. You should remember from earlier in the course that the speed of sound in air is affected by the temperature of the air. In particular, when the instrument is being played, the temperature of the air that is blown out of your mouth is higher than the normal air temperature.

In Activity 21, you saw that altering the length of the air column changed the pitch of the note produced by our simple wind instrument. Shortening the air column resulted in notes of higher pitch. Lengthening the air column resulted in notes of lower pitch.

Most, but not all, wind instruments have some means of altering the length of the air column to enable notes of different pitches to be played. In brass instruments this is commonly done either by depressing valves to introduce new sections of tubing, or by pulling a slide in or out to adjust the existing tube length. In woodwind instruments, the effective length of the air column tends to be altered by covering or uncovering holes along the length of the tube called side holes. A large side hole acts like an open end and defines the position of a pressure node. A smaller side hole will make a partial reduction in the effective length of the air column (Figure 18).

Figure 18
Figure 18 Effect of a side hole on the effective length of an air column

Activity 22 – Optional

This optional activity requires the following additional materials:

Parts 1 and 2: a descant recorder.

Reattach the recorder mouthpiece to the main body of the recorder, making sure the line of holes is lined up with the top of the mouthpiece.

Play the instrument with all the holes covered (use the first three fingers and thumb of your left hand on the top three holes and the hole underneath, and the four fingers of your right hand for the remaining holes; blow very gently). Slowly shorten the effective air-column length by uncovering the holes one by one, working from the bottom of the instrument up. You should hear the pitch of the notes rise.

The window (the slit in the mouthpiece positioned above the U-shaped gouge) ensures that when the recorder is played there is a pressure node at the tip of the mouthpiece. Using a ruler, measure the length of the whole instrument (make sure your measurement is in metres).

Part 1

Treating the instrument as a cylinder with two open ends, estimate the first resonance frequency of the recorder's air column when all the holes are covered (don't include any end corrections). The note that the recorder produces when all the holes are covered is C5. The pitch C5 has a fundamental frequency of 523.25 Hz. How does your estimate for the fundamental frequency of the air column compare with this?

Answer

The length of the recorder is 0.325 m. The fundamental frequency of the air column is thus f1 = v/2L = 340/(2 × 0.325) = 523.1 Hz. This is very close to what one would expect for the recorder to produce the pitch C5. (You should replace the overall length in this equation with your own measurement.)

Part 2

Now measure the distance from the input end to the last finger hole (a double hole). Again, treating the recorder as a cylinder open at both ends, estimate the first resonance frequency of the air column when all but the last finger hole are covered (again don't include any end corrections). The note that the recorder produces when all but the last finger hole is covered is D5. This pitch has a fundamental frequency of 587.3 Hz. How does your estimate for the fundamental frequency of the air column compare with this?

Answer

The distance from the mouthpiece to the last finger hole is 0.28 m. The fundamental frequency of the air column is thus f1 = v/2L = 340/(2 × 0.28) = 607.1 Hz. This frequency is higher than one would expect if the recorder is to produce the pitch D5.(You should replace the overall length in this equation with your own measurement.)

Part 3

Why do you think the two values in Part 2 do not agree as closely as do those in Part 1?

Discussion

Comment

What needs to be considered is that, because the last finger hole is quite small, the sounding length will be only partially reduced by the presence of the finger hole. Therefore the effective air-column length will actually be longer than measured,making the resulting calculation much closer to the value of 587.3 Hz required to produce the pitch D5.

This also explains the function of the double hole. By uncovering one of the two holes, the effective reduction in the length of the tube is even less than when both holes are open. This allows the player to produce the note in between C5 and D5 (i.e. C#5 or D5 ).

If altering the length of a wind instrument's air column were the only way a musician could change the pitch of the note produced, the range of the instrument would be very limited. Fortunately, the range can be extended by using the technique of overblowing (which does not necessarily just involve blowing harder, but in some instruments may involve such things as adjusting the tension of the lips). This has the effect of feeding less energy into the first mode of vibration and more energy into the higher modes of vibration. The result is that notes can be produced with pitches corresponding to the frequencies of these higher modes. Many of the notes available on a brass instrument are produced by using this technique of overblowing. In the case of the woodwind instruments, the technique is used to extend the range above that available just by uncovering all the finger holes. For an instrument that has a conical air column or a cylindrical air column open at each end, the musician can produce a note an octave higher by overblowing into the second mode of vibration. If the air column is cylindrical but closed at one end, overblowing into the second mode of vibration will produce a note that is an interval of a twelfth (that is, an octave plus a fifth) higher. With some woodwind instruments, a speaker key (or register key) is used to open a small side hole to assist with the overblowing process.

There is another factor that affects the pitches of notes produced by wind instruments. This will be familiar to wind players, who I'm sure are well aware of the need to warm up their instruments (by blowing down them) prior to a performance, and is a factor you might have mentioned in your answer to Part 3 of Activity 21. If you recall, as well as depending on geometry, the resonance frequencies of an air column also depend on the speed of sound. Although we have been tending to take this as 340 m/s, it is in fact dependent on temperature. The higher the temperature, the faster the speed of sound. The faster the speed of sound, the higher the resonance frequencies. So, increasing the temperature has the effect of raising the pitch of the instrument.

Activity 23

For the purposes of this Activity, the air column of a flute can be considered to be a cylinder of length 0.65 m and radius 0.01 m, open at each end.

(a) Before the instrument has been warmed up, assume that the average temperature of the column of air is 15 °C. The speed of sound at this temperature can be taken as being 340.5 m/s. The lowest note on the flute has the pitch C4 (this pitch has a fundamental frequency of 261.6 Hz). To play this note, all the keys must be covered. Calculate the first resonance frequency when this fingering is applied. Include any end corrections. Would you expect the flute to play flat or sharp, or will it be in tune?

(b) When the instrument has been warmed up, assume that the average temperature of the air column has increased to 25 °C. The speed of sound at this temperature can be taken as being 346.5 m/s. Recalculate the fundamental frequency of the air column. How would you expect the tuning of the instrument to be now?

Answer

(a) The total end correction for a 0.65 m long cylindrical tube of radius 0.01 m that is open at both ends is 2 × 0.6 × r = 1.2 × 0.01 = 0.012 m. Therefore, the effective length of the tube is 0.65 × 0.012 = 0.662 m.

The fundamental frequency of the air column at 15 °C is therefore f1 = v/2L = 340.5/(2 × 0.662) = 257.2 Hz. However, the flute should be producing the note C4, which has a fundamental frequency of 261.6 Hz. Thus the flute will play flat.

(b) Using the same value for the effective length, the fundamental frequency of the air column at 25 °C is now f1 = v/2L = 346.5/(2 × 0.662) = 261.7 Hz, which is very close to the correct frequency for the note C4. At this temperature, the flute should therefore play in tune.

TA212_2

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