Physics-Tubes equations involving lambda, modes and harmonics

[slow loris]

Let's suppose that I have certain modes previously determined for a tube that is open/closed, how do you find the equivalent modes but for a tube that is open/open?

 

 

EDIT: I was thinking that I could use the formula of L=n x lambda/2, then isolate n and find the answer like that; however the only issue would be that I don't know the L of the new tubes

Edited March 30, 2017 by slow loris

[studiot]

I am guessing that you are referring to the resonant frequencies of pipes due to standing waves in the pipe.

 

A closed end must have a node of the standing wave

 

An open end must have an antinode.

 

So for your question the fundamental occurs when you have one node in the middle and two antinodes, one at each end.

 

The distance between adjacent antinodes (or nodes) is one wavelength, so the length of the pipe corresponds to one wavelength of the fundamental.

 

Obviously the first harmonic you can have has two nodes and so on.

 

Can you work out the relationship between the length of the pipe and the wavelength of the harmonics now?