# The reflection of resonance tube

1. Mar 5, 2013

### hds123523000

any one knows why sound wave reflects at the opening end of the resonance tube? is the air inside the tube different from those outside. from the expriment, when the incident wave interference with the reflected one, the destructive point reachs almost zero showing that the reflected one has comparable amplitude as the incident one? in another word, the reflection rate is 1? if i keep sending wave into the tube, will it be explosive?

2. Mar 5, 2013

### olivermsun

You can think of the ends of a pipe as imposing certain boundary conditions for the acoustic (pressure) wave inside the pipe. At a closed end, the velocity has to be zero but the pressure can reach its maximum. At an open end, the velocity can reach its maximum but the pressure has to be zero (or ambient, anyway). In either case, the way to fulfill the boundary condition is to have a wave traveling in the opposite direction — it just has opposite velocity or pressure phase, depending on which condition you have.

In reality there is a small leakage from the end of the pipe (which is how you hear the sound), so I imagine the excess pressure would leak out before the pipe exploded.

3. Mar 6, 2013

### hds123523000

if i look at it as a boundary condition, i wouldnt expect it is a condition that is a smooth function instead of sudden function like in the case of a closed end. is it right?

In terms of leaking, there is a interesting point i want to make. if it leaks, lets say the reflection rate is R (<=1), then for the constructive interference, the peak is (1+R), for destructive interfernce, the node show be (1-R). that means the amplitute ratio at the antinode and node should be (1+R)/(1-R) which can be seen at least 10 in experiment. so that R is about 90 percent. the reaonance tube is still gaining energy right?

4. Mar 6, 2013

### olivermsun

I'm not sure exactly what you're asking. One question that occurs to me, though, is this:

How do you plan to keep sending the waves into the tube?

5. Mar 6, 2013

### hds123523000

sorry for confusing. i just want to get the idea why the relection happens at the opening end and find that the current explaination seems not very convincing and self-explainary.

6. Mar 7, 2013

### olivermsun

I suppose you could try and convince yourself by writing out the equations governing an acoustic wave in a 1-d pipe and experimenting with the boundary conditions.

7. Mar 7, 2013

### marcusl

I didn't follow your comments about leakage or gaining energy, but the ratio that you wrote is valid. The nodes and antinodes (actually the nodes are partial because the wave amplitudes there are not zero) characterize the standing wave that is set up by the outbound and reflected waves. In electrical transmission line theory, the ratio has the official name of Voltage Standing Wave Ratio (VSWR). In terms of the reflection coefficient rho at the impedance mismatch, it is $$VSWR=\frac{1+\rho}{1-\rho}$$ just as you wrote.

8. Mar 7, 2013

### hds123523000

my theory is that, if the reflection is more than 0, or in another words, if at the boundary reflection happens, it mean the energy donot go out of the tube completely, then the tube gain/accumulate some energy. what we are supposed to see is that the peak become higher and higher instead of a stable standing wave.
by the way, if the reflection is close to 0.9, shouldnot we consider muliple reflection at each end when we were trying to calculate how long is the wavelength inside the tube?

9. Mar 7, 2013

### hds123523000

So, in your opinion, physically, whats the difference between the air inside the tube and outside?

10. Mar 7, 2013

### olivermsun

You do take into account multiple reflections. The "resonant modes" correspond to the wavelengths which have constructive interference and continue to exist in the tube.

Energy will accumulate in the tube if you keep on adding energy, but if you assume the tube starts with some amount of energy, it doesn't just keep producing energy out of nowhere because of the reflection.

The difference between the air in the tube and the air outside is that the air in the tube is confined to a narrow (finite-width) passage. If there's 1 unit of air in the tube at ambient pressure, then moving another 1 unit of air into the pipe is going to double the pressure. Add 1 unit to the air outside, and the pressure isn't going to change very much.

11. Mar 7, 2013

### marcusl

If you turn on a driver or a source, then the amplitude will build up as you surmise--until it reaches a steady state level where the power entering the resonator equals the sum of power exiting the end(s) plus that dissipated. You should see a sum of traveling and standing waves (the mixture depends on the size of the reflection coefficient).

