# Why do we hear the square of the wave?

Assume we superpose two waves of frequencies $\omega_1, \omega_2$. Then what we get are beats. Adding the two sines gives us $$\psi = A\sin(\omega_1 t) + A\sin(\omega_2 t) = 2 \sin \left(\frac{\omega_1 + \omega_2}2\right)\cos \left(\frac{\omega_1 - \omega_2}2\right).$$

Then I'd say that the frequency of beats we'd hear is $\frac{\omega_1 - \omega_2}2$. On the other hand, someone told me that it's ${\omega_1 - \omega_2}$ in fact, because what we hear is the square of the wave $\psi^2$.

Why?

• Someone told you wrong. What you are hearing is the linear signal, not its square. The difference frequency can only be heard when its frequency is very low (less than approx. 12Hz), which is probably the result of neural signal processing which allows us to determine envelopes with sub-100ms time resolution. It certainly doesn't have anything to do with non-linearities in the human ear, as any artificially distorted sound recording will immediately prove. Jan 28, 2016 at 8:06
• The intensity of sound which is what you hear is proportional to the square of the amplitude of the sound wave. Jan 28, 2016 at 8:10
• @Farcher: Your ears are (roughly) sending the logarithm of the intensity of the band filtered components of the sound to your brain. That's totally different from the square of the amplitude and it equates to an excellent reproduction of the linear signal over an extremely large dynamic range, albeit without phase information, which is why some people think it's the square of the amplitude. Jan 28, 2016 at 8:24
• So it should be what you receive before processing by the eardrum is proportional to the amplitude squared. The term which I would use for the response of the ear is "loudness". Is that wrong? Jan 28, 2016 at 16:03

The function above can be interpreted as follows. Sound of frequency $\dfrac{\omega_1+\omega_2}{2}$ with amplitude modulated by the cos function of frequency $\dfrac{\omega_1-\omega_2}{2}$.