Solving Resonance Problem: Explain Frequency of Tuning Fork

[WhackyWookie]

A tuning fork is set into vibration above a vertica oepn tube filled with water. The water level is allowed to drop slowly. As it does, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is 0.125 m and again at 0.395 m. What is the frequency of the tuning work?

The answer given on the back of the book is 635 hz. Not only do I have no idea how the book got this answer, but this answer is also is less than the fundamental frequency of a 0.125 m closed tube, which is 686 hz.

Could someone provide an explanation?

 

[Chen]

This is the 7th problem here:
http://faculty.salisbury.edu/~jwhoward/Physics123/html/ch12.htm
(Just search for 'tuning' if you can't find it.)

Good luck..

 

[Physics is Phun]

Well you can use the formula L_n = {(2n-1)/4}*wavelength where L_n is length of the air column and n is the which antinode you are measuring from probably the first and second. (I believe the term is resonance length but I am not sure) You then solve for the wavelength and find the frequence using the universal wave equation V=frequence*wavelength

 

[WhackyWookie]

Chen, I went to the link, but their solution makes no sense. They assumed that the 0.125m and 0.395m tubes' resonant frequency differs by 1 harmonics and used nothing to back that up.

Additionally, as I have said before, 635 hz is below the fundamental frequency of a 0.125 m half open tube.

Physics_is_phun, I used ur equation originally and got the answer that is wrong according to the book.