As I chew my gum, I wonder: when I blow and pop bubbles, what creates the snapping sound?

Or similarly, what causes the loud POP when one pops a balloon? Is it the rush of air from inside the pocket rushing out and colliding with the air previously outside the pocket? Or is it the immediate contraction of the surface material (gum, rubber, etc.) once it is broken?

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    $\begingroup$ Related question; why is popping gum INTO your mouth louder than popping gum OUT of your mouth? $\endgroup$ Nov 20, 2015 at 18:52
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    $\begingroup$ Related question 2. How do bullfrogs make their sounds, without the chewing gum? $\endgroup$
    – user81619
    Nov 20, 2015 at 19:01
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    $\begingroup$ Related question 3: why has nobody invented a silent explosive like in that one episode of Get Smart $\endgroup$
    – Asher
    Nov 20, 2015 at 20:05
  • $\begingroup$ A simple Google search gives you excellent answers, see for example this reddit post. $\endgroup$
    – Lith
    May 26, 2019 at 11:51
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    $\begingroup$ There are two hypotheses to consider, 1) the pressure of the air in the ball is released and creates a sound wave and 2) the material of the ball itself creates the sound when it is being torn apart. The science is non-linear and tricky, so I would not trust any simple theoretical argument such as those given by a quick google search. Empirical evidence should be presented for either case. $\endgroup$
    – Void
    May 27, 2019 at 14:58

5 Answers 5


This is discussed in Panda et al., "Why Balloons Make a Loud Noise When They Pop," International Journal of Acoustics and Vibrations (2019) and the references within. The authors conclude that the sound arises primarily from the "vibration of the balloon surface and the ruptured edge of the latex". Some of the other answers on this thread address the sound pressure resulting from immediate gas expansion. However, Panda et al. observed using shadowgraph imaging that the expansion was too slow for this explanation to be valid ("There is also no evidence of compression waves emanating from the expanding gas at the breach in the latex."). Indeed, the pressure waves were observed to be shed from the balloon skin.


Let's look first at what sound is:

Sound is a compression wave in a fluid. That means that you have local variations of pressure and particle velocity over time and distance in the fluid. In other words: air molecules are wiggling back and forth bumping into each other and locally increasing or decreasing the pressure as the get close or far apart.

That pressure variation is typically very small compared to the steady state atmospheric pressure, which is about $100\,000\;\text{Pa}$ (Pascals) or $194\;\text{dB SPL}$. The sound level at the pain threshold (jet engine, super loud rock concert, some thunder) is about $120\;\text{dB SPL}$, which corresponds to $20\;\text{Pa}$ of pressure variation, i.e the overall pressure varies only between $99\,990$ and $100\,020\;\text{Pa}$.

On to the balloon: the air pressure in the balloon is higher than the external pressure, but typically not by a lot. It's maybe 1% of atmospheric pressure, depending on the tensile strength of the balloon. When it ruptures, the pressure will equalize and air will move outwards from the balloon. Now we have local pressure and velocity variation, in other words: sound. The form of the pressure variation is short, tall spike that decays quickly over time and less quickly with distance.

It's actually easy to calculate how loud the spike will be: Assuming an internal balloon pressure of 1% atmosphere $(154\;\text{dB SPL})$ and a distance of the ear from the balloon of twice the balloon's radius, we get a pressure peak of about $144\;\text{dB SPL}$. That's loud, but it's also very short and still very much below the "danger pressure", which may rupture an ear drum $(> 174\;\text{dB SPL})$.

It's misleading to talk about a "void": there are local pressure areas, where the pressure is smaller than the steady state air pressure, but the pressure is still much, much closer to the normal pressure than it would be in a vacuum.

  • $\begingroup$ It looks like this conjectured mechanism has been disproved by a recent shadowgraph imaging study that found negligible compression waves emerging from the breach; instead, the sound was concluded to be generated by vibration of the balloon material and ruptured edge. $\endgroup$ Jul 21, 2021 at 4:43

You are asking whether this popping sound is made by the difference in air pressure between the inside and outside of the balloon. Or the contraction of the material.

Actually it is none.

It is actually a wave that makes the popping sound. You would think it is a soundwave. Well it is that reaches your ears.

But the original cause is not the air.

It is the balloon material, that as it fractures, creates a phenomenon on its surface.

As per QM, the lattice molecules in the balloon material are held together by the EM force. The covalent bond that bonds the molecules together is created by electrons that are not shared by multiple atoms and multiple molecules. These valance electrons now exist as per QM around the whole molecule or group of molecules. Now these covalent bonds are constantly pushed against by the pressure of air inside. But what is exactly this pressure?

The pressure from inside is the vibrational energies of the air molecules as the gas tries to fill all available space volume.

Now after a certain time the material will not last, and there will be a fracture, caused by the brake in the covalent bond of some molecules in the lattice.

As this covalent bond brakes, air molecules start escaping between the lattice molecules of the balloon.

Now at this point the fracture in the lattice continues because the material's covalent bond around the molecules is not able to withstand the material's sideways tear force.

This is the start of a wave, inside the lattice of the balloon material.

As this wave travels, the lattice molecules form a wave and this wave formation is hitting the air molecules, transferring the waves into the air molecules structure.

A wave is formed now in the air molecules structure, and that is traveling to your ears, making you hear a popping sound.

As per the comment, I would like to use this site, as reference, I hope that is acceptable as reliable:

Now, Regarding the sound theme: The pressure inside (depends upon how much you blow) and outside (1 atm) the balloon creates a pressure difference in that area. This difference creates a longitudinal pressure wave (I'd say that as a small shock-wave) along with elastic energy of the balloon's material (while retaining to its original shape) which is perceived as sound.

Please see here:




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    $\begingroup$ To emphasise the bounty message: have you got any reliable references which back up this claim? $\endgroup$ Jun 1, 2019 at 22:17

It's for the same reason as fireworks making that short lived, sharp noise.

During explosion, huge amount of heat is created in a fraction of second that the pressure surrounding firework rapidly increases. This pulse of highly compressed wave results in a high dB sound.

As @Hilmar pointed out, the pressure inside a balloon is higher than atmosphere's. Thus after popping the balloon, the air in the balloon pushes the surrounding air, creating a compressed wave. This is why we here a "pop".


Let's use the balloon as an example. When you inflate it the contents are under pressure because the latex wants to shrink back to its natural size. When you pop the balloon that pressure is released. It rushes outward temporarily creating a void. The void then sucks in the surrounding air. The popping sound is the air molecules slamming into each otherwhen the void is filled.

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    $\begingroup$ do you have a source for this statement? $\endgroup$
    – pentane
    Nov 20, 2015 at 21:54
  • $\begingroup$ @pentane yes, I do. Two different physics teachers at my school. The pop is created much the same way that thunder is after lightning strikes. $\endgroup$ Nov 21, 2015 at 19:31
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    $\begingroup$ You can't hear "molecules slamming into each other", referring to molecules is using the wrong scale and model to explain this effect (which is in the realm of continuum mechanics). $\endgroup$ May 26, 2019 at 15:03
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    $\begingroup$ Sorry, this is wrong. The pressure variation involved are so small, that talking about a "void" here, is really misleading, unless you assume that 99% atmospheric pressure is a "void". $\endgroup$
    – Hilmar
    May 26, 2019 at 15:16

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