How does a Western free reed work? (musical instruments)

  • #1
Freixas
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In this post, I will describe, as best I have been able to determine, how a Western free reed works. As it is unlikely I will run into a physicist here who specializes in reeds, I will just ask people to speculate on the behavior of a reed given what you know about fluid dynamics or the mechanics of springs.

As Western free reeds are used in a variety of musical instruments, they have been widely studied. My problems are in finding the research, getting access to it, and understanding it. I am not a physicist. I am looking for a lay person's description of the reed's behavior, but one grounded in physics. Instruments which use the reed include the harmonica, accordion, concertina, melodica and even the organ.

It's called a free reed because it moves freely through an opening. It's called a Western free reed because it sits above the opening—Eastern free reeds sit within the reed plate.

Let me start with an introduction to a Western free reed:
Reed Block.jpg
The entire assembly is called the reed, although the tongue is also called the reed. The tongue is the flexible piece of metal that vibrates and produces sound. It is attached to the reed plate with a rivet, screw or weld. Near its base, the tongue lies just above the hole in the reed plate. Part way down its length, the tongue bends upward so that the tip lies about as high over the hole as tongue's thickness—this is called the gap. The whole thing sits on an exhaust port. When we want a reed to vibrate, we open the exhaust so that air can flow past the tongue, through the reed plate hole and out the exhaust.

The hole in the reed plate is almost the same size as the tongue. There is an extra 0.02–0.03 mm on the sides and 0.1 mm at the tip.

I am going to focus on one instrument, the melodica. Reeds in this instrument generally sit in a large air chamber into which the performer blows. When a key is pressed, an exhaust port is opened and the airflow begins. Here's what I think happens (with questions):
Diagram1.jpg
  1. No air is flowing. The static pressure on either side of the tongue is equal (presumably its the same as outside pressure).​
  2. The exhaust port is opened. Since the air chamber is large, air speeds (dynamic pressure) in it are low and static pressure is high. Air flows through the gap. This drops the static pressure below the tongue and the tongue is forced down into the gap. I've estimated that with a volumetric flow if 0.5 L/s, the air speed is about 41 m/s.​
  3. This is where it gets interesting. When the reed moves into the hole, many sources say the airflow is blocked. In fact, there is still a small space between the tongue and the reed plate walls. It is very small, about 1.552-6 m2. Air flowing through this opening would need a speed of around 322 m/s to follow the continuity formula. To the extent that the flow is blocked, the static pressure below the reed rises to match the external pressure.​
  4. The tongue moves below the plate. This allows air to vent. The spring metal has a force that wants to move it back to its resting position, so it begins to rise.​
  5. We've blocked the flow again. The theory is that on the upswing, the static pressure below the tongue is not as low as it was at #2, so there is little pressure difference blocking the upward movement.​
  6. The spring metal now wants to move downward and, as the gap closes, the static pressure below the tongue drops. The cycle repeats.​
Let's get back to #3. Is the airflow actually blocked? If air is flowing in at 0.5 L/s, it has to go somewhere. In this case, my thinking is that it has two options: accelerate to 322 m/s to go through the small gap or push the tongue down to expand the volume of space available to the air. I'm not sure how to describe what might be happening in terms of fluid dynamics. Is there an increase of static pressure above the tongue? Is the air (which I normally treat as incompressible in this system) actually being compressed to create the static pressure increase? It's not clear to me.

In #4, I'm guessing that the flow past the tip is somewhat aimed toward the final exhaust (which is toward the right) and actually serves to lower static pressure above the tongue and begin raising the tongue. I haven't studied springs at all, but I'm pretty certain that the further the metal bends, the greater the force to return it to its starting position.

#5 seems a little iffy if we accept the arguments made for #3. Air is still flowing in and still wants to exit. It should still want to push the tongue down. The static pressure below the reed, however, should be higher than in #3, so it won't assist this movement. But the main force driving the reed upwards would seem to be the spring.

These are the places that I'd love to hear what people think. Now, let me add some bonus questions:
Diagram2.jpg
This is a variant of #4 in which the exhaust walls (at least, near the reed plate) have been narrowed. I think with this design, the reed might function poorly since a region of low static pressure might be created below the tongue tip and would resist the upward movement. I would think #4 would be the better design.
Diagram3.jpg
In this alternate, we seal the reed in an airtight box except for the opening to the main air chamber. Speeds in the main air chamber are low and pressure is high. By channeling the air through a narrow tube, we increase the dynamics pressure and lower the static pressure in this reed box. Whereas before, the static pressure was acting on all parts of the tongue equally, we can now focus dynamics pressure right on the tip. Since the tongue also acts as a lever, this may make it easier to push the tongue down. Conversely, it might make it hard to get it to move up.

