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?
Best 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.