Wave Reflection: Newsletter Challenge (11/04/08)

To misquote (the shame of it), "any discontinuity will do". Of course, the proportion reflected will depend on the wavelength and the diameter of the tube - the larger wavelength (relative to tube diameter), the smaller a proportion that will escape. As we approach the long-wavelength limit, minimal energy is coupled to the spherically radiating wave, so nearly all the energy has to be reflected. If the tube diameter is comparable with the wavelength, much of the energy will be radiated, and the reflection will be small.

This assumes a compressional acoustic wave. There are of course other types of wave (including light), but the closest equivalent conditions would be described somewhat differently.

Pipes can either be open on both ends or on only one end. The open ends act as free-end reflectors (producing antinodes) and the closed ends act as fixed-end reflectors (producing nodes).

I can remember reading somewhere that a layman's explanation was as the high pressure portion of the wave exits the pipe, it creates a suction effect at the open pipe end which effectively sends a low pressure wave back into the pipe. This is why the reflection from a open end is inverted while the reflection from a closed end is not.

The phenomenon was well thrashed out in the "Playing the Didgeridoo problem"

http://cr4.globalspec.com/blogentry/3313/Playing-the-Didgeridoo-Newsletter-Challenge-09-25-07

GA.

This effect is used in many ways, but one of my favorites is its use in expansion chambers for racing two-stroke engines. In my misspent (relative) youth, I used to custom-build expansion chambers for two-stroke motorcycles. The diverging cone acts as a series of open end pipes (returning a negative wave for each positive pulse) and the converging cone returns a positive wave for each positive pulse. If the chamber is constructed to put this returning pulse in the right place at the right time, the engine's horsepower can be substantially increased. By making the cones steeper the engine could be made peakier (higher hp at the expense of a narrow power band) and vice versa.

Not so. The special effect discussed for the Didgeridoo was a net reduction in pressure at/near the open end of the pipe. It required the reflection to exist in order to set up the appropriate conditions, but was not itself the cause of the reflection. Indeed, the net suction is a non-linear consequence of the high velocity of the acoustic wave resulting from the resonance. It usually causes circulation at the end of the pipe, and this includes both continuous (DC) and second harmonic components.

The resonance we have here still occurs at low amplitude, and is caused by an impedance mismatch. Under linear conditions, the low acoustic impedance of the diverging wave means that any pressure variation at the open end will be quite small compared with the variation at the blocked end; as the pressure inside but near the open end of the pipe is the sum of the pressure in the waves travelling in opposite directions, the pressure in the return wave needs to be inverted compared with the outgoing wave. On the other hand, the velocity components near the end of the pipe will add.

N.B. Automotive scavenging assistance is often a primarily linear resonant phenomenon that uses the low pressure generated at a particular time of the cycle. In principle it could also use the non-linear pumping. This is not my field, but I find it hard to see how to make the non-linear suction into an efficient process - even if you could match the resonant pumping rate to the required rate of gas flow; conversely, I can see (at least in principle) how to make a multi-pole resonator that maintains the required low-pressure timing over a significant range of frequencies.

Not sure what exactly you are referring to when you say "not so".

But anyway, I managed to track down the layman's explanation that I gave in post #2, which I think is an excellent answer to the question as posed here.

Go to the following webpage and scroll down to view the "Air motion animations" section

http://www.phys.unsw.edu.au/jw/flutes.v.clarinets.html#endeffect

My reference to the topic being well thrashed out in the didgeridoo problem was basically to state that whether we look at the standing wave example (In the didgeridoo problem) or a single wave, the reflection at the tube end is caused by the same phenomenon - I.E. A pulse of pressurised air escapes from the end of the pipe and a pulse of vacuum (or rarefacted) air is reflected back up the pipe.

