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Looking at a crazy Multitube shotgun microphone and trying to understand the authors

open end tube calculations. (I wonder if it is actually open ended.)

https://worldradiohistory.com/Archive-Poptronics/60s/64/Pop-1964-06.pdf

The author says,

"To calculate tube length, first find wavelength by dividing the speed of sound the frequency. (1100 feet per second for practical purposes.) For example,the wavelength of 256 cycles equals 1100 ÷ 256, or 4.296 feet. Tubelength, however, is half this, or 2.14 feet, since tubes open at both ends resonate at a Wavelength twice as long as their length."

Now my experiment;

I have 22" (.559m) tube. 1100/300Hz = 3.66 ft, but it's open ended so divide by 2 = 1.833ft =22 inch = 0.559 meters.

With a mic setting at the open end I get resonance at 300hz. (Small, like I'm just off the node)?

(this is close to what the author says, but far from what the hyperphysics calculator finds.)

If I remove the tube the signal drops.

However, If I put the mic inside the pipe at the 1/2 length position I get 10 times the signal.

If I hold the mic in place and remove the tube, the signal drops to 1/40 of what it was with the tube.

I'm using the calculator at,

http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/opecol.html

My measurements say I have a node at the open end and an antinode at the center of the tube. But.

the hyperphysics calculator says /opposite/ about open ended tubes.

Can anyone shed some light on this?

         Thank, Mikek

BTW, I have both an aluminum tube and PVC of the same ID, the PVC has about 1/3 more signal at mid tube.

Not what I expected.

Bonus points!

Is the tube open at both ends, he closes it of with a funnel and Microphone. (see the article)

Does a closed of tube need a perfect seal?

If it is a perfect seal, do the 36 other tubes act as an open, making it open at both ends?

It does have an opening 36* times the size of the tube all pointing away from the mic end of the tube.

  • the other 36 tubes in the assembly.
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1 Answer 1

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Microphones measure fluctuations in air pressure. At the open end of the tubes the pressure is constrained to be atmospheric pressure. In the middle of the tube there is a pressure antinode. Diagrams for standing sound waves in tubes are often drawn showing displacement nodes and and antinodes perhaps because this matches with standing waves we can see in a string but that is not what we hear. I first learned this when I had a chance to crawl through a giant tube with a standing wave in it at a science centre. I assumed someone had marked the standing wave pattern incorrectly with nodes at the end until I heard it with my own ears.

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  • $\begingroup$ Thanks for the answer, I found a Khan academy page that explained this to me. It all became clear. And now I'm off to try a closed end 1/4 wavelength tube with a mic at the closed end. I expect to get the same amplitude as a 1/2 wavelength tube mic'ed in the center. This would take a lot of mic's to built this shotgun mic, and I haven't figured out what it takes to add electret mics together. (maybe you can help?) Just poking at ideas, not sure which way to go yet. $\endgroup$ Commented Dec 18, 2021 at 14:54

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