Open pipe end correction distance for waves

Bolter · Mar 12, 2020

Printed problem text about an open pipe with a piston, a 342 Hz tuning fork, resonance lengths of 240.3 mm and 744.2 mm, asking for the end correction in mm.

Struggling on how to do this Q

I have tried drawing a sketch of it and had come up with this

Handwritten sketch of a standing wave in an open pipe with end corrections e at both ends, length L marked, and derivation lines: lambda = 2(L+2e), f0 = c/lambda, giving e = (1/4)(c/f0 - 2L).

This is probably not right, as the speed of sound is not given here as well the length of the open pipe tube?

Any help would be really great! Thanks

Merlin3189 · Mar 12, 2020

One end is not open. There is a piston. That will affect both end correction and resonant length.

Ps. Why do they give you two resonances?

Bolter · Mar 12, 2020

Merlin3189 said:

One end is not open. There is a piston. That will affect both end correction and resonant length.

Ps. Why do they give you two resonances?

If it is a close end pipe, then the sketch should look like this then

Handwritten sketch of a closed-end pipe with a node at the left closed end and an anti-node at the right open end, with lengths L and e marked, plus derivation lines for lambda = 4(L+e) and e = (1/4)(c/f0 - 4L).

I'm not so sure why myself of why they give 2 resonance here

What do I make out of the 2 resonant lengths given here?

Merlin3189 · Mar 12, 2020

Well, what will the diagram look like for the longer tube resonance?

What do you know about the difference between the two lengths?

Bolter · Mar 12, 2020

Merlin3189 said:

Well, what will the diagram look like for the longer tube resonance?

What do you know about the difference between the two lengths?

The resonance for the longer tube would look something like this?

Hand-drawn standing wave diagram in a rectangular tube, with two curves showing a node at the left end, antinodes at the top and right end, a node in the middle, and a length arrow labelled 744.2 mm.

And the difference between these 2 lengths is half the wavelength in mm

Merlin3189 · Mar 12, 2020

So there you are.

Bolter · Mar 12, 2020

Merlin3189 said:

So there you are.

Ok thanks I’m sort of getting it but when I do this

Handwritten working on paper: λ/2 = 744.2 - 240.3, then λ/2 = 503.9 mm, so λ = 1007.8 mm or λ = 1.0078 m, then λ = 4(L+e), λ/4 = L+e, and e = λ/4 - L.

What do I choose my pipe length, L to be?

Bolter · Mar 12, 2020

I must let L be 240.3mm to get a end correction, e to be 0.01165m right?

Handwritten working on paper: e = λ/4 − L; then e = 1.0078/4 − (240.3 × 10⁻³); then e = 0.01165 m.

Merlin3189 · Mar 12, 2020

Yes. You could have used the longer one, but that would have been 3λ/4 rather than λ/4

So now you can calculate the speed of sound. Perhaps that's why they didn't mention it?

BTW
If this had been a real experiment, they wouldn't have given the same answer of course, because neither would be exactly right. And you wouldn't know which was better. I'd choose the longer one, because the ±1mm or whatever your measurement error is, would make less % error on the longer measurement.

I seriously can't believe anyone could possibly locate these resonances as precisely as they claim.
What any normal person would do, is to, preferably find more resonances and also take several measurements at each one to get an average and an estimate of the random error.

Open pipe end correction distance for waves

Discussion

Similar threads