In class we came across a problem that is essentially the following:

Suppose a source of sound has an intensity of 70 dB. Suppose 9 additional sources produce the same sound in unison (so 10 identical sources of 70 dB sound). What is the resulting sound intensity in dB?"

The solution uses the fact that the resultant intensity is 10 times the intensity of the original source, getting a result of 80 db.

I do not understand this. We are told that intensity is proportional to the power of the wave which is proportional to the amplitude squared. A superposition of 10 identical waves yields an amplitude of $$10A_0$$ where $$A_0$$ is the amplitude of one wave. Thus, we would get an intensity 100 times the original and a resulting 90 db.

Why is this not the case? Any help would be appreciated!

In a similar question, the good answer https://physics.stackexchange.com/a/714653/219989 shows a calculation of how when you have N sound waves with identical frequency - but uniformly distributed phase (not the same phase) - it is actually the intensity (the integral of square of the amplitude) which becomes N times the intensity of one wave.

The answers in Why do two coherent sounds add up $+6db$? make another important observation: If the N sound sources are coherent (same phase) but don't come from exactly the same place, then depending on your position you may hear them at 90db (fully constructive interference) or at 0db (destructive interference) or anything between, and 80db would be an average of what you would hear.

• Wouldn't fully destructive interference be $-\infty$ dB rather than 0 dB? Commented May 1 at 23:06
• Yes, indeed. "dB" measures the log of the intensity, not the intensity itself, so zero intensity (which is what I meant) indeed translates to minus infinity dB. Commented May 2 at 10:17
• "depending on your position" – You can easily try this out at home: Set up two speakers ~1–2 meters apart and play a sine wave of wavelength ~1m from both of them, close one ear and move around a bit. E.g. 440 Hz sine waves should be easily available in Youtube.
– JiK
Commented May 2 at 13:11

The conceptual reason is that different sources have uncorrelated phases, so that at each frequency, they will randomly either add or subtract from the sound pressure. Adding these sources together is like taking steps in a random walk. Imagine that for each step, you randomly choose to move to the left or the right by a distance of 1. After $$N$$ steps, the expected change in your position is 0. Correspondingly, for sound, the average perturbation in the air pressure is 0. However, for the random walk, after $$N$$ steps, the expectation of the square of the change in position is $$N$$. That is, the squares of the step lengths are additive, and correspondingly for sound, the squared amplitudes are additive.

Mathematically, we can write this as follows. Let $$P_i$$ be the sound pressure perturbation contributed by each source $$i$$. The sound intensity contributed by that source is $$I_i=\langle P_i^2\rangle$$, where the angle brackets denote the statistical expectation. However $$\langle P_i\rangle = 0$$, since sound equally contributes positive and negative pressure perturbations. Also, since different sources are uncorrelated, $$\langle P_i P_j\rangle=0$$ for $$i\neq j$$.

Adding together the contributions of all of these sources, the expectation of the total sound pressure perturbation is $$\left\langle \sum_i P_i\right \rangle = \sum_i\langle P_i\rangle = 0.$$ However, the expectation of the squared sound pressure perturbation is $$\left\langle \left(\sum_i P_i\right)^2\right \rangle = \sum_{i,j} \langle P_iP_j\rangle = \sum_{i=j}\langle P_iP_j\rangle + \sum_{i\neq j}\langle P_iP_j\rangle.$$ The second term vanishes since different sources are uncorrelated, while the first term is just the sum of the sound intensities. That is, $$\left\langle \left(\sum_i P_i\right)^2\right \rangle = \sum_{i}\langle P_i^2\rangle = \sum_i I_i.$$

When you say the amplitude of the total sound is ten times the amplitude of one of the sounds, you are assuming all sounds are in phase, in which case your answer is correct.

The question you are answering uses the words "in unison," but doesn't explain what that phrase means. If by "unison" they mean in phase, then you are correct.

But if "unison" is supposed to mean "at the same time," the answer they give is correct if we assume that all the sounds are uncorrelated, as in white noise. In that case, we must add the energy of each of the separate sounds, because there will be just as much constructive as destructive interference among all the separate sounds. The addition of the energy results in the addition of the Intensity, which is the energy flowing per unit time and per unit area in the path of propagation.