Specifically, I am trying to roughly determine the sound produced by a ball when it hits the floor and bounces. If the ball exerts a pressure onto the floor, then certainly this pressure will go on to create a sound wave in the floor (infinite block) and the air. I was just wondering if there is any way to find the sound wave (or the energy of the sound wave) released by the ball, thus finding the energy lost and the ball's approximate rebound height. Is the velocity of the sound wave given as follows?:

or is it more complicated than the basic equations provided in most physics textbooks?

  • $\begingroup$ This seems complicated because we usually think of "sound" in terms of just the frequencies we hear (roughly 20 to 20k Hz) but the ball will release energy vibrations outside of this range too. You'll need to specify what you mean better. $\endgroup$ Commented Aug 12, 2014 at 20:53
  • $\begingroup$ Also, it will be non-trivial to determine how much energy goes to heat versus vibrations. The material it strikes probably matters a lot. $\endgroup$ Commented Aug 12, 2014 at 20:54
  • $\begingroup$ I don't mean sound, just longitudinal waves in general. And I understand that the material matters a lot, I'm just assuming we know the bulk modulus and density, and perhaps other properties of the material. I figured that a more compressible ball (like a superball) would have more area to distribute its force and thus have a smaller pressure than a more rigid ball. Thus, the sound of the compressible ball hitting the ground is not as loud as the more rigid ball. But is there a way to actually calculate the approximate energy loss due to this? Like examining the pressure wave from the ball? $\endgroup$
    – Roman
    Commented Aug 12, 2014 at 21:01

1 Answer 1


There is one limit in which this computation is easy to do.

Let us consider a massive, perfectly rigid ball striking a a perfectly rigid floor. In this case, there is nothing oscillating, so that we can neglect sound generation by oscillations in the ball or in the floor. Yet there will be sound, because the ball displaces air in its fall, and the air displacement is a small-amplitude effect (so long as the ball fall speed is subsonic) and the air motion can be Fourier-decomposed into sound waves of appropriate amplitude.

The total amount of energy released in this form is the amount of kinetic energy missing from the ball, as it plows its way through a viscous medium. This is generally given by Stokes' law (for a spherical ball). So, assuming the ball falls from rest at height $h$, and reaches the floor with speed $v < \sqrt{2gh}$ determined by Stokes' law, the total energy $U$ released into the form of air motion, and thus of sound (never mind whether perceptible to the human ear or not), will be

$$ U = M ( gh - \frac{v^2}{2} )\;\;\;\;\;\;\; (1) $$

where is $M$ is the ball's mass.

For this to be a good approximation to the real world, it is necessary that the ball and floor are rigid, i.e. that time scale for displacement of the surfaces of the two be much shorter than all timescales under consideration.

The ball is deformed on a timescale $R/c_s$, where $R$ is the ball size and $c_s$ is the sound speed inside the ball. Obviously, for this timescale to be shorted than a typical frequency for sound generation (over the same length), it is necessary that the sound speed inside the ball be much larger than in air. For normal solids, this is surely true.

Under a slight displacement in the surface position, the floor will execute free oscillations, determined by a typical lenghtscale of the geometrical shape of the displacement divided by the internal sound speed. If we can take the first one to be of order $< R$, the ball radius (a good approximation if the floor has density not inferior to the ball), then the condition of floor rigidity is once again equivalent to having a sound speed in the floow much larger than in the air.

Thus, for eq. 1 above to be a good approximation, we need: $$ c_{ball,floor} \gg c_{air} $$ and $$ \rho_{floor} \gtrsim \rho_{ball} $$

Best I could think of, on the spur of the moment.

  • $\begingroup$ +1 Not really what the OP was thinking, but some cool lateral thinking about the problem! $\endgroup$ Commented Sep 13, 2014 at 10:34

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