12. Mar 7, 2013

### sophiecentaur

The difference is where the air happens to be. The fact that it is constrained by the sides of the tube alters the speed of sound through it. At the open end there is a change of wave speed, which must involve some reflection.
The reflected sound will 'stay in the tube' and build up in level if the excitation frequency is right - resonance. The maximum level of resonance is governed by how much of the acoustic power is lost through the end.

13. Mar 7, 2013

### hds123523000

In terms of the energy dissipating, another interesting point will come out. if you have done this experiment, you will found that the ratio of the amplittude at the antinode site and the node side can easily reach 10 as i suggested in my previous post. that means the reflection can reach 0.9. in another word, 90 percent of the energy dissipate from other way apart from transmission through the opening end. the only way here i can think about is the tube wall. if this is the case, then the sound wave losea energy along going through the tube. for example, when the speaker is put at the left side of the tube, then the incident sound wave has higher amplitude at the left region of the tube than the right region. in the same way, reflected wave has higher in the right region than the left. is that the case?

14. Mar 7, 2013

### hds123523000

in you opinion, when we try to work out the resonance frequency, we should use the sound speed in the tube rather than outside the tube, right?

15. Mar 7, 2013

### AlephZero

Do you have a reference for the statement that the speed of sound changes? (which is a polite way of saying "I think that's wrong" ).

What changes is the acoustic impedance of the tube, not the sound velocity. You will get partial reflection and partial transmission of energy at any change of diameter of the tube. The change to "zero diameter" at the closed end, or "infinite diameter" at the open end are just extreme limiting cases.

http://www.animations.physics.unsw.edu.au/jw/sound-impedance-intensity.htm

16. Mar 7, 2013

### marcusl

The reflection coefficient that you calculated is an amplitude value, so the fraction that is reflected power for the case you mention is 0.81. Thus 19% of the power is lost from the end of the pipe.
No. At equilibrium the speaker puts the same power into the pipe as is lost, and virtually all of the loss is through the open end.

17. Mar 7, 2013

### hds123523000

i think the sound wave is mechanic wave, like water wave, varying slow enough for microphone intead of light whose intensity is time averaged. so the signal that detected by the microphone is the amplitude directly.

even though 81 percent isnt negligiable, is it? so we are back to the begining of the circle, when the standing wave reaches the stable/equilibrium state, how can the energy gets out completely and with 81 percent reflected at the same time?

18. Mar 7, 2013

### marcusl

What is your point? Only 19% of the power is lost, and that much must be replaced by the source if the amplitude is to be maintained.

Last edited: Mar 7, 2013
19. Mar 7, 2013

### hds123523000

i mean, 81 percent is kept in the tube at the wavefront. the tube is accumulating energy even at its stable state and the peak should get higher and higher.

20. Mar 7, 2013

### sophiecentaur

A fair cop guv. Thanks for spotting it.
I was in two minds whether to refer to Impedance or Speed variation. I should have used impedance because that is true for narrow tubes. The effect on speed is only relevant when the tube is wide and the multiple path lengths start to be relevant to the wave speed. That would be more like 'cavities' than tubes.

21. Mar 7, 2013

### sophiecentaur

A maximum level is reached when the proportion of what gets out from the end and is dissipated in the source is equal to the power input from the source. The impedance of the source is very relevant here because the "81%" can either be all reflected or all dissipated or anything in between, depending upon the match.
In the quoted case with a totally mismatched source, the maximum will be very nearly reached after about five transits of the wave. The Q factor of the resonator would be about 5.

22. Mar 7, 2013

### AlephZero

Actually I have seen it in print once, in a book written by a the head of a very well known pipe-organ building company, in about 1900. His explanation of the difference between the actual length and effective acoustic length of a pipe was that sound travels faster in pipes with smaller diameter.
But that doesn't make sense, because a careful measurement shows that the speed in a very narrow pipe was the same as the speed in free air, but the speed reduces as the pipe diameter increased. I think Spock would call that "illogical"

23. Mar 7, 2013

### AlephZero

24. Mar 8, 2013

### sophiecentaur

Not if you acknowledge the longer possible path length due to reflections off the side.
Or do you mean that you think it's illogical that a manufacturer could produce something wonderful without knowing how it works?

25. Mar 8, 2013

### olivermsun

That is interesting. Would you provide references for the book and also the measurement?