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.)
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,
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?
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.
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.