Resonating frequencies of tubes

In summary, the brass tube has an actual resonating frequency that is 1687 Hz, which is different from the calculated 847 Hz. The fundamental frequency is not resonating because the tube is held at a node where the first overtone is heard. It is possible that the brass tube has its own resonance frequencies, but it is unlikely that they are so close to an resonance frequency of the air inside.
  • #1
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Greetings, didn't know where to post this. To which category it fits, if even any.
Anyway, I'm studying sound production and am doing a work on resonating frequencies of a certain tube.
Please bare in mind I'm not a physician or in the studies of physics, just a musician trying to understand resonance in acoustics. Please move thread if in wrong part of forum

I've done a test on how to calculate the actual resonating frequency of a certain tube with OPEN ends.
I'm using a brass tube of 200mm with a diameter of 0.7 cm.
In the calculation I came to the conclution of 847 Hz, but the actual frequency that's resonating while hitting the tube is 1687 Hz. I hit the tube while holding approximately 25 % of the tube to hold it on a node. So I don't disturb the resonance. It's the same frequency that's heard when dropping the tube on the ground without holding it. So it shouldn't be that I'm manipulating the frequency response by holding it on 25%. The deviation from 843.5 is small but I think what I'm hearing is the second harmonic (first overtone) and not the fundamental. But why does not the fundamental resonate? In all literature I read it says that it should be the fundamental frequency that is heard.

So the fundamental frequency should be 843.5, why am I then only hearing 1687 while making it resonate?
f1 =843.5 =fundamental frequency = first harmonic
f2 =1687 =first overtone = second harmonic
f3 =2530.5 =second overtone = third harmonic
and so forth.
What I then began to think was that when hitting the tube the "first overtone" is the one that's heard it must be that, but I can not explain why. When the calculus tells me the fundamental is 843.5 and the literature says it's the frequency that should be heard.

I'd be glad if someone has a answer for this.
 
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  • #2
The motion of air in the tube won't change the tube itself so much, so the position where you hold it should not matter much. At 1/4 the length you do not have a node for the fundamental frequency.
fregor said:
but I think what I'm hearing is the second harmonic (first overtone) and not the fundamental.
Probably.

How do you start the oscillation? What hits the tube where?
How do you measure the frequency?

The brass could have its own resonance frequencies, but it is unlikely that they are so close to an resonance frequency of the air inside.
 
  • #3
fregor said:
Greetings, didn't know where to post this. To which category it fits, if even any.
Anyway, I'm studying sound production and am doing a work on resonating frequencies of a certain tube.
Please bare in mind I'm not a physician or in the studies of physics, just a musician trying to understand resonance in acoustics. Please move thread if in wrong part of forum

I've done a test on how to calculate the actual resonating frequency of a certain tube with OPEN ends.
I'm using a brass tube of 200mm with a diameter of 0.7 cm.
In the calculation I came to the conclution of 847 Hz, but the actual frequency that's resonating while hitting the tube is 1687 Hz. I hit the tube while holding approximately 25 % of the tube to hold it on a node. So I don't disturb the resonance. It's the same frequency that's heard when dropping the tube on the ground without holding it. So it shouldn't be that I'm manipulating the frequency response by holding it on 25%. The deviation from 843.5 is small but I think what I'm hearing is the second harmonic (first overtone) and not the fundamental. But why does not the fundamental resonate? In all literature I read it says that it should be the fundamental frequency that is heard.

So the fundamental frequency should be 843.5, why am I then only hearing 1687 while making it resonate?
f1 =843.5 =fundamental frequency = first harmonic
f2 =1687 =first overtone = second harmonic
f3 =2530.5 =second overtone = third harmonic
and so forth.
What I then began to think was that when hitting the tube the "first overtone" is the one that's heard it must be that, but I can not explain why. When the calculus tells me the fundamental is 843.5 and the literature says it's the frequency that should be heard.

I'd be glad if someone has a answer for this.

How are you calculating the allowed modes? Are you modeling the vibration of the air column (like a flute or organ pipe) or the brass tube (like chimes) ? I don't think it is the former, since the value for f1 is way too low.

If the latter (which I suspect), you should be modeling the vibrations of a "bar" with one end clamped. For these vibrations, you will not get a harmonic series, like you would for a string clamped at both ends.

Something is odd with your numbers...
 
  • #4
How long is the tube?
 
  • #5
CWatters said:
How long is the tube?
200mm according to the first post.

The calculated resonance frequencies look fine for air, but if the brass tube itself oscillates we will get completely different frequencies.
 

What are resonating frequencies of tubes?

Resonating frequencies of tubes refer to the natural frequencies at which a tube or pipe will vibrate. These frequencies are determined by the length, diameter, and material of the tube.

How do you calculate the resonating frequency of a tube?

The resonating frequency of a tube can be calculated using the formula f=nv/4L, where f is the frequency, n is the number of resonating nodes, v is the speed of sound, and L is the length of the tube.

What is the significance of resonating frequencies of tubes?

Resonating frequencies of tubes are important in various fields such as music, physics, and engineering. They can be used to create musical instruments, measure the speed of sound, and design efficient ventilation systems.

How can resonating frequencies of tubes be manipulated?

Resonating frequencies of tubes can be manipulated by changing the length, diameter, or material of the tube. By altering these factors, the resonating frequency can be shifted higher or lower.

What happens if a tube's resonating frequency is matched with an external frequency?

If a tube's resonating frequency is matched with an external frequency, the tube will begin to vibrate and produce a loud and clear sound. This phenomenon is known as resonance and is commonly used in musical instruments and sound amplification systems.

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