Can anybody explain to me why sound waves get reflected from the end of an open pipe???

I have no answer at present. Do you have an example?

a stationary wave set up in a wind instrument like a flute which has an open end...

Well in case of both ends open it is not reflected , it is that you have 2 waves traveling in opposite directions to begin with forming standing waves that for natural frequencies must have maximum variation at points where there is no obstacle and 0 amplitude where there is a wall.

See more here;

https://en.wikipedia.org/wiki/Standing_wave

and

http://www.physicsclassroom.com/clas...nd-Air-Columns

http://www.physicsclassroom.com/clas...nd-Air-Columns

the second link shows a pipe open at both ends with a stationary wave pattern in so must have reflection at both ends (and is). (why would it reflect at an open end with one end closed but not with neither ends closed?)

My question is if somebody has a simple explanation of the reflection at the open ends.

To make my requirements a bit clearer I am a Physics teacher and know why reflection occurs in terms of pressure but I find it difficult to explain this idea and was wondering if somebody had a more accessible explanation.

Thanks

Any time a wave passes from one medium to another, some of the wave gets reflected and some of it gets transmitted. In this case, the pipe is one medium and the air is another.

Actually i should have replied much better than what i did but the issue is rather complicated if you want to do it all carefully and it depends on how you generate the waves to begin with. I may have been even largely wrong to completely wrong in what i said for particular real life (open ends) instruments. It is still not clear to me without studying it in true math detail how important all components are (reflection plus original vs 2 originals in opposite direction - it depends on the exact pipe/geometry origin of sound waves situation). But from what i have read since then it appears that the reflection may be the main thing even in open air ends. I thank trolly from prompting me to study this better with the different medium argument.

Technically this is not a different medium as the reflection (impedance) is seen in most cases in physics you see these things (that you do have a radically different medium indeed like air to water or one rope with some density to another rope of another density etc) as you have air both in and out of the open end but effectively it becomes different medium because of the alteration of its properties near the end that the air flow causes (locally changing the 2 regions of air just before the end of the pipe and just after on the outside) as the geometry opens completely after the end of the pipe creating essentially a different density/speed medium locally ( as the flow of air spreads out ). It is a fluid mechanics problem really at this point that will explain the reflection as the transition of the wave to a different medium (moreover it is still air it has slightly different properties than the inside the pipe air). Notice how also sounds waves expand spherically just outside as opposed to the inside region.

I will try to find a better video than this but so far i got this one;




Now look also here; (read the text and play videos including some that may require activating flash plugins etc)

http://newt.phys.unsw.edu.au/jw/flutes.v.clarinets.html

and http://www.animations.physics.unsw.e...edance.htm#ref

eg from the above '

" 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. (This is what physicists call an 'arm-waving argument': it's neither rigorous nor quantitative. If you'd like a formal explanation, see Reflection at an open pipe http://www.animations.physics.unsw.e...edance.htm#ref .) "

Look at the pressure one for an open tube.

How does the stationary wave inside flute affect the sound the flute makes for the listener outside the flute?

PairTheBoard

The player can make the effective length of the pipe change by covering holes I think, not sure of the exact mechanics of that. A particular length of tube can only set up a stationary wave at a particular frequency which depends on the length of the pipe (see the diagrams above, the wavelength is a certain multiple of the length) and the speed of the wave, so different lengths or different numbers of holes covered gives a different frequency, which is the pitch the listener hears.

"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. (This is what physicists call an 'arm-waving argument': it's neither rigorous nor quantitative. If you'd like a formal explanation, see Reflection at an open pipe"

This is the explanation I use was just wondering if anybody else could confirm it. When I searched online previously I couldn't but an old colleague had once explained it to me that way.

The only explanation I could find in books was in terms of the end being a pressure node because you couldn't change the pressure there without changing the pressure of the entire room. I didn't like this because why can't you change the pressure locally for some time? I can find the exact wording but that bit always confused me...

That was the qualitative explanation my link above gave. At the end of it there was another link to a more rigorous treatment that kind of explained the math behind it.

The pure truth here is that this is a complicated boundary physics problem. You need to see how the pressure wave that comes at the end is released outside and spreads (fluid mechanics) and what is the reaction (effectively a reflected wave) to this essential change in medium from that of the inside the air column geometry. I suppose only if one put some numbers to it to see exactly how high/low the pressure gets to compare to atmospheric you can get an idea why the end region is essentially feeling it is all a different medium (and the suction created) and all this is non trivial but an essential distinction between inside and outside the pipe (even if still open) .