Here is the wording of the article which validates what I said in post #2:-

Let's send a pulse of air down a cylindrical pipe open at both ends (such as a flute, shakuhachi etc). It reaches the end of the tube and its momentum carries it out into the open air, where it spreads out in all directions. Now, because it spreads out in all directions its pressure falls very quickly to nearly atmospheric pressure (the air outside is at atmospheric pressure). However, it still has the momentum to travel away from the end of the pipe. Consequently, it creates a little suction: the air following behind it in the tube is sucked out (a little like the air that is sucked behind a speeding truck).

Now a suction at the end of the tube draws air from further up the tube, and that in turn draws air from further up the tube and so on. So the result is that a pulse of high pressure air travelling down the tube is reflected as a pulse of low pressure air travelling up the tube. We say that the pressure wave has been reflected at the open end, with a change in phase of 180°. In the open-open pipe, there is such a reflection at both ends.

...and a further nice little applet to understand "Standing waves in pipes"

http://www.physics.smu.edu/~olness/www/05fall1320/applet/pipe-waves.html

We can see that there is no need for projection of material (at least not in the sense that is usually understood) by considering a compressional wave travelling in a thin solid rod as it reflects off the end of the rod.

We can consider this as follows:
When the wave reaches the end, the only way we can conserve the acoustic energy is for a wave of equal amplitude to be reflected down the rod;
an additional constraint is that the force that can be applied to the end of the rod is zero (else the end of the rod would experience infinite amplitude), so any compression of the rod in the outgoing wave must be matched by extension in the reflected wave.

N.B.1 You can check how the amplitudes etc. balance near the end of the rod using
acceleration=pressure-gradient/density

N.B.2 I believe I have found a valid phenomenological way to describe the effect at the open end of an air pipe. I'll work at expressing it over the next few days.

I look forward to reading your posts regarding the open end of an air pipe and I don't doubt there are some marvellous mathematical equations that describe the reflections and conditions which cause this effect.

However, from all I have read so far ( and yes I do agree with the impedance mismatch as a perfectly valid explanation for why the reflection occurs), I still like the simplicity of the explanation given by the webpage link in my post #16.

Remembering back to my early physics classes, I always had difficulty understanding why the reflection from an open pipe should be inverted and that from a closed pipe was not - the suction effect explanation of the exiting pressure wave has at last answered this conundrum for me at least.

I was fortunate in my teachers (though I didn't necessarily think so at the time), so I never really had a problem with the phase - whether blocked or open.

Before starting, we looked at the phase relation between the pressure and the movement of particles, and showed that, for a given time dependence of the pressure, reversing the direction of travel also reversed the related direction of the movement. That means that a waves travelling on opposite directions that have particle velocities in phase have the pressure in anti-phase, and vice versa.

Then, for each case, we looked at the constraints:

The blocked pipe allows no movement of the air. So the movement must be out of phase. Because the wave is travelling in the opposite direction, that means that the pressure in the reflected wave must be in-phase with the input.

The assumption for the open ended pipe is that the outside world offers no resistance to the motion (this is more apparent for for a compressional wave hitting the cut-off end of a solid rod). So the pressure at the end of the pipe is identically zero. That requires the pressure of the reflected wave to be in anti-phase with the input. Equally, the direction of travel of the wave means that the movements must be in phase.

Perhaps what I should have said was that you would have to be very selective in your reading of the sucking didgeridoo challenge, because the fundamental solution to the didgeridoo would would take the existence of the resonance as the starting point, and work on why that resonance would cause a reduction in average pressure over the cycle. The observed suction results from the Bernoulli effect - the reduction in pressure observed from the side of a moving stream is (first term) proportional to the square of the velocity. The problem with trying to extract anything sensible from the didgeridoo contributions is that too many of the postings appeared to believe that the basic resonance was the cause of a net suction, whereas the reality is that pure resonance effect alternate between pressure increases and pressure reductions.

