The observer is outside the water; the stones are in water (say, 1 m below the surface). This produces a higher-pitched sound for the observer than if both the observer and the stones are in air.
Is this because it takes more energy for the sound waves to travel through water than through air, so that the ones that we hear from outside are the ones that had higher frequencies after the collision to begin with?
Does the density of the medium that is disturbed by a rigid body collision have any effect on the frequency distribution of the sound waves that are generated? For example, does the higher "stiffness" of the cage of water molecules surrounding the stones that are vibrating mean higher frequency normal-modes?
Finally, does the refraction at the water-air interface play any role?
Best Answer
The frequency of a sound wave cannot change as it crosses the water-air boundary. The wavelength can, and does, change but the frequency cannot because if it did there would be no way to match the two waves at the interface. This means that the higher frequency is not some quirk of the sound propagation, but that the colliding stones emit a higher frequency when in water.
To see why this is we need to consider how the sound is generated. When the two stones collide this generates a shock wave that propagates into the interior of the stones and makes them vibrate. At the surface of the stones the vibrations propagate into the surrounding medium as a sound wave, and that's ultimately what we hear. A stone will have some set of normal modes, and the shock wave will transfer energy into these normal modes exciting them in some probably rather random way that will depend on the details of the impact. The sound we hear is the combination of the frequencies and amplitudes of all these normal modes.
The reason that the sound of the impact is different in water is simply that the normal modes of an object in a medium like water are different to the normal modes of the same object in air. This is because water has a (much!) high bulk modulus than air and the surrounding water resists being moved by the surface of the vibrating stone far more than air does. It would be a brave physicist who would predict exactly how the sound would change because this will be complicated. Vibrational modes that cause lateral displacement of the water will tend to be slowed because the water has a higher density than air. On the other hand vibrational modes that cause compression of the water will shift to a higher frequency because water has a higher bulk modulus than air.
It might be possible to calculate exactly how the bulk modulus and density of the medium affect the normal modes for some idealised object like a perfectly elastic sphere. I've Googled for such calculations but with no luck - if anyone finds a relevant link please feel free to edit it in or add it as a comment. For now all i can say is that experiment shows the normal modes that shift to higher frequency dominate the sound we hear.