Did you go to the next page after the quoted blue statement? You will see there that continuity of flow and pressure results in a reflection (much like there is nowhere to go for molecules but in such directions or redistribution that effectively generate a new wave back). It is air inside and outside and no real wall in between at the end but it does behave like some kind of wall due to the different geometry and the redistribution of air this causes at the boundary to satisfy essential things like conservation of mass (no more air is generated or lost in the whole system) and lack of discontinuity in pressure (smoothness in gradient no infinite forces).


See here about the reflection of a wave when it changes medium. http://www.acs.psu.edu/drussell/Demo...t/reflect.html

Imagine that someone is blowing air inside the pipe that needs to get out eventually so this is a substantial source of alteration of local properties and deserves full fluid mechanics analysis of the boundary.

Read what the link says;

Earlier in http://www.animations.physics.unsw.e...-impedance.htm

"The specific acoustic impedance z is the ratio of sound pressure to particle velocity, and z = ρv , where ρ is the density and v the speed of sound. (See Acoustic impedance, intensity and power to revise). So for our duct with cross sectional area A, provided that the wave is strictly one dimensional and traveling in one direction, the acoustic volume flow is just U = Au .

For this very special case, we define the characteristic acoustic impedance Z0, where

Z0 = p/U = p/Au = z/A so

Z0 = ρv/A
Of course, there usually are reflections from the other end of the pipe, whether open or closed. So there is a sum of waves traveling to the right and left and quite often these give strong resonances, which is after all how musical wind instruments work. For ducts however, the characteristic acoustic impedance Z0, is the dimensional quantity that scales the acoustic impedance. To see what the impedance spectra of simple ducts and musical instruments look like, see What is acoustic impedance.

It is possible to have no reflections, however. If the pipe length L is very long, then it takes a long time for the reflection to return. So, for the time 2L/v, the duct is effectively infinite. Further, when the echos do eventually return, they are much attenuated by losses at the wall. (I mention this because, in our research laboratory, we use acoustically infinite acoustical wave-guides as calibrations for acoustic measurements. See this web site http://newt.phys.unsw.edu.au/music/ for the research lab, this page http://newt.phys.unsw.edu.au/jw/acou...asurement.html for measurement techniques or this scientific paper http://newt.phys.unsw.edu.au/jw/repr...SmithWolfe.pdf for some technical information, if you’re interested.)
"

"Let's suppose that we send a sine wave along the pipe towards the right. Let the pressure due to this wave be p = p>sin (kx − ωt) , where the subscript in the amplitude is to remind us that the wave is traveling to the right (See The wave equation for sound). The flow due to this wave is U = U>sin (kx − ωt) and, as we saw above,

p>/U> = Z0.

Outside the pipe (at x = 0 but looking out, which we'll write as x = 0+), there is sound radiation away from the pipe. Let's write that the pressure* looking into the radiation field (at x = 0+) is p(0+) = prad sin (− ωt).g(x) , and the flow due out of the pipe due to this wave is U(0+) = Urad sin (− ωt).h(x) . As we also wrote above

prad/Urad = Zrad ,

remembering that Zrad is almost entirely imaginary, so that prad is almost 90° ahead of Urad in phase—so at least one of them has to be a complex number.

* We can't write a simple one dimensional wave equation here, because the radiation diverges. The geometry very near the pipe end is actually somewhat complicated, but it's not important to this argument. Away from the end, i.e. for kr >> 1, where r is the distance for the end of the pipe, the amplitudes of both prad and Urad go like 1/r and they are in phase: the local geometry approaches that of a plane wave.

Continuity of pressure. Now exactly at x = 0, at the end of the pipe, the pressure just inside the pipe must equal that just outside. (If we didn't satisfy continuity of pressure, we'd have an infinite pressure gradient and so infinite force.) We also have

Continuity of flow: the flow coming out of the pipe (at x = 0−) must equal the flow radiating outwards (at x = 0+). (Continuity of pressure is required because we are not creating or destroying air.) Of course, we can't satisfy those conditions with these two waves: the magnitude of Z0 is much greater than that of Zrad so, if we set the flows equal, the pressures can't be equal.

What happens is no surprise: there is a reflection at the open end: let's write the reflected wave, which travels to the left, as p = p<sin (kx + ωt) and the flow due to this wave is U = U<sin (kx + ωt). So, at x = 0, we have to satisfy the continuity of pressure and flow conditions:

p> + p< = prad and U> − U< = Urad .