However, we can add to that that the reference you provide in post #16 is at best misleading. The description treats the acoustic wave as a moving pulse of air that exits from the pipe. However, the distance that the molecules move with the wave is proportional to the amplitude of the wave; that means that this explanation would imply that quiet sounds in the pipe are not reflected, which we know to be untrue. (It may clarify what I write if you consider the ideas of "projection" in forced ventilation systems)

I have tried to come up with an explanation that handles the wave in the manner that the reference tries to treat a pulse of air, but so far I have failed miserably.

Regards

Fyz

I'm with MPM on this one.

I find the reference of post #16 convincing and not at all misleading. It also demonstrates the end correction required to relate the physical length of the tube into a fraction of the wavelength as stated in my post #13.

I don't agree that the explanation means that quieter sounds are not reflected, it seems to me that they're just reflected quietly.

I suggest you try to use this reference to generate a numerical basis for a crude model. My take on this is as follows:

The size of the reflection will be proportional to the product of the pressure change and the distance over which the change can be observed. As this is considering the movement of material, both the distance and pressure change are proportional to the amplitude of the movement when the movement is small compared with the diameter of the tube). So for small amplitudes the predicted effect would be square law, not linear.

I accept that the reflection may not be linear with amplitude, but I'm afraid I lack the time, motivation and probably the ability to make a meaningful model of what is happening. I see, however, from post #25 that you have at least the beginnings of one.

The following comments may help you to advance your model

I expect pressure changes are adiabatic in nature

I come back to the well recognized "end correction" which places the pressure node a significant distance from the tube end. I suggest any model should locate a "virtual impedance interface" ¼ diameter away from the pipe end.

Regards SLR

Just on the subject of end corrections: the actual value depends on the diameter of the pipe (measured in wavelengths) and the effective extent of any flange (the thickness of the pipe can be treated similarly to a flange).

Unsurprisingly, the smallest end correction is found with thin-walled pipes. At low frequency the correction for a thin-walled pipe is about 0.3 pipe diameters; the correction for pipes with infinite flanges is approximately 0.43 pipe diameters. For small diameter pipes, the approximate theories from the 1950s (and even earlier - e.g. Lord Rayleigh) give results that are very close to measurements. However, there is some departure once the diameter of the pipe approaches the values that are used for musical instruments*, and a value of 0.28 pipe diameters is generally used as a rule of thumb for initial design of organ pipes and the like.

You can find more details, including some theory, and graphs extracted from theoretical and practical references here.

*I am unable to say whether this is due to the approximations made or to the presence of adjacent detuned pipes, or reflection from the ground, etc.

The wave is reflected back because the open end is an impedance mismatch. Enough said.

Sorry to disturb.

The challenge is quite undefined.

The tube wall are not solid?

What's the difference between this "two type of walls"?

If the intention was to compare solid "bar" with tube, how are the tube wall ends geometry?

The "same" wave is sent through the tube wall in longitudinal direction? or through the hole? or through the tube wall in radial direction?

What wave type we are talking? mechanical one (sonic waves) or electromagnetic ones?

If acoustic ones, which type longitudinal, shear, surface waves? It behave different.

If electromagnetic ones, which wavelength? visible? X-ray? microwaves? Some of them can be reflected at first solid surface and some other can penetrate.

Too much open challenge for me!

Kind regards

The refelection depends on the type of the wave,Frequency,medium,energetic wave or non energetic,diameter of the tube,material it is made of..etc..so your question to be specific.

From

V.Ram Mohan

It is all about impedance matching. If a water wave hits a stone wall, it bounces back 180 degrees out of phase. If the same wave hits a beach, (beaches were designed to match the impedance of waves) work is done, and energy is dissipated. To my knowledge, that's true regardless of what kind of wave you are dealing with.

To my knowledge, that's true regardless of what kind of wave you are dealing with.

Sorry again but that's anything but true.

