Phys 1240 Fa 05, SJP 12-1 DRAFT Chapter 12: Wind instruments When we studied strings, we saw that the fact that a string has no vibrations at the two ends leads to very important consequences, basically everything we've talked about for the last several weeks! (The "zeros" at the ends mean that only special wavelengths "fit", the harmonics. Think back to Chapter 10, when we drew pictures of all the harmonics on a string. The fundamental, n=1, was the simplest (longest wavelength) wave we could draw which went to zero at the two ends. The "boundary" (forcing zero motion at the two ends) is what determines the allowed wavelengths, which in turn determines the sounds we hear! Now let's turn to wind instruments. The story is similar but a tick more complicated. That's partly because the vibration inside a wind instrument is not so easy to visualize as the vibration of the string. When I draw my string picture (above) you can picture the actual metal string vibrating in the pattern shown. It's a transverse wave of the string atoms. But in a wind instrument, we have longitudinal pressure oscillations in the air. It's the pressure that varies sinuosoidally (wave-like) as you move along the tube. Now, it's true, you COULD think about the longitudinal displacement of air molecules, again a wave-like pattern in there, but we've mostly tried to think about sound as a pressure wave, so that's the language we'll often stick with. Let me imagine a tube which is OPEN at both ends. It could be a flute (just an empty cylinder, basically), if you want something concrete to visualize. You apply some pressure variations (by blowing), and it sets up "pressure waves" which travel down the inside of the tube, bounce off the ends, interfere with themselves, and set up a standing wave. To be very precise, I'm talking about OVERpressure waves here (the air starts off at "atmospheric pressure", and then you get some overpressure or underpressure which travels along the tube) I could use those EXACT same ideas, by the way, to describe the string wave shown above (and earlier): you apply some "displacement variation" (by plucking a string), and it sets up transverse waves which travel along the string, bounce off the ends, interfere with themselves, and set up a standing wave. So the story is very nearly the same, we're just talking about motion of a string in one case, and air pressure in a tube in the other. With a string, the motion is zero at the ends because we clamp down the string there. In an open tube (which means open at BOTH ends), it is the OVERpressure which is zero at the open ends - because the ends are just open air, which tends to like to be right AT atmospheric pressure. (If there's solid pipe all around you, it's a lot easier to set up "high pressure" or "low pressure" than if you're just in open air. Of course, you CAN still have high pressure in the room, that's how sound travels, but it's just much easier and strong, inside the tube.) So this "end condition" of the overpressure being zero is NOT quite as "rigorous" as clamping a string down, where it's strictly zero. Here in the pipe, it's just that, compared to the overpressures experienced inside the chamber, the ends (and outside) are much much closer to being zero overpressure all the time, the variations are much smaller. But the main point is, that pressure inside an open tube looks very much like displacement of a string: It's a standing wave, zero on the ends, vibrating in the middle. In the end, what you get if you graph pressure versus position inside the tube, is a graph which looks JUST like that string picture at the top of the page! You can have "overpressure" or "underpressure" inside the tube, but you're pretty much at zero at the two ends, and so the simplest possible standing wave will be the one we've drawn before for strings.
