Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I've seen a few questions about sound waves and sound travel here on Physics SE, so I'm hoping this question is a good fit for this site.

During my internet research on soundproofing, I've come across many acoustics gurus who say that, when soundproofing a wall, you have to account for every single little opening such as a wall or ceiling even if the opening is as tiny as the shaft of a screw. If you don't, you may as well not soundproof the wall at all.

Since I have approximately zero understanding of the physics of sound waves, my incredulity brings me to doubt those words. I imagine the contact of sound waves upon a wall being much like taking a handful of pebbles and slinging them against the same wall. When the pebbles leave your hand, they spread out. If I were to sling a handful of pebbles at a group of people from a distance, the chances are high that everyone's going to get hit. If you put up a solid wall in front of them, however, no one will feel a thing as the wall will stop 100% of the pebbles. If you cut a hole 2' in diameter and sling the pebbles, perhaps only a few people will be hit since only the pebbles that make it through the hole will have a chance of striking anyone. If I were to apply the "weakest link" argument here, it would suggest that slinging a handful of pebbles at a wall with a small hole in it would be the same thing as slinging the same pebbles as if the wall weren't there. The only way this could be true is if all the pebbles spread out after leaving your hand, magically coalesced enough to fit through the hole, then spread out again to hit the group of people.

Now back to sound travel. How could a small, screw-sized hole totally negate the entire soundproofing effort? While I can understand that a little sound would still make it through that hole, I am not seeing how that little leak would make it as if the wall isn't there at all.

Can someone explain how a small opening in a well-soundproofed wall affects sound travel?

share|improve this question

3 Answers 3

The sound gurus exaggerate when they say that an opening as tiny as the shaft of a screw would let all the soundproofing go to waste. But although this is an exaggeration (and we all know how purist sound gurus are) there is some truth behind their argument that I share with you here:

To understand the physical argument behind their exaggeration one has to look up the concept of diffraction of waves.

Wave diffraction through a slit

When the sound waves encounter the wall their energy is absorbed by it (and if the soundproofing is good most of it won't reflect back nor will refract to the adjacent room).

Now consider that the wall has a narrow opening (like the shaft of a screw). When the crest of the wave reaches this opening all its energy is absorbed by the wall but a new wave is generated in the wall opening, with an initial energy equal to the energy that the original wave had when it reached the wall. The difference is that the new wave shows an interference pattern so that in the new room you have areas where you won't here anything and areas where you will hear what came out of the soundproofed room.

Think for example of the camera obscura which is the same only with light waves instead of sound waves. You are able to generate the image of what's in front of the camera just by letting the light pass through the small hole. But the image that is generated inside the camera is very faint. This happens because with such a small opening only a certain number of photons can pass through. The same happens with sound and pebbles: only a certain number of phonons will get through so the sound you are able to hear in the adjacent room is not quite what you can hear in the soundproofed one.

This does have an obvious impact in the soundproofing, but of course it's not so much that all soundproofing goes to waste. You were able to stop all the incoming phonons except for the ones that got through the hole.

share|improve this answer

Exaggeration. You hear much less sound through a door which is ajar than through one which is closed. Obviously, the more you seal, the less sound gets through, and sealing every little orifice is subject to obvious diminishing returns.

The only way that the slightest aperture could be as "bad" as a large one would be if the room's interior propagated sound waves without any loss, and if the walls likewise reflected all sound without loss. Noise would therefore build inside the room, similarly to pressure. A larger aperture would leak less intense sound. A small aperture would allow more sound to build up, therefore it would leak a narrow beam of more intense sound.

But real soundproofing works by absorption, not only containment. Sounds do not reflect without loss. The soundproofing materials are supposed to absorb sounds (ideally without any reflection at all).

However, something to consider is whether you are trying to keep sound out, or keep sound in. If you are trying to keep sound out of the room, then the more successfully you sound-proof it, the quieter it becomes inside (its noise floor drops, so to speak). The human hearing system has quite a dynamic range. The quietest sound you can hear is many orders of magnitude less intense than the loudest that your hearing can withstand. A sound that you would not even hear if the room were not as well soundproofed as it is can suddenly turn into an annoyance. We can be annoyed by relatively small sounds, like a dripping tap.

On the other hand, if you're keeping sound in a room, then you have to consider that the environment of the rest of the building is not silent. At some point, the room is so well sealed that any remaining escaping noises are well below the ambient noise level of the surrounding building.

share|improve this answer

Building off something mentioned in @Kaz's answer: Your ears are roughly logarithmically sensitive to the amplitude of sound waves. That is why we often measure sounds in decibels. This means that increases in sound energy at the low end have a far more profound effect than equivalent (in power) increases at the high end.

Plugging in some numbers... Say $P_0$ is the power of the minimum audible sound to your ears. If your ears are the same as the "standard," this corresponds to $0\ \mathrm{dB}$. Let's say the uninsulated wall is $10$ square meters in area and transmits enough sound to register as $100\ \mathrm{dB}$, which is rather loud. This is $10^{10}$ times more power than the threshold $P_0$. In terms of "sones,"1 this is about $2^{100/10} \approx 1000$ times as loud as the quietest sound you can hear, where you should note that human perception is not perfectly described by these simple laws.

Suppose you now insulate the wall with perfect sound-proofing, but leave a hole with area only 1 square centimeter. This is one part in $10^5$ of the area, so $$ \frac{10^{10}P_0}{10^5} = 10^5 P_0 $$ power will get through. This is equal to $50\ \mathrm{dB}$, so it will sound roughly $2^{50/10} = 32$ times louder than the quietest sound you can here. Equivalently the $100\ \mathrm{dB}$ sound will be reduced by a factor of $2^{(100-50)/10} = 32$ in perceived loudness.

So insulation helps, but not as much as you would expect. In this particular example, insulating $99{,}999$ parts in $100{,}000$ of a wall stops $31$ parts in $32$ of the perceived volume.

1 Thanks to @EnergyNumbers for pointing out that one needs this extra conversion.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.