Reflection and refraction behaviour of "mechanical" waves when hit two media interface depend on the impedance mismatch of both media. Acoustic impedance being defined as density times the wave speed. The behaviour depends also on the incident angle, if the angle is 0º or close to 0º, the wave is reflected and transmitted. the fraction of energy or pressure going in each direction being a function of both impedances. If the incident angle is greater, the incident wave can result (depending on the media) in two reflected waves and two refracted ones. If the angle is greater than that called "second critical angle" just reflected waves exist. If the angle has just that critical value, you have reflected and refracted waves, but refracted (or transmitted) ones go just by the surface (surface, Lamb, Rayleigh... waves).

If you take a thick walled tube and introduce a surface wave in circumferential direction and no discontinuities are present, the wave will never be reflected, it will travel continuously attenuating till disappear.

If the wave is electromagnetic, the behaviour depends on the energy (or wavelength, both are proportional) just try to protect yourself with a mirror from a high power red or any colour you prefer laser beam and from a high power 20 MeV gamma ray beam and you'll see by your own the differences.

Kind regards

At high power density, conventional wave behavior does not apply. A huge wave can knock a wall over. A high pressure water jet can cut a hole in a wall. Similarly, a high power density laser beam can melt or vaporize any material, even the mirror you are trying to protect yourself with. The physics of the interaction of energy and matter changes at very high power density. I don't believe that's what the original challenge question was about.

The question is cryptically worded, but I still think we can resort to Occam. Take part of the first sentence: "when a wave hits a solid wall it is reflected back". "In the absence of qualification, "reflected" should be taken to mean "near 100% reflected", and the generality of this statement would at most apply to acoustic waves in air and to surface waves on a fluid. (As you have pointed out, other acoustic waves would generally have high reflection losses due to imperfect impedance mis-matching, and the effect on EM waves depends on the nature of the wall).
. So we can assume acoustic waves in air or on the surface of a fluid - so they will not be waves in the wall of the tube, but waves that are contained within its interior.
. Now we have a question we can (presumably) answer. If we are looking at a sound wave in air, efficient reflection would require the tube diameter to be small compared with the wavelength (and the phase of the reflection would be inverted compared with reflection from a wall). If we are looking at a surface wave propagating in a liquid in the horizontal tube, a change from shallow fill to deep water would have a similar effect (though the initial wave would probably be lossy, and anyway I personally doubt that this is what was meant).

Try a flat piece of paper and a spool that thread comes on. Blow with force thru the tube in the spool and the paper will be stuck to it. It can't come off no matter how hard you blow. Suction is created by the air moving thru the tube. Too basic, sorry that's me.

You are describing a second-order effect which is only significant at large amplitude. Don't you think it is reasonable to assume that the challenge is intended to be about the basic reflections that affect the wave regardless of its amplitude.

If you are considering height (or pressure) a water wave hitting a wall bounces back in phase. A water wave constrained between two walls reaching an opening bounces back in antiphase.

Acoustic waves are regions of compression or rarefaction passing through the transmitting region (assumed to be air)

One important point that has not been addressed in the previous postings, is that at the reflection from an open ended pipe, a compression is reflected back as a rarefaction, and vice-versa

This give us a clue as to what is happening. Consider a rarefaction to travel through a tube. As it reaches the open end, and passes into the outside environment, molecules of the ambient air rush into the low pressure rarefaction to annihilate it. But these molecules have mass and inertia, and they over-pressurize the region. Some of these annihilating molecules crowd into the pipe exit and the result is a compressive wave is initiated and it travels backward down the tube. In this way, a rarefaction is reflected as a compression as noted.

The fact that the "reflecting process" takes place outside the tube also explains the need for Rayleigh's correction factor which is necessary to convert the tube length into a wave length. (I learned at school that L_effective = L + d/4 where d - tube diameter: but was warned that this was an approximation; the length of the tube and the shape of the open end change this value.)

Sorry - I missed MPM's comment in #2 which is essentially the same

Actually, to add to the jibberish, if the combination of a "steep" slope and "minimal" friction is sufficient, the flow could be supercritical. Then a wave could not travel back upstream.

