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
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
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.
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.
Ron E said:
If you look at my example, the waves should easily diffuse around the obstruction (no reflection, and I'm not sure why you mentioned refraction. From my fundamental understanding, refraction applies to waves interactions as they transition into different mediums), but how wide (relative to wavelength, just as wavelength >obstruction size for diffusion vs reflection to occur) must the aperture be?
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...
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...
Andy Graddon said:
How small is small enough for air to pass through? The size of the air molecules (thats pretty insane). Whats a good practical value for the smallest size?
thadman said:
The math behind this is not very simple.
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).
thadman said:
A pin hole in very thin material will pass sound, its just that the amplitude is GREATLY reduced, maybe below audible levels.
The maths involved is something you will have to find for yourself if you want exact values.
Andy Graddon said:
I dont care if the amplitude is reduced on the other side of the obstacle, I'm just trying to figure out how wide the aperture has to be for no reflection of energy to occur on the front side of the obstacle.
thadman said:
NO reflection - Infinitely wide
A little bit of reflection - rather wide
A lot of reflection - pretty narrow
Basically, there will always be ....
refraction, reflection, diffraction, absorption and transmission at any corner, point, hole or surface.
the amounts or each depends on MANY things, including....
density of medium, reflective index, absorption index, width, frequency, density of the air...... and other aspects I really can't remember at this time.
in other words.. there is no real way of answering your question without a whole lot more data, and a whole lot more mathematics and applied physics.
I did touch on this stuff very briefly about 30 years ago while studying optics and optical patterns etc, but, sorry, there is no way I can recall it now without considerable work.
refraction, reflection, diffraction, absorption and transmission at any corner, point, hole or surface.
the amounts or each depends on MANY things, including....
density of medium, reflective index, absorption index, width, frequency, density of the air...... and other aspects I really can't remember at this time.
in other words.. there is no real way of answering your question without a whole lot more data, and a whole lot more mathematics and applied physics.
I did touch on this stuff very briefly about 30 years ago while studying optics and optical patterns etc, but, sorry, there is no way I can recall it now without considerable work.
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