Phys 1240 Fa 05, SJP 12-2 DRAFT Here's a graph of PRESSURE vs POSITION inside the pipe: I've drawn the "pipe" on top of the graph in grey, to show how the wave "fits". The story is completely analogous to the strings: If you want a wave that starts and ends at zero (overpressure), then the simplest wave you can draw has exactly HALF a wave fitting in the pipelength L, which means (λ/2)= L. I interpret this picture as saying that at the very center of the pipe, the pressure "swings" from very high (overpressure) to very low (underpressure), over and over... (It's an antinode) At the ends, it's very close to atmospheric pressure all the time. The next possible standing wave that "fits" will be one where TWO half-waves fit, i.e 2(λ/2)= L. That graph (without drawing the pipe) is just this: Again, I'm drawing OVERPRESSURE on the y-axis, and position down the pipe along the x-axis. So the way I interpret this second mode (n=2), is that there is no overpressure at the two ends OR at the midpoint, but the pressure is swinging from very high above normal to very far below normal at TWO spots (these are "anti-nodes", ¼ and ¾ of the way down the tube.) This story continues exactly in parallel to the string story: we will have harmonics (higher possible allowed vibration patterns, the more complex standing waves). They will have wavelengths that "fit", which means n (λ/2)= L (where n is an integer, that counts how many half-waves fit, it's the harmonic number, or the "mode" number, just exactly analogous to the string story from last chapter) And once again, we have standing waves, which satisfy my favorite equation: λ f = v Here, v is the speed at which pressure waves travel down the pipe. That's just the speed of sound! These waves are themselves basically just intense sound waves "trapped" inside the pipe. So just like strings, we get a series of allowed harmonics if we blow on an open pipe: the fundamental frequency, f 1 = v/(2l). and the higher harmonics (labeled by an integer "n"), with f n = n f 1 The sound that you hear is kind of "leakage". There's oscillating air inside the tube (standing waves of pressure), and although the pressure at the end is CLOSE to zero, it might vary just a little - which is enough to send a traveling wave out into the room, and to your ears. If the standing wave has frequency f, then that's the frequency that the pressure varies ANYWHERE in the tube. So the tube itself will be wiggling at frequency f, as will the air (VERY slightly) at the ends. So you're going to generate sound waves in the room with that same frequency f... If you were to change the speed of sound (v) in the air inside the tube, then λ f = v tells us that either λ or f will have to adjust. Which? Well... remember, wavelength λ is determined by the length of the tube. It's geometry! So that really CAN'T adjust. Wavelength is just what "fits" in the tube, it has nothing to do with how fast the waves move. So it's the f that will adjust. Thus, as the air in a wind instrument like a flute warms up (increasing the speed of sound waves, v) the frequency that you hear will ALSO shift. This is different from a string instrument (where the frequency is determined by the length of the string, and the tension of the string, and the mass of the string, but NOT the speed of sound in air!)
Phys 1240 Fa 05, SJP 12-3 DRAFT If you look at your book, Figure 12.2, you will see some graphs that match what we've been drawing. They draw the first three modes, f1, f2, and f3, going down the page. The graphs on the right match exactly with what I've been drawing on the previous page, OVERpressure p vertically, and position along the tube horizontally. They're time dependent, which is why we draw several lines in each graph. The solid and dashed lines, e.g. represent the pressure at slightly different times. Visualize an animation, a standing wave wiggling up and down, which means that at the center of the tube (for the n=1 fundamental wave) the pressure is VERY high, then VERY low, repeating at frequency f... Let's think about how this could happen. How can the pressure get higher at the center of the tube? High pressure means that more molecules of air are squeezed together there. Air must have rushed into the tube from the outside, squeezing to a high density (and thus pressure) in the middle. But then the high pressure at the center "pushes back", causing those air molecules to rush back outside. Indeed, just like a mass on a spring, there will be an overshoot. As the molecules all rush away, they CONTINUE to head out for awhile, and there will be a "partial vacuum" in the middle for a short while. That's the time which I draw as "dotted", where the wave has NEGATIVE overpressure at the middle of the tube. So air is rushing out of the tube, then into the tube, at a frequency f of the oscillation itself. Now let's think about the motion of individual air molecules. Right at the VERY center, molecules just sit still. (Air rushes in and pushes them from BOTH sides, then rushes back out but the molecule SMACK in the middle can just sit there, by symmetry, which way would it go?) So if you were to try to draw DISPLACEMENT OF MOLECULES as a function of where you are in the tube, you would find that the molecule at the very center stays put the whole time. It gets squeezed for a while, then "sucked on", but it doesn't budge. That looks like a NODE, a position where there is always ZERO motion. On the other hand, molecules at the openings at the end are moving like crazy. That's where the molecules FLOW in and out. So you get a displacement ANTINODE at the two ends. Take a look at the top of fig 12.2 in the text, shown also here: I'm showing you EXACTLY the same mode as the top figure on the previous page (the fundamental, the SIMPLEST wave!) But now I'm not showing PRESSURE on the vertical axis, I'm showing "displacement away from where you're supposed to be" of air molecules. A slightly odd idea. It's like each air molecule has a little tag that says "you belong at x= such-and-so". If there's no standing wave, that's where the molecule lives. But when you introduce this (fundamental) wave into the pipe, the molecules jiggle back and forth, and at any snapshot in time, some molecules are "off to the side from where they belong". What I'm graphing is "how far off to the side is the molecule that belongs here". So for the solid curve, you see that an air molecule on the left side of the tube has a big positive displacement, i.e. it's shifted to the RIGHT, which means it's "deep inside" the tube (helping make that high pressure at the center we know is there!) Look at an air molecule at the right side of the tube, over at x=l. The solid curve has a big negative value, which means THAT molecule is shifted way to the LEFT, again "deep inside" the tube (again helping make high pressure inside!) My dashed curve is a moment later, when all displacements are zero. At this moment, the air molecules are all in their proper positions, but they're MOVING - they're rushing outwards. Another moment later you have my dotted curve. Now the molecule at x=0 has a large NEGATIVE displacement (which means it's shifted left, outside the tube. It's left behind a vacuum, a negative pressure at the middle of the tube where it's feeling evacuated). Over at x=l, the dotted curve is positive, THAT molecule is shifted RIGHT, outside of the tube also.