This might be a difficult question to answer. I can model airflow with SimScale, but only with a static geometry. I could build a reed box and try it out, of course, but whether the reed sounded or not, it wouldn't help me understand what was happening.

There's no penalty for guessing on any of this. Hopefully, some of you will make higher quality guesses than I'm making.

By the way, the Western free reed makes a sound only when air flows in one direction. If you study the six steps again, but imagine the flow in the opposite direction, it should be clear why—there will never be a point at which the air is blocked (or mostly blocked). The reed will probably bend upwards a bit, but won't vibrate.
 

Answers and Replies

  • #2
sophiecentaur
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Air flowing through this opening would need a speed of around 322 m/s to follow the continuity formula.
I'm not sure that air would have to flow continuously. The local 'DC' flow could be interrupted and bidirectional over the cycle.
 
  • #3
Freixas
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35
I'm not sure that air would have to flow continuously. The local 'DC' flow could be interrupted and bidirectional over the cycle.

Thanks. Could you elaborate?

If the input flow is a continuous 0.5 L/s, what does it mean to say that the flow is not continuous? Where does the air coming into the system go if an equal amount is not flowing out?

By the way, I've been going through research papers and seen simulations that enforce the idea that, when the tongue enters the gap, air speeds drop. Unfortunately, the graphics just showed air speeds and not pressures, which would have completed the picture.

I can only imagine two options for the extra air: compression (more air in the same space) or bending of the tongue (increasing the air chamber space to accommodate more air). I suppose there's a third option: the air flow stops, but one of the other two options seems more likely.

The term "DC" means Direct Current to me. I don't think I know what you mean by the term. By "cycle", I assume you mean the reed vibration from 1-6. But I don't understand how you would have a bidirectional flow during this cycle. I'm having problems just understanding a unidirectional flow. :-)
 
  • #4
sophiecentaur
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Where does the air coming into the system go if an equal amount is not flowing out?
It's a gas and the pressures could be high when the vane passes through the gap. Constant mass is not necessary.(?)

Edit: I was being sloppy in my use of DC in the electrical sense. The instantaneous motion of a small volume could be either direction, even if the mean flow is one way. The peak velocity of a resonating reed would be much higher than the mean flow velocity so 'on the way back through the hole, it could be taking a layer of air back with it.
 
  • #5
Freixas
279
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It's a gas and the pressures could be high when the vane passes through the gap. Constant mass is not necessary.(?)

Edit: I was being sloppy in my use of DC in the electrical sense. The instantaneous motion of a small volume could be either direction, even if the mean flow is one way. The peak velocity of a resonating reed would be much higher than the mean flow velocity so 'on the way back through the hole, it could be taking a layer of air back with it.

I think the reeds velocity is way slower than the air speeds. Let's assume that tip travel is about 4mm each way. At 440 Hz (concert A), that means 8mm/.002272 seconds or (if my math is right) 3.52 m/s. I've already calculated airflow through the gap starting at 41 m/s and going up from there.

Air speeds vary throughout the volume and over time, so it's difficult to say exactly what's going on at each point. I can model the airflow using a static model of the geometry, but I'd like to include the forces from the tongue (functioning as a spring). When the tongue is going down (#3), I doubt there's a bidirectional flow. When it's going up (#5), having the tongue reverse the airflow will just slow the tongue down, so it would still seem the spring tension is the the main driving force at this point--at that point, the tongue is fighting against the airflow, which would seem detrimental to getting a nice, even vibration.

When you say constant mass is not necessary, I translate this to mean that the fluid density could increase (locally, at least). This sort of follows my guess that the pressure above the reed increases (because the fluid density increases). This forces the tongue down, which increases the volume of the air chamber and probably drops the air pressure. Again, there's a lot happening and I'd like to be able to model the tongue along with the air.

If you know of any free tools to do that, let me know. :-) I'll keep looking to see if I can find any existing research that models the entire cycle. Someone must have done this already.
 