(Why the minus sign? Because U> and Urad are flowing out of the pipe while U< is flowing in.) We can use the equations above for Z0 and Zrad to rewrite the equation above right as p>/Z0 − p</Z0 = prad/Zrad. This and the equation above left are two equations that can be solved to give the reflection coefficent at the end:

Reflection coefficient = p</p> = (Zrad − Z0)/(Zrad + Z0) = − (1 − Zrad/Z0)/(1 + Zrad/Z0)

where we remember that the reflection coefficient is in general a complex number (Zrad is almost completely imaginary). Now, except at very high frequencies, |Zrad| << Z0, so the reflection coefficient is almost −1: we have reflection with a change in phase of π in pressure, but no change in phase in the flow. This gives us a pressure node (two waves out of phase add approximately to zero) and flow antinode at an open end (two waves in phase add to give a maximum).

So, we have a reflection at the open end of a pipe: very useful the standing waves in musical instruments, the voice and elsewhere. Examples, explanations and animations are given in Open and Closed Pipes and Pipes and Harmonics.

Reflection at a closed end is somewhat easier. If the pipe is ideally closed, then the conservation of flow must mean that U> and U< add to zero, which means that they are out of phase by π. Their pressures are in phase, so the closed end of a pipe has an antinode in pressure and a node in flow.


"

I just put all this here to motivate you to go and really look at the link for a cleaner attempt to analyze this than the hand waving arguments about why reflection is unavoidable for essential physical reasons from fluid mechanics.

Think of a wave as a hand waving. It is moving back and forth and (overall) not actually going anywhere.

Sound is just what you hear when the air pressure is vacillating in the immediate area of your ear.

I get that a wave is stuff moving up and down. So what is moving forward?

The signal, the disturbance (energy). Local disturbance based on fluid mechanics analysis or whatever local Physics you have in place (eg in solids - elastic models) obeys the wave equation. https://en.wikipedia.org/wiki/Wave_equation (see here to recognize what is happening in some examples eg with springs obeying Hooke's law etc) (eg see "The wave equation in one space dimension can be derived in a variety of different physical settings. Most famously, it can be derived for the case of a string that is vibrating in a two-dimensional plane, with each of its elements being pulled in opposite directions by the force of tension.[6]") .

https://en.wikipedia.org/wiki/Sound#...ansverse_waves

"Sound can propagate through a medium such as air, water and solids as longitudinal waves and also as a transverse wave in solids (see Longitudinal and transverse waves, below). The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The sound source creates vibrations in the surrounding medium. As the source continues to vibrate the medium, the vibrations propagate away from the source at the speed of sound, thus forming the sound wave. At a fixed distance from the source, the pressure, velocity, and displacement of the medium vary in time. At an instant in time, the pressure, velocity, and displacement vary in space. Note that the particles of the medium do not travel with the sound wave. This is intuitively obvious for a solid, and the same is true for liquids and gases (that is, the vibrations of particles in the gas or liquid transport the vibrations, while the average position of the particles over time does not change). During propagation, waves can be reflected, refracted, or attenuated by the medium.

----

Sound is transmitted through gases, plasma, and liquids as longitudinal waves, also called compression waves. It requires a medium to propagate. Through solids, however, it can be transmitted as both longitudinal waves and transverse waves. Longitudinal sound waves are waves of alternating pressure deviations from the equilibrium pressure, causing local regions of compression and rarefaction, while transverse waves (in solids) are waves of alternating shear stress at right angle to the direction of propagation.

"

see also https://en.wikipedia.org/wiki/Acoustic_wave_equation

http://www.feynmanlectures.caltech.edu/I_47.html

Eg see here in the above Feynman link;

"

We shall give a derivation of the properties of the propagation of sound between the source and the receiver as a consequence of Newton’s laws, and we shall not consider the interaction with the source and the receiver. Ordinarily we emphasize a result rather than a particular derivation of it. In this chapter we take the opposite view. The point here, in a certain sense, is the derivation itself. This problem of explaining new phenomena in terms of old ones, when we know the laws of the old ones, is perhaps the greatest art of mathematical physics. The mathematical physicist has two problems: one is to find solutions, given the equations, and the other is to find the equations which describe a new phenomenon. The derivation here is an example of the second kind of problem.

We shall take the simplest example here—the propagation of sound in one dimension. To carry out such a derivation it is necessary first to have some kind of understanding of what is going on. Fundamentally what is involved is that if an object is moved at one place in the air, we observe that there is a disturbance which travels through the air. If we ask what kind of disturbance, we would say that we would expect that the motion of the object produces a change of pressure. Of course, if the object is moved gently, the air merely flows around it, but what we are concerned with is a rapid motion, so that there is not sufficient time for such a flow. Then, with the motion, the air is compressed and a change of pressure is produced which pushes on additional air. This air is in turn compressed, which leads again to an extra pressure, and a wave is propagated.