Phys 1240 Fa 05, SJP 12-4 DRAFT You really have to think hard about this one, reading it probably won't make sense, but if you can just "run the movie" in your head, this should make more sense to you. You're just visualizing the motion of air in and out of the tube that CAUSES the standing wave pattern we're looking at. See box 12.1 and fig 12.1 for the book's attempts to help you put this all together. Perhaps after reading these notes, that figure will make just a little more sense? Give it a try! In any case, if you look at the fundamental wave pattern shown on the previous page (for displacement, rather than for pressure), you'll see that some things are the same when you compare the "displacement" graph with the "pressure" graph, and others are different. Differences: There is a displacement node at the middle of the pipe (right where the pressure antinode lives). That's a general rule: pressure antinodes correspond to displacement nodes. The places where the pressure gets really high (or low) is also the place where the particles actually never move. A little weird, but it does make sense. Think about right next to a wall - the molecules there can't move ('cause there's a wall there!) but you can build up really high or low pressure! There is a displacement antinode at both ends (right where the pressure nodes are!) Again, a general rule: pressure nodes correspond to displacement antinodes. The places where the pressure stays atmospheric all the time is the place where the particles are rushing in and out. Again, a little weird, try to wrap your head around this. These rules make sense if you can just visualize "pressure" and "displacement" of air molecules, and how those will relate to one another. Similarities: Most important, the WAVELENGTH of this wave is exactly the same whether you draw it in the displacement way, or the pressure way. Stare at the two graphs and convince yourself. In BOTH cases you have λ =2L. (In the case of the displacement graph, you're going from a "high" to a "low", which is half a wave, fitting in the length L. In the case of the pressure graph, you're going from a "zero" to the next "zero", which is again just half a wave) Since λ is the same, it means the frequency is the same. You can think about displacement waves or pressure waves - it changes the picture you draw, but does NOT change what sound comes out! Look now at the pattern of the left graphs in Fig 12.2 of the text. Convince yourself that the wavelength is the SAME for the corresponding graph next to it (count quarter wavelengths if you want to compare). Convince yourself that antinodes and nodes have all switched positions. Convince yourself that there is always a pressure node at the two ends, and always a displacement ANTInode at the two ends... This just about finishes all that we need to know about OPEN pipes. Couple of summarizing comments: Open pipes have harmonics just like string instruments. The formula f n = n f 1 where f 1 = v/(2l) is the same as strings, for the same reasons. The only difference is that "v" now represents the speed of sound in air (rather than the speed of waves on a string) The formula is a little bit more approximate for pipes than for strings, because in reality you don't have exactly zero overpressure at the ends. It's also generally the case that that pressure node is a little bit beyond the end of the pipe, which means "L" is just a hair longer than the pipe length. In fact, as the DIAMETER of the pipe (which we've totally ignored so far) gets bigger, that effective length gets a little bigger still. That's called the "end correction", it usually won't matter much, except for really fat organ pipes. (The text says you can add 0.3*diameter to your effective pipe length, as a rule of thumb)
Phys 1240 Fa 05, SJP 12-5 DRAFT Closed pipes: When we talk about closed pipes, we really mean "half closed", i.e. open at one end, and closed at the other. (If it was closed at BOTH ends, no air could go in or out, and you wouldn't have a very good musical instrument, because it would be hard to hear the sound that resulted. There is an exception to this - when you stick a closed tube up to your ear, sealing it around your ear. Now you DO have a doubly-closed tube, and you CAN hear the harmonics. This will be a hw problem for you to think about - you'll find the situation is pretty similar to the doubly open tube. Try to work it out, make sense of it yourself. Are there nodes at the end? pressure or displacement nodes? What does that tell you about the allowed wavelengths? But we'll stick to "closed => half-closed" for the rest of these notes) So let's zoom in on a pipe that's open at the right end, but sealed/closed at the left end. Your text draws the pressure graphs for the various allowed standing