How to explain the difference of attenuation of different pitch of sound by kinetic molecular energy ? I know that «sound of low frequencies travel farther in air than sound of high frequencies». But since the energy of low pitch is less than high pitch. So logically, the high power mean more molecular collision so more volume where sound can disperse (travel) and less power mean less collision so attenuation so fast (sound travel less volume).


1 Answer 1


You cannot, really. Your question is based on an incorrect assumption that it is the energy of the sound waves that cause attenuation. Instead, let's have a closer look at the three main mechanisms of attenuation in free air:

  • Attenuation due to viscosity (proportional to $f^2$). Simply put, this is higher at high frequencies because the fluid is then expanded and contracted more rapidly.
  • Attenuation due to heat conduction (proportional to $f^2$). Simply put, this is higher at high frequencies because the sound wave peaks (warmer) and troughs (colder) are closer together, so that the heat conduction occurs faster.
  • Attenuation due to molecular relaxation (peaks at a specific relaxation frequency). This is related to the speed of the transfer of energy between the gas molecules' translational kinetic energy (i.e., connected to fast they move) and their vibrational energy. If the sound wave period is long compared to how long this process takes, the attenuation is low, and if these two times are about the same, the relaxation attenuation is at its peak.

Thus, none of these attenuation mechanisms are really related to the amount of kinetic energy that a sound wave has, and all have unrelated reasons for being stronger at higher frequencies.


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