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## Homework Statement

A tuning fork is set into vibration above a vertical open tube filled with water, as seen in the figure below.

http://img207.imageshack.us/img207/5817/dgian1686mm6.gif [Broken]

The water level is allowed to drop slowly. As it does so, 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 L1 = 0.100 m and L2 = 0.450 m. What is the frequency of the tuning fork? Assume that the speed of sound in air is 343 m/s.

## Homework Equations

f1 = [tex]\frac{\upsilon}{4L}[/tex]

where v = 343 m/s

## The Attempt at a Solution

I know that when the distance between the tuning fork and surface of the water is 0.1m. This means it is the first harmonic. Since it it a closed tube, i use the formula for the first harmonic of a closed tube, but my frequency comes out to 858 Hz which is incorrect.

I'm sure that i have to use the second length they give to us, but i'm not sure what its for..

cheers

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