How Does Air Pressure Behave in Vibrating Wind Instruments?

AI Thread Summary
In a tube open at both ends, air pressure is equal to atmospheric at both ends and oscillates with maximum pressure at the midpoint, creating a pressure antinode. The fundamental frequency has a wavelength twice the tube length, while the second overtone has a wavelength equal to the tube length with two pressure antinodes. When one end of the tube is blocked, it becomes a pressure antinode, and the fundamental frequency's wavelength changes to four times the tube length. The discussion clarifies how pressure behaves in vibrating wind instruments, emphasizing the differences between open and closed tube configurations. Understanding these principles helps demystify the behavior of wind instruments compared to string instruments.
TheLil'Turkey
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In a tube (open at both ends) that's vibrating at its fundamental frequency, is the air pressure the ambient air pressure at both ends and higher in the middle, lower in the middle, or can it be either higher or lower in the middle? Why?

If the same tube is vibrating at the frequency of its second overtone (1/3 the wavelength of the fundamental frequency), can the pressure be the ambient pressure at one end and either higher or lower at the other end? If you blow on one end, does that end become the one with the regular pressure or the higher/lower one?

If one end of the tube is blocked, does its fundamental frequency then become half of what it was before?

String instruments seem so much easier to understand than wind ones.
 
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TheLil'Turkey said:
In a tube (open at both ends) that's vibrating at its fundamental frequency, is the air pressure the ambient air pressure at both ends and higher in the middle, lower in the middle, or can it be either higher or lower in the middle? Why?

If the same tube is vibrating at the frequency of its second overtone (1/3 the wavelength of the fundamental frequency), can the pressure be the ambient pressure at one end and either higher or lower at the other end? If you blow on one end, does that end become the one with the regular pressure or the higher/lower one?

If one end of the tube is blocked, does its fundamental frequency then become half of what it was before?

String instruments seem so much easier to understand than wind ones.

For a tube open at both ends, the pressure is equal to atmospheric at both ends. It oscillates inside the tube when excited at the appropriate frequency. The lowest mode will see the maximum pressure oscillations at mid-length (this is called a pressure antinode). So the lowest-mode (fundamental) wavelength is twice the tube length. The next higher mode with have two antinodes, at the 1/4 and 3/4 length points, and will have a wavelength equal to the tube length. The pitch will be an octave higher than the fundamental. The third mode (second overtone) will have three antinodes, at the 1/6, 1/2, and 5/6 length points. The open ends are still pressure nodes (equal to atmospheric).

If you block one end of the tube, that end becomes a pressure antinode. So the wavelength of the fundamental will be 4 times the length of the tube, the next mode will have two pressure antinodes, at the 1/3 point and the closed end (assuming the closed end is at 1), the next mode will have three pressure antinodes, at the 1/5, 3/5, and closed end (again closed=1).

BBB
 
bbbeard said:
For a tube open at both ends, the pressure is equal to atmospheric at both ends. It oscillates inside the tube when excited at the appropriate frequency. The lowest mode will see the maximum pressure oscillations at mid-length (this is called a pressure antinode). So the lowest-mode (fundamental) wavelength is twice the tube length. The next higher mode with have two antinodes, at the 1/4 and 3/4 length points, and will have a wavelength equal to the tube length. The pitch will be an octave higher than the fundamental. The third mode (second overtone) will have three antinodes, at the 1/6, 1/2, and 5/6 length points. The open ends are still pressure nodes (equal to atmospheric).

If you block one end of the tube, that end becomes a pressure antinode. So the wavelength of the fundamental will be 4 times the length of the tube, the next mode will have two pressure antinodes, at the 1/3 point and the closed end (assuming the closed end is at 1), the next mode will have three pressure antinodes, at the 1/5, 3/5, and closed end (again closed=1).

BBB
Wow. I was very confused before, but this cleared a lot of things up. I think you made one mistake though; the bolded 1/3 point should be 1/2, right?
 
TheLil'Turkey said:
Wow. I was very confused before, but this cleared a lot of things up. I think you made one mistake though; the bolded 1/3 point should be 1/2, right?

No, it's 1/3. See attached.
 

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That makes sense. Thank you!
 
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