Helmholtz Resonator

  • Last Post 20 December 2017
  • Topic Is Solved
emirdegirmenli posted this 12 December 2017

I'm working on finding modes of the cavity, especially Helmholtz mode by using acoustics ACT. I designed a basic model and I got some results but they do not match results which I got using Helmholtz resonator formula (below). I found some document about Helmholtz resonator in Acoustics ACT docs but it is not detailed

To find Helmholtz resonance, I defined surface of the hole as a 101325 Pa for 1 ATM air pressure condition. but I am not sure if this approach is correct. Because, pressure travel through the neck and in the air. We know neck length but don't know the distance traveled in the air.  (l attached SpaceClaim folder (18.0) of my model)

Best, Emir



maybe this model can be worked ??

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peteroznewman posted this 14 December 2017

Hello Emir,

I downloaded your geometry, and measured it to come up with the frequency that the equation gives.


peteroznewman posted this 17 December 2017


I am gradually learning the ACT Acoustics Extension and worked most of the workshops that came with the documentation.


I am reading “The Science of Sound” 2nd Edition by Thomas D. Rossing. Example 4.3 is to find the first three modes of vibration of a pipe 750 mm long with one open end and one closed end (neglect end correction). The answer in the book is f1 = v/4L = 343/4(0.75) = 114 Hz, f2 = 3(343)/4(0.75) = 343 Hz and f3 = 5(343)/4(0.75) = 572 Hz.  These frequencies are calculated from knowing the standing wave patterns of the first three modes. I built a Modal Acoustics model in ANSYS 18.2 and got excellent agreement.




Rossing provides a formula to calculate the frequency of a Helmholtz resonator in Section 2.3 and gives an example of a flask with a 980 mm diameter sphere, a 30 mm diameter neck that is 100 mm long. 

Rossing formula f = 207 Hz 
Acoustic model  f = 232 Hz.


Your formula         f = 262 Hz
Acoustic model    f = 299 Hz 
Rossing formula  f = 346 Hz


Rossing says that open pipes have an “end effect” that adds to the length of typically 0.61D, while the formula you gave seems to have an “end effect” correction of 0.75D.  I find it interesting that if I zero out the end correction in your formula, I get the same value as the Rossing formula.

If I use a correction of 0.35D, then I get f = 298 Hz, which is very close to the acoustic model.

I hope some of this is useful for you.

Best regards,


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emirdegirmenli posted this 20 December 2017

Thak you Peter, You researched very detailedly. Maybe we can define new hole length of the acoustic model by regarding "end effect" because we defined the end of the neck as a constant pressure (0 Mpa). Or we use original neck length and we define other body has compressible air condition for the open end. ı will test them, thank you again. Best, Emir