Frequency of sound in an open-open tube

AI Thread Summary
The discussion revolves around confusion regarding the calculations for parts C and D of a problem involving sound frequency in an open-open tube. Participants are unclear about the correct formulas for harmonics, particularly the fundamental frequency and its relationship to the length of the tube. There is a specific debate about the effective length of the tube when partially submerged, with emphasis on distinguishing between the total length (L) and the submerged portion (h). Clarification is sought on how to determine the wavelengths and harmonics in this context. Understanding the relationship between wavelength and the effective length of the tube is crucial for solving the problem accurately.
NP04
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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
Screen Shot 2019-05-13 at 5.34.09 PM.png

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!
 
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NP04 said:
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?
NP04 said:
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?
 
haruspex said:
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.

haruspex said:
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.
 
NP04 said:
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.
NP04 said:
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.
 
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##?
 
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