Solving Open Pipe Problem with Tuning Fork Frequency

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The problem involves a tuning fork vibrating above a vertical open tube filled with water, with resonance occurring at two water levels: 0.125 m and 0.395 m. The difference in these lengths is calculated to be 0.270 m, which is interpreted as half the wavelength of the sound wave due to the node at the water's surface. To find the frequency of the tuning fork, the speed of sound in air (343 m/s) is used in conjunction with the wavelength. The relationship between speed, wavelength, and frequency is crucial for solving the problem, leading to the conclusion that the resonances are indeed successive. The discussion emphasizes understanding the physics behind sound resonance in open tubes.
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Hello. I'm having some trouble with this problem:

A tuning fork is set into vibration above a vertical open tube filled w/ water. Water level is dropping slowly. As it does so, the air in the tube above the water level is heard to resonate w/ the tuning fork when the distance from the tube opening to the water level is 0.125 m and again at 0.395m. What is the frequency of the tuning fork?

So I subtracted the two distances and got
(n+2)L - nL = .395 - .125 = .270 m

Are 0.125m and 0.395 successive though ?
Now I'm stuck.

Any help would be appreciated.
 
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From the statement, it would appear that the resonances are successive.

What is the relationship between the speed of sound, wavelength and frequency. If one knows the wavelength, one can find the frequency.
 
Since there is a node at the surface of the water the difference between the two lenghts is equal to \lambda/2. I assume one would then use the speed of sound in air, 343 m/s, to calculate the frequency.
 
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