# Pressure variation during resonance?

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1. Sep 3, 2017

### Hamish Winter

Consider a hypothetical pipe of length L and diameter d with both end closed. The pipe is filled with a some fluid. Let there be a ultrasound source of frequency f=v/2L at one end of the pipe and a sensor to measure pressure at the other end of the pipe, where v is the speed of ultrasound wave. Clearly the energy will be trapped in the cylinder and resonance phenomenon will be observed. The question is what pressure measurement will be observed at the sensor? Specifically we know the pressure at the ultrasound source at the other end. What is the relation between pressure at the source and sensor?

More importantly, I want to know if a "portion" of the fluid in the if the pipe moves across the diameter (not in the direction of length) , this means fluid is moving perpendicular to the wave propagation: in this case how the pressure at the sensor would differ compare to earlier case where there the fluid is stationary?

Any hint how to think of this problem would be highly appreciated.

2. Sep 8, 2017

### tech99

It seems equivalent to an electrical transmission line with both ends open circuit. For resonance to occur, the line must be a multiple of half a wavelength long. The pressure at the sensor will then equal that at the source.

3. Sep 8, 2017

### Hamish Winter

The question is fluid moves perpendicular to the wave is there going to be delay at the sensor?

4. Sep 8, 2017

### tech99

I presume the pipe is much smaller than the wavelength. It appears to me that if the source is smaller than the pipe, there will be sideways waves. But these will be reactive (evanescent) in nature, as no net energy is travelling, so may de-tune the source. This will create a phase shift which might appear as a delay.

5. Sep 8, 2017

### olivermsun

If you put a point source at one end of the pipe, say, in the center, then I imagine you are going to get mode formation in the pipe where the waves have to interfere constructively considering both the side reflections and the reflection off the far end (and at the source end when the waves return). Unless you are considering the leakiness of the pipe or some kind of loss in the fluid, then as tech99 points out there is no net energy transfer in any direction.

Also, there will certainly be some phase delay from one end of the pipe to the other due to the length and mode speed (which will be different from the free sound speed), but I can't think off the top of my head why the frequency would be de-tuned.

6. Sep 9, 2017

### tech99

If the pipe is much smaller diameter than the wavelength then mode formation cannot occur. The pipe needs to be at least half a wavelength in diameter for this to occur.
The side reflections will present a reactance to the source. In mechanical terms, they will add mass or compliance. If the source is a resonant type, this reactance might cause de-tuning.

7. Sep 9, 2017

### olivermsun

You are right. I read your post exactly backwards, as the wavelength being much smaller than the pipe diameter! You were talking about the pipe being very skinny and a resonant source being detuned. Got it!

Last edited: Sep 9, 2017
8. Sep 9, 2017

### sophiecentaur

I don't think there is enough specified about the (hypothetical?) experiment. It sounds to me like a resonant cavity and the Q factor will depend on the match at each end (ignoring any loss through the walls of the pipe). For a high impedance source and detector, the Q could be a matter of ten(s), which would mean that the energy in the standing wave would be ten times the input energy per cycle.

9. Sep 9, 2017

### sophiecentaur

I thought that the heart rate monitors used doppler. Can that be improved on? It's a pretty rugged system and can be used under 'inconvenient' conditions.

10. Sep 9, 2017

### sophiecentaur

I was thinking about foetal heart rate monitors which are not large, although bigger than a watch. Otoh, one on the wrist would only need to penetrate ten mm or so. The foetal monitor gives the mother's rate and volume flow too. And for £10 on eBay you can get a clip-on finger monitor as a really basic facility.
I haven't understood where the resonance would come in. What would resonate? If you were to use ultrasound transducers I don't think they are easily tuned. If you are looking for a cavity then would the boundary of the bone / soft tissue be well enough defined to give a 'resonant cavity'. What sort of power levels would you envisage? That could be problematic with the sort of battery you get in a watch.
But on finger! You can get a phone app,even, that uses the change of colour as the blood pulses but pulses are read in hospital these days with a monitor with a probe that fits on the patient's finger. That uses a small light source, I believe.
You have obviously done some reading round. Is the FitBit based on passive detection?

11. Sep 9, 2017

### Hamish Winter

Thank you for your response. It is not about getting the heart rate, it is about from where and how. Imagine wearing a foetal heart monitor and do your usual day to day work? Imagine wearing that infra red monitor in your finger and work.. It is simple totally inconvenient. Existing wrist monitors are not good, it is a big fake, use those in your tennis racket it will show pulse rate...

12. Sep 9, 2017

### Hamish Winter

By the way in my opinion resonance is when energy is trapped within a region. So one can try to create a resonance on a bounded structure, only need to chose frequency of excitation appropriately.