Anyways, assuming the slope and friction are small enough so the the celerity (c) is greater than the velocity (v) then the wave will travel up stream

As many others have stated, the wave is created because of the transition at the end of the tube from steady and uniform flow to unsteady flow. The acceleration term is becomes relatively large. This causes a loss of energy which means that the the depth of water will change in either a negative wave or postive wave.

dH/dx is being modified. Therefore, dy/dx and dv/dx will change. The equilibrium will oscillate thus creating waves.

Same effect as sending an electrical impulse down a pair of wires. You get a reflection whether the circuit is shorted or open. The only time you get no reflection is when the impediance is exactly the same as the wire pair.

The tube looks like one impediance and the open water is another.

Let's try starting over. 90% of the disagreement is not disagreement at all, but the use of different terms to describe physical phenomenon. A wave travels down a tube of unknown length. Because we don't know anything about the tube except that a wave is traveling toward an open end, we can't construct a resonator. In fact, we don't even know the polarity of the wave. About all we know is that when a wave reaches the open end of a tube, part of the wave will exit the tube, and the balance will be reflected back toward its source. Because the tube has an open end, the polarity of the reflected portion will be 180 degrees out of phase with the incoming wave. The ratio of incident to reflected pressure or amplitude will depend on the impedance mismatch. I hope this is a description most of us can agree with.

Yes, I think we agree the effect of an impedance mismatch.

The difficult part is to explain the source of the impedance mismatch. And (I think) it all falls apart if one tries to explain the effect in terms of replacement of material (e.g. "like air sucked by a speeding truck" - which is a direct comparison with a known-square-law effect if ever I saw one).
I can model it OK, but I can't offer a simple explanation - I think we need something deeper than likening an open end to a void.

Well, I think we agree the air outside the tube is at atmospheric pressure. If the wave compress the air inside the tube, the pressure difference implies a density difference (we are with compressible fluids) and a density difference implies an impedance difference/mismatch. When a pressure wave reaches an "interface" between two media, the pressure fraction reflected equals (Z₂-Z₁)/(Z₂+Z₁). Where Z is impedance and suffix 1 and 2 are related to the first and second media.

When Z₂<Z₁ the result is negative which means the reflected pressure wave is 180º (opposite phased) to the incident wave. If Z₂>Z1 then the reflected wave is "in-phase" with the incident one.

Could this be applicable to this case?

Obviously we haven't a definite "interface"

Kind regards

In the absence of the wave, the air inside the tube has the same density as the air outside. Very small amplitude waves give very small density changes, which would mean negligible proportionate reflection. Clearly, it is not the density change due to the wave that is responsible for the impedance mismatch.

The only way I have obtained a linear small-signal answer is by calculating velocities, displacements and pressure gradients (including for example dV/dt proportional to dP/dx), but I can't seem to convert that to a verbal explanation.

Here is an example from real life.I had trenched two pipes into the same trench, and one of them was 50 feet long, the other was 125 feet long.My labels fell off after several rainy days and I had already filled in the trench.

To determine which was which, I blew down the pipes.The pressure wave I exterted was followed by a return series of pressure waves,each decreasing in amplitude,but the rate was the same.After blowing down both tubes it was easy to determine ths short one from the longer one by the return time.

My theory is that the pressure wave exited the tube,and there was more air in motion than I had introduced into the tube.Some of this extra volume followed the original pulse out,leaving a low pressure behind it.This low pressure was quickly filled by the atmospheric pressure, and the resulting wave propagated back up the pipe to the other end.

This seems to agree with some of the other posts.

does it matter whether the tube is cylindrical or square, or some other shape?

If it is a sound wave in air, not significantly - so long as no dimension is so small that the transmission down the tube becomes noticeably lossy, or both dimensions are not sufficiently large that radiation loss from the end becomes important.

For you young'uns who've never seen a exponential tube (the question doesn't say constant diameter tube, now does it?), look at old record players (or early hearing aids). Essentially no reflection if manufactured well.

