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Statinary waves in open pipes

  • Thread starter leena19
  • Start date
  • #1
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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[tex]\lambda[/tex]


The Attempt at a Solution


[tex]\lambda[/tex]/2 for the 1st open pipe= l+2e ,therefore f1=v/2(l+2e)
[tex]\lambda[/tex]/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
 

Answers and Replies

  • #2
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
 
  • #3
370
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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.
 
  • #4
186
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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[tex]/l+2e[/tex]

Chewy0087 Re: statinary waves in open pipes

--------------------------------------------------------------------------------
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?
[tex]\lambda1[/tex]/2=0.5+2e therefore [tex]\lambda1[/tex]=1+4e
[tex]\lambda2[/tex]/2=1+2e therefore [tex]\lambda1[/tex]=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?
 

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