Frequency of sound in an open-open tube

[NP04]

Homework Statement

A hollow tube of length L open at both ends as shown, is held in midair. A tuning fork with a frequency f initial vibrates at one end of the tube and causes the air in the tube to vibrate at its fundamental frequency. Express your answers in terms of L and fo.

C. Determine the next higher frequency at which this air column would resonate.

D. Determine the height h of the tube above the water when the air column resonates for the first time. Express your answer in terms of L.

Relevant Equations

Harmonic numbers for open-open and open-closed tubes

v = lambda x f initial

I do not understand parts c and d.

For part C I have no idea where to start. And for part d, I got 1/4λ = L (open-close tube fundamental harmonics) so λ = 4L. But the answer key says it is 1/2 L. Help!

 

[haruspex]

For part C I have no idea where to start.

What is the formula for the nth harmonic of the tube? Which harmonic are you given, and which are you asked for?

for part d, I got 1/4λ = L

In that standard formula, what is L? Is it the same as the L in the question?

 

[NP04]

What is the formula for the nth harmonic of the tube?

Well...I am not sure what formula you are referencing, but would it be just adding 1/2λ to the fundamental 1/2λ(which is given)?
Then L = λ, so v = λf, which does not make sense. The harmonic asked for is the 2nd one. But since the speed in the tube is already 2Lf I am confused.

In that standard formula, what is L? Is it the same as the L in the question?

In part D, the length of the new tube would be L + h = L + 4L = 5L, I don't understand what this is used for.

 

[haruspex]

I am not sure what formula you are referencing

In an open-open tube, the fundamental has half a wavelength in it, so 2λ₁=L.
How many wavelengths, λ₂, are in the tube at the second harmonic.

the length of the new tube would be L + h

No, the tube is partly submerged. The question is ambiguous, but it must mean the height of the top of the tube above the water. Its effective length is less than L.

 

[RPinPA]

The symbol ##L## is already taken. It's the length of the whole glass tube, which hasn't changed. You are being told not to change the meaning of ##L##.

When the tube is partially submerged, the resonating chamber is something less than L. You were told to call it ##h##.

So what is the relationship between ##\lambda## and ##h##?