In https://www.britannica.com/science/sound-physics there is a statement about the relationship between the attenuation of sound waves in air and the wave frequency in the part "Sound Absorption"(great article for beginners like me strongly recommend). It is stated that, the attenuation rate is proportional to the square of the frequency.
I want to understand the logic behind this statement and I have a guess, which one of you may check.
My guess is that with increasing frequency, the number of contact between air molecules per second increase but the contact time per interaction decreases. Overall, since each contact produces heat, the number of contacts per second dominates and there is attenuation.
Still, I do not understand the square of the frequency relation. Overall, I need my theory to be tested and to understand this process.
Edit: grammar
Best Answer
The biggest effect causing attenuation is the viscosity of the air. The size of the viscous forces depend on the rate of change of the air velocity with position.
If the frequency of a sound is doubled, the wavelength is halved. Therefore the velocity changes from maximum to minimum in half the distance, and the viscous forces are doubled. That removes twice as much energy from the sound wave during each oscillation of pressure.
But if the frequency is doubled, there are also twice as many oscillations per second, so four times the amount of energy is removed per second.
Of course the viscous forces arise in some way from the interactions between the individual air molecules, but molecules are very small compared with the wavelength of audible sound, and the pressure changes in sound waves are tiny compared with atmospheric pressure, so the sound wave makes no significant difference to the number of collisions between the air molecules.