waves in Fig. 12.3. Let's try to make sense of those. We'll start with the "pressure" graph, I find it the easiest to think about. So, at the right end, it's open, same as what we just discussed. The pressure will be "atmospheric" where it's open to the atmosphere, so you'll have zero overpressure there. So when we try to graph "allowed standing waves", we MUST have our pressure graph go to zero at x=l. (Look at Fig 12.3, the right hand graphs are the pressure graphs. Notice they all go to zero at the right hand end) What about the left end, that's the closed or sealed end. Now, if a tube is sealed, there is NO reason to expect the pressure will stay at atmospheric all the time. If air rushes into the tube, it'll get HIGH pressure there next to the seal. And if air rushes out, it'll go to low pressure. In fact, at the left end (x=0), the molecules can't move, which (as we discussed on the previous page) corresponds to the MOST over and underpressure possible, it's a pressure anti-node there. So here's the puzzle. Can you draw a wave which has a node at x=l, and an anti-node at x=0? That's our goal. If you can draw it (if it "fits"), then you have a standing wave! Let's start with the simplest, the longest possible wave... So we should just have one QUARTER of a wave in our picture, because a quarter of a wave is all you need to go from a zero (node) to a peak (antinode). That's the picture here on the right. It's the fundamental (lowest frequency) for a closed tube, n=1. Look: we have λ/4 = L, one QUARTER of a wavelength fits. That means (since f λ = v), that the fundamental frequency is also different than before: f 1 = v/λ = v/(4l). It's divided by 4, rather than 2 (like before), so we have a LOWER fundamental frequency for closed tubes. Your ear canal is a closed tube (open to the atmosphere on one end, closed by the eardrum on the other). The length is about 3 cm, which means f 1 = (344/(4*.03m)) ~ 3000 Hz. This is the lowest frequency where your ear itself "resonates", and that's why the Fletcher-Munson curve is a minimum at about this frequency - you are most sensitive to waves of this frequency, because your ear canal actually resonates with the sound!
Phys 1240 Fa 05, SJP 12-6 DRAFT Now, what's the NEXT harmonic that's allowed? You might want to just assume that it's going to be like before, 2f 1, but that's not right. We have to draw the picture to find out why! Once again, I challenge you to try to draw a real sin wave that starts at zero, wiggles past the first max, goes back to zero, and then gets to the NEXT maximum. (Because that's what we need, it has to go from zero at one end to max at the other end. The fundamental was where it just went directly, and now we're drawing the NEXT possibility). Here's the picture: (focus just on the solid curve first, convince yourself it's what we want. Zero, to the SECOND max possible...) Now you have to think about what the wavelength of this funny wave is. You can do it, try! I claim that we have THREE quarter wavelengths fitting in this picture. (Don't take my word for it. Look at the picture, trace out the quarter waves!) So (3/4)λ = L, which means the corresponding frequency will be f=v/λ = v/(4l/3) = 3 (v/(4l)) I did a little algebra there. Don't let it "slide by", work it out, see how all those 3's and 4's work out! In the end, notice how it compares to what we had for the fundamental, which was v/(4l). This new one, the NEXT allowed harmonic, is not twice the fundamental, it is THREE times. And if you look at the bottom right picture in Fig 12.3, you should be able to convince yourself that the next allowed wave that fits will be FIVE times the fundamental. So we have a different harmonic pattern with closed pipes. We get the fundamental (which is LOWER than expected, by a factor of 2). And then we only get the ODD harmonics, i.e. 3f 1, 5f 1, etc. Some instruments are closed tubes, like a clarinet or an oboe. You have a reed which sticks into one end, but the END is still basically sealed. (The OTHER end is of course open, that's where most of the sound energy "leaks out"!) If you look at the spectrum of those instruments, you'll see that every other harmonic is absent (or nearly so). It makes for a rather characteristic timbre for those instruments! Organs have a mix of open and closed pipes. They will used closed ones for the very lowest tones (because they can be half as long to get that same low note!) Notice also that the NEXT higher frequency where your ear canal will resonate is three times the fundamental, about 9 or 10 khz, which is the next dip in the Fletcher-Munson curve! (Cool, this stuff really works) Bottom line: Half open pipes (also called "closed") have fundamental f 1 = v/(4l) And then the only allowed harmonics are ODD multiples of that.
