# Statinary waves in open pipes

1. May 26, 2009

### leena19

1. The problem statement, all variables and given/known data

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]

2. Relevant equations
1. v=f$$\lambda$$

3. 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

2. May 26, 2009

### 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

3. May 26, 2009

### 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.

4. Jun 13, 2009

### leena19

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

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?