Sound waves adding up If there is a swarm of bees, why is it louder than a single bee? This question may sound stupid but ... if they are all buzzing at the same frequency with the same loudness and the phases of the sound waves are random, should they not cancel out when they arrive where the listener is? Where is the catch?
 A: You can compare this to taking the mean of a random sample of mean zero random variables. Some observations will be positive, and some will be negative, and so there will be a lot of canceling. However, the standard deviation will increase the larger the sample size. The canceling will cause it to grow as the square root of the sample size, rather than linearly, but it will still grow.
Similarly, when you have $N$ bees randomly making noise, the amplitude will grow as $\sqrt N$. Intensity is amplitude squared, so it will grow linearly, i.e. it will be proportional to $N$ rather than $\sqrt N$.
In general, when you have waves adding together, the power on average will add together. Sometimes there will be peculiarities of the geometry such that particular spots have destructive interference, and/or spots have constructive interference, but the average will be additive.
You can think of the space of all possible waves as being a vector space, and each particular wave is a vector in that vector space. Two waves being uncorrelated means that the vectors that represent them are orthogonal, so the length of the sum follows the Pythagorean theorem: the square of the sum is the sum of the squares. The length of the vector represents amplitude, and the square of the length represents intensity or power, so the power of the sum is the sum of the power.
A: Since the phases are random, the waves do not add coherently... but neither do they cancel coherently. Furthermore, loudness is really a measure of the intensity of the sound. Thus, if we consider a wave generated by a person $i$ to have amplitude
$$
A_i(x,t)=A_i\cos(\omega t - kx +\phi_i),
$$
then the total amplitude is
$$
A(x,t)=\sum_{i=1}^N A_i(x,t)=\sum_{i=1}^N A_i\cos(\omega t - kx +\phi_i),
$$
whereas the intensity is
$$
I(x,t)=|A(x,t)|^2=\left|\sum_{i=1}^N A_i\cos(\omega t - kx +\phi_i)\right|^2 =\\
\sum_{i=1}^N A_i^2\cos^2(\omega t - kx +\phi_i) + 
\sum_{i=1}^N\sum_{j=1, j\neq i}^N A_iA_j\cos(\omega t - kx +\phi_i)\cos(\omega t - kx +\phi_j).
$$
If we now average over phases, by assuming them homogeneously distributed in interval $[0,2\pi]$, then the second term vanishes (no interference), but the first term gives non-zero contribution
$$
\overline{I(x,t)}=\frac{1}{(2\pi)^N}\int_0^{2\pi}d\phi_1...\int_0^{2\pi}d\phi_N I(x,t)=\sum_{i=1}^N \frac{A_i^2}{2}.
$$
If all the bees buzz at the same loudness, then $I=NA^2/2$ (in practice this is not the case, since the sound is attenuated, i.e., its amplitude depends on the positions of each bee.)
