Open pipe end correction distance for waves

[Bolter]

Homework Statement

See below

Relevant Equations

frequency = speed/wavelength

Struggling on how to do this Q

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

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]

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]

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

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]

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

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

 

[Bolter]

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?

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

 

[Merlin3189]

So there you are.

 

[Bolter]

So there you are.

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

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

 

[Bolter]

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

 

[Merlin3189]

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.