Accoustic Resonance of an open tube - help please

In summary, the conversation discussed the formula for determining the frequency of a stainless steel pipe with air pushed through it, and the factors that affect it such as length, wall thickness, air flow, and inner diameter. It was also mentioned that overblowing a pipe can cause the frequency to jump an octave, but it is difficult to predict when this will occur. The role of the resonator in determining frequency was also noted. Additionally, the conversation touched on the accuracy of measurements and the effect of openness or closedness on the overtones of a pipe.
  • #1
faytmorgan
33
0
Accoustic Resonance of an open tube -- help please

So I am trying to make sense of the different varying formulas that I have found. Some of the formulas are not fully explained hence my questions. This is a music related topic.

I am attempting to figure out what the formula is to determine the frequency of a stainless steel pipe (the one constant) with air pushed through it with these following variables.

(A) Length of pipe
(B) Wall thickness of pipe
(C) CFM going through the pipe
(D) Inner Diameter of pipe

Also I have read that overblowing a pipe will cause the frequency of the note to jump an octave. At what point is it considered to be overblowing a pipe? Is there a formula for the max cfm prior to the said overblowing?
 
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  • #2


I am working with 57132.3958333 cfm btw
 
  • #3


I am working with a 336.6 CID sbc. At 7000rpm VE is at 83.8. Peak VE is at right around 5500rpm being at 92.1

RPM is a factor of cfm, it doesn't make sense to me that cfm is not a factor in the sound and the rpm is seeing that they are related. The higher the rpm the higher the cfm and vise versa.

The formula should technically be able to find the frequency giving the cfm comming out of the engine. valve sizing as far as I can see and all other factors of the engine cause the ve to be where it is at, so ve is the only number that should matter there. which again is a factor of the cfm.

Also one should be able to even out the pulses of the exhaust running first through an x pipe back into a single pipe. That single pipe is the one that matters in this formula.

As for the formula given I understand that "N" is an integer being "1,2,3" what I don't get is what it is supposed to represent. It says the resonant node, are they trying to say the frequency that we are attempting to achieve, or something else that I am missing here?

For example. if f=nv/2(L+.3d) and the req we are looking for is say middle c (261.626 hz) at 20c, which would make speed at 343 mps (meters per second) the length is say about 1 meter and the diameter is .0254 meters (1 inch) then f would equal 44546.0087328hz which makes no sense to me that appears to be very high. A 3 foot open pipe with 1 inch diameter openings. Unless that happened to be the limit of the pipes ability to make x amount of HZ with max amount of allowable cfm to travel through the pipe. Which this does not tell us when it is too much cfm for the pipe and then the octave raises.

Show me where I am wrong here, because I feel like I am missing something.
 
  • #4


The frequency relates essentially to the length of the pipe. You have an integer number of half-waves in one pipe length, and sound propoagates over one wavelength within one period, which is the reciprocal of frequency.

Overblowing can't be predicted easily. It doesn't depend just on throughput.
 
  • #5


I am a musician. I am an expert of nay. ( a kind of end-blown reed flute)
an about an inner radius of 16mm and an about a length of 72cm pipe gives nearly A (440 hertz) when end blown.
however when side blown it is different.
different wind instruments have different mouth pieces and give different frequencies at the same pipe length.
therefore, the resonator has an important role on the frequency.
 
  • #6


faytmorgan said:
I am working with 57132.3958333 cfm btw

WOW! What kind of meter did you use to get that many digits?

Or is that just calculator vomit?

Please post measurements with a reasonable number of significant digits.
 
  • #7


The frequencies of the overtones of a pipe will depend upon whether it can be regarded as open ended or closed ended.
Look at this link to see what I mean. The first overtone can be at around 2f or 3f, depending.
In a real situation, the openness or closedness will be a bit approximate and will depend on what is connected to the ends and the overtone may be a long way off from a harmonic frequency. Is this to do with a car exhaust?
 

1. What is acoustic resonance?

Acoustic resonance is a phenomenon that occurs when a sound wave matches the natural frequency of a resonator, causing the resonator to vibrate at a higher amplitude.

2. What is an open tube?

An open tube is a type of resonator that is open at both ends, allowing air to flow freely through it. This type of tube is commonly used in wind instruments such as flutes and clarinets.

3. How does sound travel through an open tube?

Sound travels through an open tube by causing the air particles inside the tube to vibrate at the same frequency as the sound wave. This creates a standing wave pattern, which produces the characteristic sound of the instrument.

4. What factors affect the acoustic resonance of an open tube?

The length and diameter of the tube, as well as the speed of sound in the medium it is filled with, all affect the acoustic resonance of an open tube. Additionally, the placement and size of any openings or valves can also impact the resonance.

5. How is the acoustic resonance of an open tube used in music?

The acoustic resonance of an open tube is used in music to produce different pitches and tones in wind instruments. By changing the length or opening of the tube, musicians can create a range of notes and melodies. It is also used in the design and construction of musical instruments to achieve specific tones and sound qualities.

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