I was reading about flash bang grenades, and I was really surprised to learn that in an M84 grenade there is only 4.5 grams of actual explosive (flash powder made from magnesium and ammonium nitrate). The specs claim that the explosion creates between 170 and 180 decibels of noise within 5 feet. That seems like a lot from such a small payload, so I was wondering if there was math to verify it. I found the Sadovsky equation which calculates the increase in air pressure at a given distance for a known type and mass of explosive. Intuitively that seems like it would relate to the noise produced, but I don't have the physics chops to translate into decibels. Any brilliant insights would be appreciated. Thanks!
[Physics] formula to calculate the sound produced by an explosion
acousticsexplosionsnoise
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I will go mostly with Chad's argumentation. Larger equals slower, which should excite relatively more low frequency sound. Also note that the larger the blast (hopefully) the further away the observer is. And air is not a perfectly elastic acoustic medium, some energy is lost, and the higher frequencies attenuate quicker than shorter, so distance will selectively filter out the higher frequencies.
Also, note, explosion usually means detonation. A detonation is an exothermic reaction which spreads by the compression (adiabatic) heating from the shockwave, and the chemical energy maintains the shockwave. A shockwave is essentially a highly nonlinear soundwave, and as the overpressure decays with distance from the source is will grade into a soundwave. Fireworks (pyrotechnics) are not explosives, but are the (relatively) slow reaction of chemicals (combustables, and an oxidizer) due to heat. Fireworks may generate shockwaves in air, if the package ruptures at sufficiently high pressure. Likewise volcanic blasts are not detonations, but shockwaves formed by the escape of high pressure gas.
If an explosion is fast compared to the sound frequencies the detector is sensitive to (probably human ears in your case), then we might be able to model the explosion as a delta-function in time. A delta function should equally excite all frequencies, so it should be a simple matter of distance attenuation of sound waves.
An explosion close to a solid surface, creates an amazing effect I've heard called a mach-stem (although wikipedia does not produce anything useful for this term). In any case, at fixed distance the near surface shock wave is much stronger than the shockwave at height above ground. In essence the ground effect part of the shockwave weakens roughly only as 1/R rather that the 1/R**2 one would expect for a free air spherical blastwave. I don't know what effect this has on the sound spectrum (pitch), but you need to be at a much larger standoff distance from a near groundblast than from a samesized high altitude blast because of it.
The wavelength of sound waves in a solid is much greater than the dimensions of atoms. For example the speed of sound in steel is 6100m/sec, so the note middle C (262Hz) has a wavelength of about 23 metres. Sound waves are collective motions of a vast number of atoms, and it isn't especially helpful to think of them as being generated by atom scale phenomena.
Because, as you mention in your question, atoms don't have a sharp edge, in a solid like steel atoms can be pushed together slightly and pulled apart slightly. This gives the solid some elasticity, and this elasticity allows compressions waves (i.e. sound waves) to travel through the solid.
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OK, I did some more research on this and I think I have an answer, but I would still appreciate someone who actually knows what they're doing checking my math.
First, I used the Sadovsky equation to calculate the increase in pressure in atmospheres. The equation is here:
https://www.metabunk.org/attachments/blast-effect-calculation-1-pdf.2578/
The explosive mass has to be multiplied by a relative effectiveness factor (REF) to account for the type of explosive. For example, the REF of ammonium nitrate is .42, because it has 42% of the explosive power of TNT.
Second, I found that pressure can be converted to decibels with the equation:
Db = 20*log(P/Pref) where:
P=The pressure caused by the explosion
Pref-The reference pressure for 0 decibels, which is the threshold for human hearing. It's 20 microspascals, or about 1.97 EXP-10 atmospheres.
So using a distance of 1.5M, a REF of .42, and a mass of 4.5 grams, the Sadovsky equation tells me the air pressure will be increased by about .091 atmospheres. Plugging that into the equation above yields a decibel level if 173.3, which is right in line with the specification.
So I guess the takeaway is that it takes very little explosive to make a major noise if you're close enough to it. Thanks everyone for your help.