So, as far as I understand, sounds are ripples through a material. If you strike a rock, then (ignoring thermal motion, if we may) the molecules in the rock start off stationary, then wiggle back and forth as the waves pass through them, then return to a stop. How far do they move? I assume it varies based on material and loudness, but is there a table and equation somewhere? I found no good leads, searching the internet.

Also, is this motion at least roughly equal to the motion at the surface of the object? If a sound is produced in a perfectly smooth steel cube, how much displacement would you see at the surface? Would it be on the order of nanometers, micrometers, or what? (For, say, a solid tap of the hammer, if you can factor out the resulting net motion of the cube.)


1 Answer 1


Audible sound in air ranges over twelve orders of magnitude in power, six orders of magnitude in sound pressure. It also depends on frequency, which ranges over three orders of magnitude for the audible but ultrasound is often used in applications (in tissue or other media).

To generate sound in water etc one can use piezoelectric or magnetostrictive transducers. At maximal power, amplitudes are on the order of microns. Exact specifications are in data sheets.

It would not be too difficult to measure the amplitude of the ends of a "singing rod", but I have not been able to google a laboratory report.

  • $\begingroup$ Ok; "microns" sounds like the answer I was looking for; thanks. Would you happen to have a reference or link for further information? $\endgroup$
    – Erhannis
    Sep 17, 2020 at 21:25
  • $\begingroup$ The specifications of the displacements of such actuators vary over tremendous ranges, depending on the power needed and the frequency at which they operate. But look for example at en.wikipedia.org/wiki/Terfenol-D with a displacement of up to 0.002 m/m. (Of course, the generators are never one meter thick.) $\endgroup$
    – user137289
    Sep 17, 2020 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.