Solving Open Pipe Problem with Tuning Fork Frequency

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SUMMARY

The discussion centers on calculating the frequency of a tuning fork based on its resonance with a vertical open tube filled with water. The distances at which resonance occurs are 0.125 m and 0.395 m, leading to a difference of 0.270 m, which represents half a wavelength (λ/2). Using the speed of sound in air at 343 m/s, the frequency can be determined by the formula: frequency = speed of sound / wavelength. The assumption that the resonances are successive is confirmed, allowing for accurate frequency calculation.

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  • Calculate the frequency of the tuning fork using the formula: frequency = speed of sound / wavelength.
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Koyuki
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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 [itex]\lambda/2[/itex]. I assume one would then use the speed of sound in air, 343 m/s, to calculate the frequency.
 
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