Why does higher frequency sound dampen faster in air? I know as a general fact that higher frequency sound dampens quicker in air so when music is heard from a distance only the bass part is audible.But I don't know what the physical reasoning behind this is. I couldn't find an answer anywhere on the internet (or they were too technical for me to understand). I would appreciate any insight on this topic.
Note: I'm a second year physics undergraduate and I know a little wave mechanics and acoustics.
 A: Recall that wave equations will usually have a damping term and acoustic waves are no different.  Wave damping is usually modeled with a velocity dependent term.  The faster you try to distort the medium, the higher the damping.  The viscosity of the fluid through which the sound wave is traveling plays a large role in the damping.  The link here supposedly gives an interactive player so you can model attenuation for various parameters such as humidity and temperature.  I can't seem to get it to work, but nonetheless the plot shows how absorption (damping) increases with frequency.  He also mentions relaxation processes as a factor in sound attenuation.  Hope this helps!
A: When sound waves dissipate the energy turns to heat.
If you have an insulated room, and you generate sound in that room, the influx of energy (in the form of sound waves) will end up as a temperature rise that matches the influx of energy.
My understanding is: the higher the frequency of the sound, the faster the rate of dissipation.

Some years ago someone suggested that it might be possible to transfer energy in the form of ultrasound. The receiver then has to turn sound energy back to electric energy.
(There are in fact microphones that can do that: those microphones do not consume power, the energy for the electric signal emitted from that microphone comes from the energy of the sound. The effciency is low, but it has it applications: since you don't need external power it is a very robust setup.)
But trying to transfer energy with ultrasound is a dead end. Among the many problems: the efficiency is low, so you would need to blast a lot of acoustic energy. Because of the rapid dissipation the range would be  limited to a couple of meters. Even if used at that short range: efficiency is so low that the emitter would have to pump out hundreds of watts of power. In effect that is a big heater in the room.
So: to learn more about sound energy dissipating to heat I suggest you look up information about the idea of transferring energy with ultrasound. In technical articles debunking that idea there will be discussion of why the energy of high frequency sound dissipates faster.
[Later addition]
To a good approximation the compression and rarefaction in sound propagation is adiabatic. As we know: if the process would be perfectly adiabatic there would not be any dissipation.
I assume the process of sound energy dissipation is loss of heat from compressed volume to rarefied volume. I will refer to that as 'leaking heat'. The shorter the wavelength of the sound, the less distance the leaking heat needs to travel, so I expect shorter wavelength sound to have faster dissipation.
