Tubes and wavelengths

[Lee T]

I’m confused about what the reading and notes say about sound waves traveling through a tube. I understand the concept of the fundamental wavelength of the tube, but I don’t get why the other wavelengths that fit have frequencies 3, 5, … times the fundamental. What about all the even multiples of the wavelength? Are they too trivial to mention or do they not work for some reason?

[Simon]

A tube that is closed at both ends will have resonances at all (both odd and even) multiples of the lowest resonant frequency. The sound waves travel as follows

1. start at closed end (vocal folds)
2. travel to other closed end
3. reflects back
4. travels to back to other closed end
5. meets next pulse from vocal folds, is in phase with it, and so gets bigger

The wave has to travel 2 times the length of the tube before meeting the next pulse, in order to be “in step” with it.

For a tube that is closed at one end and open at the other, it’s a little different. The sound wave reflects from the closed end, just as in the other case, but when a wave is reflected from the open end, it is inverted. Therefore, the wave needs to travel as follows before it will be “in step” (i.e., in phase) with the next wave:

1. start at closed end (vocal folds)
2. travel to open end
3. reflects back, but gets inverted in the process
4. inverted wave travels to closed end
5. is reflected and remains inverted
6. travels to open end
7. reflects back, and is inverted again (so is back to ‘normal’)
8. travels back to closed end
9. meets next pulse from vocal folds, is in phase with it, and so gets bigger

You see that the wave needs to traverse 4 times the length of the tube this time.

The higher frequency resonances of these tubes occur by putting pulses in more frequently. Try to work that out using the same reasoning as above.

[Shubha G]

Could you explain what is meant by the wave inverting at the open end of the tube? What physical properties of the opening cause this inversion?

[Author]

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