The Fundamental Frequencies of Open Pipes

[leena19]

Homework Statement

could someone please explain :-
1. why the fundamental frequencies of 2 open pipes of the same diameter and of lengths l and 2l are not exactly in the ratio 2:1?
2. given that 2 open pipes of the same diameter but of lengths 50cm and 100cm are found to produce fundamental frequencies in the ratio 1:95,what can be concluded from these figures? [ans:endcorrection 1.31cm]

Homework Equations

1. v=f\lambda

The Attempt at a Solution

\lambda/2 for the 1st open pipe= l+2e ,therefore f1=v/2(l+2e)
\lambda/2 for pipe2 = 2(l+2e),and f2=v/2(2l+2e)

f1 divided by f2 gives 2:1. since the frequency here depends on v,length and the endcorrection only,shouldn't the ratio be 2:1?
seems like I'm missing something but i can't figure out what & i haven't got a clue as to how to workout part 2.
any help would be much appreciated

 

[earlofwessex]

I don't know but i'd say that the end correction isn't doubled when the tube length is doubled, since it changes as a function of the radius not the length.

lambda by 2 for pipe 2= 2(l) + 2e

 

[Chewy0087]

Hmmm, An open tube resonates at the same frequency as a closed tube of half its length for open tubes, the equation you're using is for closed tubes, try it out again halving the length, you've practically got it though.

 

[leena19]

I don't know but i'd say that the end correction isn't doubled when the tube length is doubled, since it changes as a function of the radius not the length.

lambda by 2 for pipe 2= 2(l) + 2e

Yes.Thanx for pointing out my mistake.So after correcting it I get,
2l+2e/l+2e

Chewy0087 Re: statinary waves in open pipes

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Hmmm, An open tube resonates at the same frequency as a closed tube of half its length for open tubes, the equation you're using is for closed tubes, try it out again halving the length, you've practically got it though.
May26-09 05:41 PM

I think I've used the right equation

As For part 2)I get an endcorrection of -0.503m which is way off the answer given ,1.31cm.Can someone please check my working?
\lambda1/2=0.5+2e therefore \lambda1=1+4e
\lambda2/2=1+2e therefore \lambda1=2+4e

v/f1=1+4e
v/f2=2=4e and according to the given data f2=95f1,so after substituting and dividing one by the other I get
95(2+4e) = 1+4e
190+380e = 1+4e
e = -0.503m
My other problem is, can the endcorrection of a pipe have a negative value?