  • #6
sophiecentaur
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. I've already calculated airflow through the gap starting at 41 m/s and going up from there.
I'm not saying you're wrong about that but what are your assumptions about air flow through the instrument? The air flow through the gap is varying. If you put a shutter in the way for a short time, the mean flow wouldn't change noticeably so can you assume that some air has to be flowing all the time in order to get the flow rate you're using? So the air flow is modulated - is that a problem?
(I'm only using intuition here.)
 
  • #7
Freixas
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I'm not saying you're wrong about

Actually, it's fine to say I'm wrong. :-) It's quite likely I have something wrong. I'm new to all this stuff. You're the one with the Science Advisor label.

but what are your assumptions about air flow through the instrument

I'm basically relying on the continuity equation: if 0.5 L/s flows in, then 0.5 L/s flows through any cross-section of the system.

If you put a shutter in the way for a short time, the mean flow wouldn't change noticeably so can you assume that some air has to be flowing all the time in order to get the flow rate you're using?

Well, if 0.5 L/s of air is flowing in, it's got to go somewhere. Although it's for a short time, what happens during this short time is exactly what I'm interested in. The mean flow is not useful.

It is possible that the input airflow is blocked temporarily. It's one of three possibilities: the flow is blocked, the air is compressed or the air chamber expands (or some combination of any of these). I don't know of other possibilities.

I have an intuitive idea about the fact that it's the weakest of these that occurs first. For example, if the flow can't be stopped (imagine that it's driven by a piston backed by 400 horsepower) and if the air chamber can't expand, the air will be compressed. When the air can not be compressed further, the air chamber will expand, even if it has to blow up to do so. If the air chamber is made of indestructible material, the 400 HP piston will seize up.

Of all the possibilities, bending the tongue to increase the air chamber size seems like the first thing to go. But air has a lot of empty space; maybe a little bit of compression would occur first, until the extra pressure bent the tongue. Least likely would seem to be the flow stopping. These are just intuitive guesses, though, not calculations. Intuition is not reliable in this kinds of things.
 
  • #8
sophiecentaur
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Well, if 0.5 L/s of air is flowing in, it's got to go somewhere.
This is your problem. If air is flowing into the chamber at a constant rate, that doesn't mean it's flowing out at that constant rate. There is a chamber with finite volume involved and the mass of air in that volume can change temporarily because the pressure can change; it doesn't have to 'go somewhere'. The electrical equivalent is a Capacitor across the input of a load, fed from a constant Current source (that's your 0.5l/s) into a constriction. There will be a certain mean voltage (pressure, in your case) and that voltage will go up or down, according to the instantaneous resistance of the load.
So your assumption needs to modified. (You couldn't blow hard enough to maintain constant flow through the gap.)
I have a harmonica and I can 'bend' notes by sucking harder. Your post says all this, effectively and your high power compressor would correspond nearer to a constant current source - but you'd still have the Capacity of the chamber there.
 
  • #9
Freixas
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This is your problem. If air is flowing into the chamber at a constant rate, that doesn't mean it's flowing out at that constant rate.

Fair enough. When analyzing airflow through an open system, I learned that the air is treated as incompressible. However, I've also know that this is an approximation.

However, I think you are simplifying the analysis quite a bit when you focus on just this aspect. There is a lot going on and the compression of some of the air above the reed is just a small part of it.

I asked for speculation, but I'm beginning to think the question is a bit unfair. Even the experts run simulations to analyze the behavior of a reed and, if I can't run a simulation myself, I may just have someone who has. Sadly, many of the interesting research papers are behind paywalls.

As for harmonicas and bending, that requires a much more complex analysis as there are two reeds with different tunings and the mouth/throat gets involved. I need to learn to walk before I try to run. :smile:
 
  • #10
256bits
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I've estimated that with a volumetric flow if 0.5 L/s,
Isn't that a bit high for a person to blow
 
  • #12
Freixas
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Isn't that a bit high for a person to blow

No. 0.1 L/s to 1 L/s is a reasonable range. You can sustain 0.1 L/s much longer than 1 L/s, obviously. See https://newt.phys.unsw.edu.au/jw/air-speed.html

As most of the effects we're examining would occur stablize well within 1 second, we could use even higher flow rates. A brief burst of 10 L/s is not impossible for those with big lungs.

Examining the behavior of the reed with different volumetric flows would add another interesting dimension to what is already a complicated problem, though.
 

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