Acoustic reflections in air-filled tubes

As stated elsewhere, the reflection can be viewed as being caused by a sudden impedance change. This is indeed a valid basis for calculating the size of the reflection; however, to my mind it is an intermediate step in calculations rather than an explanation of the cause of the reflection.

Now, I find the full case of an open ended pipe is conceptually rather difficult, so I propose to work with a transition between two pipes P1 and P2, both having cross-sections that are very much smaller than a wavelength. The pipes have cross sectional areas A1 and A2.

Consider now a wave travelling in pipe
In order to calculate the reflection, we require the following:
a) Conservation of energy: Pin = Prefl + Ptrans
. (using suffixes refl for reflected and trans for transmitted)
b) Conservation of displaced volume: we define the volume displacement in any wave as the linear displacement multiplied by the cross-sectional area. This is in fact the volume that of material that would have crossed a notional surface. This must be the same for the sums of the waves in each of the two pipes, otherwise there will be accumulation of material at the interface (and therefore infinite pressure variations). (This is true only where the interface between the two pipes can be approximated as instantaneous)

So, let us consider a wave with peak displacement V_in and power P_in propagating initially in the pipe P1. The linear displacement is Vin/A1.

For a given instant in a wave, the velocity will be proportional to the linear displacement, so we can write:
Pin = = K.(Vin/A1)².A1
Prefl = K.(Vrefl/A1)².A1 ; and
Ptrans = K.(Vtrans/A2)².A2
. (K is a constant that takes the air density, wave velocity and frequency spectrum into account)

Conservation of energy gives: Vin2.A1 = Vrefl2.A1 + Vtrans2.A2
Conservation of volume displacement at the instantaneous interface gives:
Vin + Vrefl = Vtrans

Solving for Vrefl gives:
Vrefl=(A2-A1).Vin/(A1+A2); therefore
Reflected Power = ((A1-A2)/(A1+A2))².(Input Power)

A2/A1=0 corresponds to a perfect blockage, and we can see that all the energy will be reflected. We can see that the constraint on the displaced volume means that there is no movement at the interface. That means that the acceleration and hence the pressure gradients will be in anti-phase. Because the direction of travel of the wave is reversed, this corresponds to the absolute pressure being in phase. (N.B. that the limitation on the accuracy here is the density and/or rigidity of the blockage and of the pipe walls)
At the opposite extreme, this simplified theory would predict perfect reflection from the end of an open pipe. It also predicts that the material velocity in the reflected wave will be in phase with that in the incoming wave. The same arguments as above lead to the pressure at the reflection being in anti-phase. N.B that the limit on accuracy here is that the constraint on displaced volume becomes less and less accurate as the dimensions of the transition region between the pipes approach a modest fraction of a wavelength - and this is the reason both for incomplete reflections and the requirement for an end correction.

This is of course true. There are two reasons I can think of. 1- The wave will cause a change in pressure as it exits the pipe and 2- turbulence will occur due to the pressure drop interacting with the medium surrounding the area at the end of the pipe.

Hi I would like to explain this as follows .

The wave in the tube looks like a cylinder , the length of the cylinder depends on the length of the wave pulse .

The cylinder wave travels true the tube , from the moment that the cylinder wave comes out the tube the wave spread immediately to all directions .

By doing so it forms a membrane , this membrane travels slower than the cylinder wave .

The membrane becomes stronger and finally shoot the cylinder wave back where it came from .

If you could measure the length of the wave you will see a significant loss but it would be strong enough to measure .

Kind regards

Van de Velde Willy

termination shock from change of energy direction. It causes what is basically a ripple, kind of like a raindrop falling on water.

I'm not an expert in this field, but it occurs to me that if the wave fills the tube as it flows, the water moving along the top of the tube will fall when it exits the open end. It would then strike the water exiting at the bottom, causing the reflection just as the wall does.