## The Return of the Resonant Tube: What If We Cut It In Half? - Question of the Week 2014 Summer Girls Special Part 2

Shown in the photograph below is an open tube about 50 cm long. If you blow at an angle into one end of the tube you can make a sort of musical tone.

Click your mouse on the photograph above to hear the fundamental frequency of the tube on an mpeg video.

Shown in the photographs below are (top to bottom) a closed tube the same length as the original tube, an open tube with half the length of the original tube, and a closed tube with half the length of the original tube.

(a) tube a: (b) tube b: (c) tube c: What is the relationship of fundamental frequency of each of the three tubes above to the frequency of the original tube?

Case (a): Relative to the original tube, the frequency of tube (a) is:

• (a) higher in frequency by a factor of two.
• (b) lower in frequency by a factor of two.
• (c) the same frequency.

Case (b): Relative to the original tube, the frequency of tube (b) is:

• (a) one octave higher.
• (b) one octave lower.
• (c) the same frequency.

Case (c): Relative to the original tube, the frequency of tube (c) is:

• (a) one octave lower.
• (b) two octaves lower.
• (c) the same frequency.

We will take the three cases in order.

Case (a): The answer is (b): lower in frequency by a factor of two. Click your mouse on the photograph below to hear the answer. Note that in this case the wavelength is exactly twice as long. Case (b): The answer is (a): one octave higher. Click your mouse on the photograph below to hear the answer. Well, almost one octave higher. The difference has to do with the difference in end correction for the two tubes. The actual loop length for the original tube is L + 2e, where L is the length of the tube and e is the end correction. For the shorter tube, the end correction is the same, so its loop length is L/2 + 2e. Note that the end correction for the shorter tube has a greater relative effect on the loop length

L + 2e < 2 (L/2 + 2e) = L + 4e ,

making the wavelength of the shorter tube a bit longer than half that of the longer tube. So the frequency is a bit low - not quite an octave above the original lonnger tube. In fact, the frequency is about 6% low, one half-step in musical terms. Case (c): The answer is (c): the same frequency. Click your mouse on the photograph below to hear the answer. Note that in this case, the length of the original (open) tube is L + 2e, and the length of the shorter tube is L/2 + e. Thus the loop lengths of the two tubes, respectively, are L + 2e and 2 (L/2 +e) = L + 2e, exactly the same. However, because the shorter tube is a closed tube, it supports only odd harmonics while the longer tube supports all harmonics, so the shorter tube has a timbre that is a bit like that of a square wave. Note that for case (a) above the loop lengths are in the ratio of exactly 2 to 1, including the end effect, so the frequencies are exactly a factor of two, or one octave, apart.