# Why do we add intensity instead of pressure for sound waves? [duplicate]

If two sound sources cause sound levels of 50dB and 53dB independently at a certain point, to measure the sound level at that point we have to first convert each sound level (dB) to its intensity ($$W/m^2$$):

$$\text{Sound Level} = 10 * log{ I \over I_0} \;\;\;\; \text{with reference intensity equals } 10^{-12} \;{W / {m^2}}$$

Then we add up their intensities before converting the value back to sound level:

$$\text{Sound Level} = 10 * log {{I_1 + I_2} \over I_0} = 10 * log { {10^{-7} + {2*10^{-7}}} \over {10^{-12}}} = 54.8dB$$

That's how it was written on my textbook.

The part that I don't understand is: shouldn't be the amplitudes of waves that add up when there is constructive interference, or cancel out for destructive interference, instead of intensity? If there are two identical traveling waves $$A*Cos(kx+wt)$$ traveling in opposite direction, the two will superimpose and we'll end up with a standing wave $$2A*Cos(wt)$$. What add up are the amplitudes $$A+A = 2A$$. And since amplitude is pressure ($$N / {m^2}$$ or Pascal) for sound waves, it should be their pressures (or amplitudes) that add up instead of intensity. We should convert intensity to pressure before adding their magnitudes.

$$I = { {\text{pressure}\;^2} \over {\text{density of air}} * \text{speed of sound}} = { {p^2} \over {\rho}c} \approx { {p^2} \over 400} \implies p = \sqrt{400I}$$

• 50dB sound level = $$10^{-7} \;W/m^2$$ intensity = 0.00316Pa pressure
• 53dB sound level = $$2*10^{-7} \;W/m^2$$ intensity = 0.00894Pa pressure

If we add up their pressure (wave amplitudes) 0.00316Pa + 0.00894Pa = 0.0121Pa, convert it back to intensity $$I = (0.0121Pa)^2 / 400 = 3.66*10^{-7} \;W/m^2$$, we'll get sound level = 55.635dB!

If I add sound intensities, I get 54.8dB. But if I add sound pressures, I get 55.65dB. It makes more physical sense to me to add the sound pressure (amplitude of wave), but everywhere I read always add their intensities and am not sure why.

• Does this answer your question? Why do we add intensities for coherent sound sources? Jun 27 at 9:04
• @VincentThacker I'd read that post before asking the question. I had a hard time reading through the answers to be honest and am not sure if that's even related and hence my questions here laying out the simplest equation with values. I don't get why adding pressure is not more accurate than adding intensity.
– KMC
Jun 27 at 9:44