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It's bonfire weekend here in the UK, and everybody is intent on shooting fireworks.

I was wondering, what determines the pitch of the sound produced by an explosion?

The sound has clearly a dominant mid-range pitch, around a few hundred hertz, and bigger fireworks also have a very low, but large component at around 20-50 hertz.

What determines this pitch? The amount of explosive energy? The shape of the vessel? Something else?

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Do balloons of different sizes make different sound when popping? I think so. This is a pretty simple way to model it. –  endolith Nov 13 '10 at 18:49
    
why would explosions have different sizes? how do you define the size of an explosion, and what properties determine it? –  Sklivvz Nov 13 '10 at 19:05
    
i think if you have bigger blast then you will have high intensity. pitch in related with the time of explosion. if it exploded quick then it will have low pitch. though not sure. –  Santosh Linkha Dec 31 '10 at 7:37
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Ask this at Audio.SE just to annoy them. :) –  muntoo Jan 2 '11 at 0:07

4 Answers 4

up vote 9 down vote accepted
+50

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.

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+1, because I think you nailed down the two factors. If you can clean up the answer a bit, I'll accept it. –  Sklivvz Jan 2 '11 at 9:12
    
Brief cleanup: Eardrums are air-pressure sensors. The blast produces a pressure profile which propagates in all directions. As it does, amplitude goes down, but frequency stay constant. The pitch, then, is just the pressure profile, decomposed into its Fourier modes. A short, sharp punch -- like the pop of a burning log -- will produce a higher pitch, a slower "push" will produce lower frequencies. For a given mass of material, it takes more energy to displace it through air more quickly, hence to produce a higher pitch. (And, attenuation favors low frequencies if you're far away.) –  Eric Zaslow Jan 3 '11 at 0:01
    
The 'ground effect': the reflection of the energy in the ground adds to the directed (from the blast to ears) received energy. A monopole antenna radiating near the ground has the same effect (and saves energy). And the delta-time is, imo, linked to the lenght of the container (cilindrical). Shorter cilinders -> shorter wavelength , i.e. higher frequencies. –  Helder Velez Sep 23 '11 at 18:55

If I had to guess at an answer, I would guess that it has something to do with the size of the blast. Lower frequency means a longer wavelength, which will be excited more effectively by a blast with a bigger displacement.

You can see the sort of thing I mean by looking at a piezoelectric transducer driving a speaker. For a fixed peak-to-peak voltage, the sound produced will be louder at higher frequencies, where the limited displacement of the speaker is closer to the wavelength of the sound wave. Speakers that are driven by PZT's tend to have some electronics to boost the amplitude of lower-frequency signals for this reason (with old-school magnetic coil speakers, you get this for free, and the displacement naturally ends up being larger for a slower oscillation). This is also why woofers for a stereo system tend to be larger-- the greater size more effectively excites low-frequency, long wavelength modes.

(We discovered this by accident one day when I was in grad school when we were doing an experiment involving a mirror on a PZT, and spent an hour or so trying to figure it out before one of the older staff scientists wandered in and explained it. It sticks in my memory as a result.)

I could be wrong, but if I had to guess, this would be my guess.

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Given similar overpressures, densities, and external pressure. I would think that velocities are largely scale invariant. So the time scale for things to change should be proportional to the linear dimension, which should scale as the cube root of the yield. This probably only applies for fixed external pressure however. –  Omega Centauri Dec 29 '10 at 22:52

Another factor: explosion is a dirac-type pressure excitation. The frequency spectrum is flat (Fourier transform $F(\delta)=1$ is unity for every frequency) so my guess is, different pitch comes from attenuation of different frequencies over distance - far away explosion is higher pitched, than the same explosion nearby; at the same distance the stronger explosion of the two is lower pitched, as low frequencies are not completely attenuated.

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You have a number of factors, and I'm not even sure I even know them all.

For explosives like fireworks, one issue is pressure -- larger fireworks tend to have thicker shells, and be wound more times. (in the manufacture of larger fireworks, they wrap them tightly with string to compress them and make them burst further). So, when the firework finally bursts, the internal pressure is higher, and so you'd end up disturbing a larger volume of air in the same amount of time. I have no idea how much this is a factor ... I assume you've have to take an identical amount of explosive, in identical volume containers of different strengths.

Obviously, using more explosives would also result in disturbing more air, but I don't know how much of a contributing factor volume is vs. pressure.

One other factor is the chemical used in the explosion -- obviously, different ones burn differently, but even in the case of balloons bursting, and we're dealing with similar pressures of similar sizes, helium and hydrogen don't make the same sounds as something filled with compressed "air" ... but I don't know if it's an issue with the molecules being smaller, and so less likely to vibrate other things they pass by (just by odds of there being a collision), or if it's simply an issue of different total mass, and so there's less stored energy at the time of release.

I'm guessing shape is also a factor, but that's going to be even harder to quantify, as with a shaped charge you're going to have gas moving at different speeds in different directions, which may make the pitch vary with the location of the observer.

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