Musical resonance of a cylinder/tube with a hole in the center

  • #1
Ben Boldt
3
1
TL;DR Summary: Resonance of a cylinder/tube with a hole in the center

The Science Museum of Minnesota has this really cool xylophone hanging from the ceiling, named "Seismofon":

seismofon.jpg


There is a computer connected that receives almost-real-time seismic activity, and it uses that information to play various chords and riffs, etc. So it is always making little random noises basically.

I wanted to reproduce 1 of the tubes, just as a personal project for my own curiosity. You will notice, it is not exactly like a normal xylophone. The resonators are horizontal, with a hole in the CENTER, and the xylophone bar is parallel to the resonator. There is a solenoid that strikes the bar. I have determined that the tubes seen in the photo are 3 inch diameter, and they are very thin-walled. It would be pretty much exactly an exhaust pipe from a pickup truck.

I found an old 3" PVC pipe to experiment with. The inside diameter is 3", and the outside diameter is 3.5". I wanted this pipe to resonate at 440 Hz (Musical Note A4).

I found this website that describes the equations to determine the correct length of resonator versus frequency:
http://supermediocre.org/index.php/2016/04/13/tuning-the-tubes/

Tube closed on one end:
f1 = c / [ 4 (L + 0.61 r) ]

Tube open on both ends or closed on both ends:
f1 = c / [ 2 (L + 0.61 r) ]

f1: fundamental frequency
c: speed of sound
L: length of resonator pipe
r: radius of resonator pipe

At my elevation, the speed of sound is about c = 344.44 m/s.
I want f1 = 435 Hz (I can sand it shorter to tune it up to 440 Hz).
r = 1.5 inches = 0.0381 meters

My assumption, which seems to be wrong, was that the hole in the middle makes it like 2 separate open-ended resonators. So I solved the length using the second equation, then I cut the pipe twice as long as that and put a 3" hole in the center.

[435 Hz] = [344.44 m/s] / [ 2 (L + (0.61 * 0.0381) ) ]
<<solved for L>>
L = 0.3727 meters = 14.67 inches
2L = 29.34 inches

This is what my pipe looks like:
seismofon_pvc.jpg


That same website from before, the guy described how he used a speaker and microphone to play a sine wave into his resonator. It starts at a low frequency, and moves higher. The resonator will amplify the sound at its fundamental frequency (f1). I tried this using this website to generate the sweep:

https://onlinesound.net/sweep-tone-generator

I found that my pipe's fundamental frequency was much lower than expected. It is about 377 Hz even though I was shooting for 435 Hz. So I do not think my assumption is correct that I can treat this like 2 open-ended resonators. I did not find any harmonic resonance lower than 377 Hz, so I am feeling like that is the fundamental frequency of this particular pipe.

Looking for clues and ideas, I noticed that if I divide 377Hz / 435Hz, I get a value very close to (√3)/2. The unit circle has the point (1/2, √3/2), seeing as how the hole is half-way down the pipe, maybe that gives us √3/2? It seems like a stretch but I am not sure what to think.

Does anyone have any insight how to how to correctly calculate this pipe length with the hole in the middle?
 
Last edited:
  • Like
Likes sairoof
Science news on Phys.org
  • #2
I wanted to add some clarification and more details to this:
  • The Seismofon sculpture was made by a man named Trimpin. I can't believe I didn't mention that.
  • In the book "Trimpin: Contraptions for Art and Sound", this instrument is referred to as "Magnitude in C#," however, at the actual museum and in the documentary "Trimpin The Sound of Invention," it is referred to as Seismofon.
  • It is basically a large xylophone. Each tube plays a different pitch note just like a normal xylophone.
  • Each tube has an object that is struck. The fundamental resonant frequency of the tube is matched to the frequency of the object, so as to amplify and prolong the sound produced.
  • There are 3 different types of objects that are struck. Metal xylophone bar, wooden xylophone/marimba bar, and woodblock.
  • In the case of a woodblock, it is cut round and stuffed into the end of a tube. It is not clear to me if or how the tube actually affects the sound of a woodblock, but they are definitely quite loud and hollow sounding in person.
  • A small box containing a solenoid and a plunger is electrified in order to strike each object.
  • The tubes vary in length and they do not have any sort of blockage in them. One of my photos vaguely shows where you can see straight through the tubes.
  • A similar exhibit by Trimpin exists named "Conloninpurple" (Conlon in purple). It uses smaller-diameter tubes which are plugged on one end and fashioned with a horn bell on the other end. The horn end is much longer than the other end. I don't know exactly what effect the horn has; I have not seen this one in person.
 

Attachments

  • conloninpurple.jpg
    conloninpurple.jpg
    22.7 KB · Views: 2
  • Metal Xylophone.jpg
    Metal Xylophone.jpg
    14.4 KB · Views: 2
  • open-ended.jpg
    open-ended.jpg
    27.4 KB · Views: 1
  • Wood Blocks.jpg
    Wood Blocks.jpg
    53.4 KB · Views: 2
  • Wooden Xylophone.jpg
    Wooden Xylophone.jpg
    44.4 KB · Views: 1
Last edited:
  • #3
Ben Boldt said:
I don't know exactly what effect the horn has; I have not seen this one in person.
The horn matches the acoustic impedance inside the tube to the external environment. That will more efficiently radiate sound, as a spherical wave, from the horn. The length of the tube that feeds the horn will act as a harmonic filter.
 
  • #4
I say that a xylophone with resonators is a marimba. They can be quite loud.
 
  • #5
Here is one additional picture of ConlonInPurple. Trimpin has provided some interesting details on this drawing. (I scanned this from the book mentioned earlier.)
 

Attachments

  • ConlonInPurple.jpg
    ConlonInPurple.jpg
    56.1 KB · Views: 1
Back
Top