Determining if a pipe is open or closed from given the resonant frequencies

[AAAA]

Homework Statement

A pipe resonates at successive frequencies of 540 Hz, 450 Hz, and 350Hz. Is this an open or a closed pipe?

Homework Equations

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The Attempt at a Solution

My assumption is that because the harmonics of an open pipe are odd number multiples of the fundamental frequency (1f, 3f, 5f), and ∵ the difference between the frequencies is the fundamental frequency (I think?). The fundamental frequency is ∴ around 100 (?). And ∵ the frequencies go up by ~100 each time, it's an open pipe, as it's harmonics are even number multiples (1f, 2f, 4f, 6f).

I don't think my assumptions are correct. Any help is appreciated, thanks.

EDIT: Is it possible to determine the fundamental frequency from the given frequencies? I'm beginning to think that the fundamental frequency is the difference between those frequencies, divided by two. Because to get from on harmonic to the next, we multiply by two to get from one freq to the next, with the exception of the even number multiples(1f→2f).

 

[AAAA]

Is it even possible to determine if it's open or closed based on the given information? I imagine it would be easier if they told you which harmonics those are, so the fundamental frequency could be determined, and from that information determine whether pipe is closed or open.

 

[haruspex]

the harmonics of an open pipe are odd number multiples of the fundamental frequency (1f, 3f, 5f)

Yes, because the number of wavelengths goes 1/4, 3/4, 5/4...

an open pipe, as it's harmonics are even number multiples (1f, 2f, 4f, 6f).

I assume you meant to write 'closed pipe' here. Anyway, that list is wrong. Consider the numbers of wavelengths.

the fundamental frequency is the difference between those frequencies, divided by two

Right, so what do you get for the fundamental frequency? What multiples of that are the given frequencies?

Is it even possible to determine if it's open or closed based on the given information?

Yes, because you are told these are successive harmonics.

 

[AAAA]

Well, 450-350= 100
100/2 = 50

50*7 = 350
50*9 = 450
50*11 = 550 ≈ 540

∵ I used odd number multiples to get that value, it must be a open pipe with a closed end (a fixed and free end system), it can't be an open pipe (with two free ends) because then the successive frequencies would be 50 apart, no?

And a pipe with two free ends is 1f, 2f, 3f, 4f... Right? Which is the same as a pipe with two fixed ends?

 

[haruspex]

Well, 450-350= 100
100/2 = 50

50*7 = 350
50*9 = 450
50*11 = 550 ≈ 540

∵ I used odd number multiples to get that value, it must be a open pipe with a closed end (a fixed and free end system), it can't be an open pipe (with two free ends) because then the successive frequencies would be 50 apart, no?

And a pipe with two free ends is 1f, 2f, 3f, 4f... Right? Which is the same as a pipe with two fixed ends?

That all looks right.

 

[AAAA]

Thanks! I had neglected to do the math for the frequencies and just looked it up online, I guess I interpreted it wrong. I just did the math for the frequencies, and now it all make sense, thanks again!

 

[StillAnotherDave]

I'm wondering that if you don't yet know if the pipe is open-closed or open-open why you would divide the difference between the two successive frequencies by two? With this example, you need frequency values of 350Hz, 450Hz etc so it makes sense. Would it not be possible to have successive frequencies where m = 1, 2, 3 and m = 1, 3, 5 would work? i.e. you wouldn't be able to conclude whether the pipe was open-open or open-closed?

 

[haruspex]

I'm wondering that if you don't yet know if the pipe is open-closed or open-open why you would divide the difference between the two successive frequencies by two? With this example, you need frequency values of 350Hz, 450Hz etc so it makes sense. Would it not be possible to have successive frequencies where m = 1, 2, 3 and m = 1, 3, 5 would work? i.e. you wouldn't be able to conclude whether the pipe was open-open or open-closed?

Not sure I understand your question.
If you know they are successive frequencies then the differences are $\Delta f=\frac c{2L}$.
For a half open pipe, $\frac{f_n}{\Delta f}=n+\frac 12$, so given three successive frequencies not all will be integers. For a doubly open or doubly closed pipe, all three will be integers.

 

[mathloverr]

Not sure I understand your question.
If you know they are successive frequencies then the differences are $\Delta f=\frac c{2L}$.
For a half open pipe, $\frac{f_n}{\Delta f}=n+\frac 12$, so given three successive frequencies not all will be integers. For a doubly open or doubly closed pipe, all three will be integers.

why did you divide 100 by 2? What does that represent?

 

[haruspex]

why did you divide 100 by 2? What does that represent?

The pipe has length L. Being half open, at each resonance it contains some number of half wavelengths plus one quarter wavelength.
$\lambda_n(\frac n2+\frac 14)=L$
$\nu_n=\frac c{\lambda_n}=\frac cL(\frac n2+\frac 14)$
In particular, the fundamental is $\nu_0=\frac cL(\frac 14)$
$\nu_{n+1}-\nu_n=\frac c{2L}=2\nu_0$