We now want to formulate such a process. We have to decide what variables we need. In our particular problem we would need to know how much the air has moved, so that the air displacement in the sound wave is certainly one relevant variable. In addition we would like to describe how the air density changes as it is displaced. The air pressure also changes, so this is another variable of interest. Then, of course, the air has a velocity, so that we shall have to describe the velocity of the air particles. The air particles also have accelerations—but as we list these many variables we soon realize that the velocity and acceleration would be known if we knew how the air displacement varies with time.

As we said, we shall consider the wave in one dimension. We can do this if we are sufficiently far from the source that what we call the wavefronts are very nearly planes. We thus make our argument simpler by taking the least complicated example. We shall then be able to say that the displacement, χ, depends only on x and t, and not on y and z. Therefore the description of the air is given by χ(x,t).

Is this description complete? It would appear to be far from complete, for we know none of the details of how the air molecules are moving. They are moving in all directions, and this state of affairs is certainly not described by means of this function χ(x,t).

From the point of view of kinetic theory, if we have a higher density of molecules at one place and a lower density adjacent to that place, the molecules would move away from the region of higher density to the one of lower density, so as to equalize this difference. Apparently we would not get an oscillation and there would be no sound. What is necessary to get the sound wave is this situation: as the molecules rush out of the region of higher density and higher pressure, they give momentum to the molecules in the adjacent region of lower density. For sound to be generated, the regions over which the density and pressure change must be much larger than the distance the molecules travel before colliding with other molecules. This distance is the mean free path and the distance between pressure crests and troughs must be much larger than this. Otherwise the molecules would move freely from the crest to the trough and immediately smear out the wave.

It is clear that we are going to describe the gas behavior on a scale large compared with the mean free path, and so the properties of the gas will not be described in terms of the individual molecules. The displacement, for example, will be the displacement of the center of mass of a small element of the gas, and the pressure or density will be the pressure or density in this region. We shall call the pressure P
and the density ρ, and they will be functions of x and t. We must keep in mind that this description is an approximation which is valid only when these gas properties do not vary too rapidly with distance."

(so keep reading the link for the real math)


See also here on this chapter;

https://www.usna.edu/Users/physics/e...1/Chapter2.pdf

The wave is propogating. The amplitude of the sound wave being drawn as being up and down on a graph is a bit confusing.

The vertical axis is changes in the density of air. It is getting pushed and pulled towards and away from the vibration caused by the musical instrument.

It is easier to understand if you actually watch a woofer playing some deep bass sounds. You can visualize (use your imaginations) it pushing and pulling the air immediately in front of it, which in turn pushes and pulls the air in front of that air, which in turn pushes and pulls the air in front of that air. Basically the high pressure caused by the air being pushed and the low pressure caused by the air being pulled causes the same effect on the next bit of air.

Thanks both for the replies. Been wondering about that one for a while, but was too afraid to ask.

You shouldn't be afraid at all. I had to revisit things in that thread too (regarding the deeper reasons/mechanism for reflection) . Just read the Feynman lecture derivation of the wave equation for sound waves i linked (using effectively Newton's second law F=m*a but relating F to pressure difference and second space derivative eventually and a with second time derivative) and you will see it is definitely not a laughing matter if someone doesnt know about its origins. It takes some care and range of approximations to derive it.

In fact the entire topic of standing waves in instruments is oversimplified and treated without deep explanations in most textbooks of introductory physics courses and is very popular with exams/tests and even lab experiments but only in terms of the results not the fundamental physics behind it. Now the advanced wave mechanics books know it is far from simple and they do not deal with it properly either to be honest. Only some direct research in the topic books eg serious study in acoustics will do a thorough job for actual geometries.


The entire topic deserves a monograph it seems (regarding the exact instruments) .

The pressure variation that propagates here in most sounds is order 10^-6-10^-8 and less of the atmospheric pressure (read about decibels in the Feynman link) . Real tiny pressure differences compared to eg what you have in explosion waves or near typical fluid mechanics flow problems/situations.

Well, I saw that there was a Physics / Maths forum here so I thought I'd test it with a pretty open question and I have to say I'm impressed. Lots of good posts here and plenty of food for thought!

And one thing to note - if in doubt, go and read Feynman!