Resonant Frequency of Open Ended Pipes with a Barrier Imposed

[Twenty_Three]

I was just wondering if anyone here had any experience with an open pipe system with a barrier imposed.

Basically, say for a pipe open at both ends, at one end we place a barrier a distance away from the pipe. How would this affect the resonant frequency of the pipe?

From experimentation, results I've gotten show that as the barrier distance decreased, the resonant frequency decreases as well. Likewise as the barrier distance increases, the resonant frequency increases until the point where the barrier has no effect where resonant frequency of pipe levels out.

Does anyone have any idea about the theory behind this? The air particle movement, radiation impedance, free fields, some key terms that I could research (seeing as I'm stuck at the moment on how to explain this phenomena).

Thanks in advance.

 

[jambaugh]

Think of the barrier as a mirror to sound waves. A barrier some distance from the end will behave like having two tubes end to end with a space twice that distance. Far apart they can be in sync without interfering. As they grow closer the waves exiting one's end interferes with the standing wave of the other. This will be a function of distance and freq. The coupling will be proportional to square of the distance and phase shift is distance times freq over speed of sound. (plus a half cycle phase shift due to reflection)

In the boundary case where the two tubes meet you have an open tube of double length at a harmonic where there is a node at mid point or equivalently your original tube becomes closed on that end.

This looks like a great physics problem. What is your interest?

 

[Twenty_Three]

Think of the barrier as a mirror to sound waves. A barrier some distance from the end will behave like having two tubes end to end with a space twice that distance. Far apart they can be in sync without interfering. As they grow closer the waves exiting one's end interferes with the standing wave of the other. This will be a function of distance and freq. The coupling will be proportional to square of the distance and phase shift is distance times freq over speed of sound. (plus a half cycle phase shift due to reflection)

In the boundary case where the two tubes meet you have an open tube of double length at a harmonic where there is a node at mid point or equivalently your original tube becomes closed on that end.

This looks like a great physics problem. What is your interest?

Hello Jambaugh,

Thanks for your reply, I'm actually doing this investigation as part of a high school assignment, although I may've bitten off more than I can chew.

The coupling will be proportional to square of the distance and phase shift is distance times freq over speed of sound. (plus a half cycle phase shift due to reflection)

^Would that be referring to N = (D^2f)/4v? What do you mean by the coupling, I'm still slightly puzzled with this. The results I've obtained was that for an open pipes of around 60cmish (excluding end correction), the resonant frequency with the barrier at a distance of 1cm varies only by 2Hz to that of 8cm (where the resonant frequency levels out).

In the boundary case where the two tubes meet you have an open tube of double length at a harmonic where there is a node at mid point or equivalently your original tube becomes closed on that end.

^That's what I was thinking at first, on our spectrum we were getting a really tiny peak at half the loudest peak, which we believe is actually an undertone as opposed to the fundamental. That's why I'm not quite sure as to whether it equates to the tube being closed at one end.

Thanks again for your response :)