Phys 1240 Fa 05, SJP 12-7 DRAFT We've been talking about the allowed harmonics in pipes. One thing I've slightly glossed over is how you get them going in the first place. How do you add energy to the pipe at many frequencies (so that the resonant frequencies can get "amplified") With a string it's easy - you just need to pluck it or bow it, that gets the string moving. Plucking is kind of obvious - you displace different parts of the string by different amounts, and let it go - all the different parts start to oscillate, it's like you created a starting "waveform" given by the shape of the initial pluck. That waveform is built out of all the harmonics, so they're all there to start with. The bow does something a little different to the string - you move the bow smoothly, but the string sticks and displaces, then (randomly) slips. This "stickslip" motion makes the string vibrate, and it's a more random/chaotic motion than the smooth motion of the bow. This is a way to generate all sorts of frequencies, and only the resonant ones (the harmonics) will build up. You're "feeding" energy in at all frequencies, and the harmonics will amplify. With a wind instrument, you also need to produce pressure fluctuations. If you just blow smoothly, you get steady flow of air, which tends not to "oscillate". You need to somehow break it up, get it unstable. You want something that is analogous to the bow. There are a variety of methods. One is to blow the air across a hard, sharp edge. The resulting turbulent flow of air contains lots of different components (different frequencies) You can also get this by having air come out of a small hole, and enter a big space, or pass through a narrow gap. All of these are "edgetones" instruments - some variant on this idea. Recorders, flutes, organ pipes are all examples. Alternatively, you can have a vibrating reed, where some physical object "buzzes" as the air flows past it. It's sort of like the edgetone, but the edge is not fixed, there's a kind of feedback between the buzzing reed and the oscillations of pressure in the air. It's a more complex kind of system, but in the end produces sound in ways quite analogous to what we've been discussing above. It's all just different mechanisms to excite those harmonics! There's yet another class of instruments, which are "reed like", but have no actual reed. These are the brass instruments, like the trumpet or trombone. Here, the thing that vibrates is YOU, your lips can play the role of the reed! Again, there's a kind of dynamic feedback - as the cavity approaches a natural resonant frequency, your lips start to vibrate at that same frequency, which increases the amount of energy being fed in. This is positive feedback, and allows you to hit and hold a note strongly, loudly, and consistently. There's yet another piece to this story - the shape of the instrument. Clearly the length is the most important, "L" determines the frequency. (Opening or closing holes, or stretching the trombone, or any of a variety of tricks can change the size of the vibrating cavity to get different fundamentals. And different excitations like blowing harder or softer might get a different harmonics, so that e.g. a bugle player can get a couple of octaves without changing the length by exciting different modes) But there's another aspect too. If, e.g. the inside of the instrument is flared (conical) instead of straight (cylindrical), you can subtly change the frequencies of the harmonics. I guess the best way to think about it is if you go back to the "aside" that the node actually occurs a LITTLE bit to the outside of the end. If the end is flared, higher harmonics might have that node shift to slightly different spots! So, the harmonics might not actually be "perfect", and this adds an interesting shift to the character of the sound you hear. We do like perfect harmonics, but a little variation can be interesting too. There's one last wind instrument we should really talk about - the human voice! Here, the vibrating vocal cords play the role of "reed", and your mouth makes the length and shape of the instrument. Perhaps we'll come back to this one if we have time!