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#1 |
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diyAudio Member
Join Date: Dec 2005
Location: west lafayette
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I'm just curious, how wide (relative to wavelength) must an aperture be for 100% of a sound wave to pass through it? F. Alton Everest specifically stated in the Master Handbook of Acoustics that if 13% of a thick steel wall was free (ie holes/slits) then 97% of the energy would pass through...Is this wavelength specific or universal to all wavelengths relative to the aperture?
Think of it like this ____ ____ ____ (3 4" obstacles with 2 .5" apertures, a 1700hz wave (~8") would easily diffuse around the 4" obstacle, but is the opening large enough for the 1700hz wave to pass through?) for reference, the sound waves wavelength must exceed the fundamental baffle width (wavelength>baffle width) before the 2pi to 4pi transition can occur |
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#2 |
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diyAudio Member
Join Date: Jun 2002
Location: USA, MN
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Is this a conceptual thing or do you have an application in mind?
It is relatively obvious that if there is any obstruction, something will be reflected or refracted and 100% of the sound will not pass through. If the obstruction is small compared to wavelength it may be essentially neglectable, but it won't pass 100% Then there is diffraction from the aperture to worry about, where the energy may mostly pass, but won't have the same directivity....and yes this is very frequency dependent.
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Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Science is a way of thinking much more than it is a body of knowledge. --Carl Sagan |
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#3 | |
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diyAudio Member
Join Date: Dec 2005
Location: west lafayette
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#4 |
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diyAudio Member
Join Date: Apr 2002
Location: Newcastle, Australia
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Ok, I'm trying to remember back 30 odd years when I did this stuff, so it all goes under the heading/disclaimer of ...... IIRC !!!
ok. sound is an air compression wave, so , so long as a small hole in a very thin material is big enough for air to get through, then the wave should re-propagate in a circular pattern from the exit side of the hole, albeit with a reduced amplitude depending on hole size. If the hole is significantly large wrt to wavelength, then the wave may pass through intact, but with re-propagation starting at the edges of the hole. (diffraction) More likely with any hole is that the depth of a hole, and the weight of air in it will cause enough friction and/or inertia to actually stop the compression wave. (this is in fact, basically how a vent in a loudspeaker works) All this assumes that the material in which the hole exists is not subject to passing the compression wave... |
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#5 | |
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diyAudio Member
Join Date: Jun 2002
Location: USA, MN
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Quote:
__________________
Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Science is a way of thinking much more than it is a body of knowledge. --Carl Sagan |
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#6 | |
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diyAudio Member
Join Date: Dec 2005
Location: west lafayette
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#7 | |
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diyAudio Member
Join Date: Feb 2004
Location: Stockholm
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Quote:
I would start with a simpler example to get a grip on what is happening. Take a single plate of 10x10 cm. let a sound wave fall onto it. The sound wave will be reflected back, regardless of wavelength. But, and here is the key to frequency dependence, there will also be diffraction of the reflected wave the at the plate edges. This is a situation very similar to the baffle step of a loudspeaker. The diffraction source at the plate edge is of opposite sign and equal amplitude as the incoming wave, and slightly delayed, by an amount corresponding to the plate size. For low frequencies, the delay is negible and the reflection is largely cancelled by the diffraction, ie near zero of the sound energy is transferred back. If on the other hand the wavelength is short, the diffraction form the different parts of the plate edges will be largely uncorrelated and cancel itself, and the reflection remains undesturbed. So sound will be reflected back, similar to the reflection of light in a mirror. So, calculating how much of the sound energy that is transferred through/around an object is not very simple in the general case. At best it can be simulated, and in order to do so, the math behind diffraction has to be understood. If you really want to dig into this, look for papers by Peter Svensson in JAES and JASA (I don't have the references here now, sorry). |
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#8 | |
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diyAudio Member
Join Date: Apr 2002
Location: Newcastle, Australia
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Quote:
The maths involved is something you will have to find for yourself if you want exact values. |
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#9 | |
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diyAudio Member
Join Date: Dec 2005
Location: west lafayette
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#10 | |
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diyAudio Member
Join Date: Feb 2004
Location: Stockholm
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Quote:
A little bit of reflection - rather wide A lot of reflection - pretty narrow |
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