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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?

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a link to a video to show the effect, pls? – Helder Velez Oct 16 '15 at 20:49
    
Can you provide any evidence to justify your first sentence (a video, WP, etc...)? (contrary to all others I do not have for granted your sentence as a truth, sorry) – Helder Velez Oct 21 '15 at 1:36
    
@HelderVelez Sorry, no video. This was a casual observation made at a beach with friends two years ago. But maybe you can try doing a small experiment at home. – Tolga Yilmaz Oct 22 '15 at 2:45
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I made an experiment at home with rocks, an audio or fft analyzer on android from google play, and I can not say that your first sentence is correct. I know that the different recipients contributed with resonances, etc. I will try at the beach, eventually. Sorry no video. – Helder Velez Oct 22 '15 at 21:43
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+100

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.

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JR's answer gets to the nub of the issue... The oscillating rock can be reduced to a single mass-spring system. Adding water is like adding another spring in parallel without changing the mass. The combined spring is therefore stiffer and so the resonant frequency goes up. – Owen Boyle Oct 16 '15 at 12:42

Let's examine and answer each of your questions as follows.

  1. 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?

The frequency of the sound generated depends on the sound source and medium, the stones colliding in the air vs the water. Sound from the stones colliding in the water will have a higher frequency then the stones colliding in the air as deduced by observing a higher pitch. The pitch of a sound (how high the note is) depends on the frequency of the wave. The higher the frequency, the higher the pitch. Since the pitch heard was higher it follows that the frequency from the stones colliding in the water has to be greater. And yes since the stones resonant frequency(s) does not change in either medium the higher frequency cage or greater density of the water surrounding the stones must account for the higher frequency normal-mode of the wave generated. {It was logically deduced since the resonant frequency of the stones colliding will be a constant in both mediums. It then follows only one variable remains and it must not be a constant the main characteristic of the medium as it pertains to sound waves, its density and propensity to change shape, water is much more rigid and denser than air.

  1. 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?

It does take more energy to create a sound wave in water than in air due to their different densities, water being more dense and stiffer so to speak. However the energy in this case should not have an impact since the amount of energy to cause the collision of the two stones in either medium would need to be constant for this comparison to make sense. If it were not the sound generated would just be louder due to the greater energy. Additionally adding energy to a sound wave will not effect its velocity, the energy only impacts its amplitude so all waves in water will travel at the same speed even if they have different frequencies and amplitudes. The higher amplitude wave will be louder.

  1. Finally, does the refraction at the water-air interface play any role?

Sound in air

In a gas like air, the particles are generally far apart so they travel further before they bump into one another. There is not much resistance to movement so it doesn’t take much to start a wave, but it won’t travel as fast.

enter image description here Sound in water

In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next. This means that the sound wave travels over four times faster than it would in air, but it takes a lot of energy to start the vibration. A faint sound in air wouldn’t be transmitted in water as the wave wouldn’t have enough energy to force the water particles to move.

Wavelength is given by: λ = v / f

where v is the velocity and f is the frequency. Since the frequency of the wave generated by the stones colliding in the water {at a 1 meter depth} is the same in both media, the wavelength will simply be proportional to the velocity of the wave. When the sound leaves the water the speed and wavelength both change. The sound wave will slow down in the air since the density of air is less than the water and the wave length will be shorter since the the velocity has slowed. But the frequency will not change therefore we must conclude that refraction does not effect the pitch of the sound one hears.

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The higher pitched frequencies have higher energy. The path of minimal time is to the surface.

Lower frequencies naturally travel further in water despite their lower energy value. Hence the use of infrasound by whales for the purpose of long distance communication.

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Energy has nothing to do with this. Assuming the rocks collide at the same speed, the energy is the same inside or outside water. Also the speed of sound is independent of the energy in the wave (or you'd hear loud sounds first). – Owen Boyle Oct 16 '15 at 10:12
    
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Jon Custer Oct 19 '15 at 0:28

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