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#1 |
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diyAudio Member
Join Date: Dec 2002
Location: Guelph, Ontario, Canada
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Does anyone know the formula for finding the resonant frequency of an enclosed space? I don't mean the resonant frequency of an enclosure with a speaker in it. The enclosed space could mean a speaker enclosure or a room.
Thanks; Doug
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#2 |
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diyAudio Member
Join Date: Oct 2002
Location: Barrie, Ontario, Canada
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Its not that simple as to apply a formula. It involves the various materials used, the geometry, and a whole lot of factors. If your interested in actually calculating this, get a good machine dynamics book.
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The *** is a muscle that MUST be rested! |
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#3 |
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diyAudio Moderator
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If you make simplifying assumptions, like perfectly relective, rigid boundaries and loss-free interior, the equations are quite straightforward once you've set the boundary conditions. What shape space do you have in mind?
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If there's a sucker born every minute, where do the rest of them come from? |
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#4 |
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diyAudio Member
Join Date: Mar 2001
Location: Melbourne, Australia
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Doug
If you have a rectangular space, then there are a series of resonances across each of the parallel walls, starting when the half the wavelenth is equal to the distance between the walls, then 1x, 1.5x 2x etc the wavelength. You can calculate the wavelength from the formula : Wavelength (m) = speed of sound in air (345 m/s) / frequency (Hz) So a frequency of 100 Hz has a wavelength of 3.45m, and the first resonance will occur across walls 1.725m apart. There are also tangential and oblique modes which occur between non-parallel walls. Have a read of the Harman white papers by Floyld E Toole, which explains all about room acoustics and should answer the question much better than I can here. See http://www.harman.com/wp/index.jsp?articleId=65 Mick |
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#5 |
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diyAudio Member
Join Date: Oct 2002
Location: Barrie, Ontario, Canada
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SY,
A simplidied equation is only that, it wont give the actaul natural frequency, it may be close enough for his purposes, but the purposes were not specified. Kanga, He wanted the natural frequency of the room or enclosure, not the modes of a given frequency within that space. Im not trying to be rude or make it seem like its impossible, its just that the theory of vibrations is no where near as simple as a lot of people think. The simplest way to calculate the natural frequency is with a signal generator and spl meter. Sweep the signal at constant amplitude, the frequency at which the spl meter is highest will be the spaces natural frequency. Jeff
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The *** is a muscle that MUST be rested! |
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#6 |
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diyAudio Member
Join Date: Mar 2001
Location: Melbourne, Australia
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Maybe Doug could clarify what he means by resonant frequency, so that we can get on the same wavelength...
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#7 |
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diyAudio Member
Join Date: Oct 2002
Location: Barrie, Ontario, Canada
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No pun intended right?????
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The *** is a muscle that MUST be rested! |
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#8 | |
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Account disabled at member's request
Join Date: Feb 2002
Location: Clifton Park, NY
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To tell the truth, I don't understand the question. But I do have a comment on the following statement :
Quote:
But I still do not understand what the original post was asking. |
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