Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The observer is outside the water; the stones are underwater (say, 1 m below surface, if that matters). This produces a higher pitched sound for the observer than when both the observer and the stones are in air.

Is this because it takes more energy for sound waves to travel through water than through air, and so the ones that we hear from outside the water are the ones that had higher frequencies immediately after the collision to begin with?

Does the density of the medium which is disturbed by a rigid body collision have any effect on the frequency distribution of sound waves that are generated by the collision? For example, does the higher "stiffness" of the cage of water molecules surrounding the vibrating stones post-collision mean higher frequency normal modes for water (than for air)?

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

share|improve this question
    
Did you consider the speed of sound in air and water? –  Bernhard Mar 24 '13 at 15:19
    
No, but how does it help? It explains why frequency should be higher in water than in air but the observer is in air in both cases. –  Tolga Yilmaz Mar 24 '13 at 16:01
    
Interesting question, but also difficult (as you almost note yourself), since the sound likely covers a huge range of frequencies, so it's not so easy to properly define everything. –  Vibert Mar 24 '13 at 17:15
add comment

1 Answer 1

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.

share|improve this answer
    
It's not clear to me what you are saying. If lower frequencies have longer penetration, why do you think this answers the question? –  Tolga Yilmaz Mar 24 '13 at 20:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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