ttonon Posted February 13, 2015 Share Posted February 13, 2015 Hi Lucasz, your observation that the tongue begins vibrating with an amplitude too small for it to enter the slot in the shoe is important. Years ago, I had noticed that, but I forgot about it when I wrote my previous post. If indeed this happens all the time during the start transient of the free reed, then the actual situation becomes very difficult to explain in detail, without further experimentation. It would be nice if someone made a slow motion video of the starting free reed in order to verify this speculation, complete with smoke entrained by the airflow, so that the air stream becomes visible. I say the actual situation becomes very difficult because I’m very sure that such “pre-vibration” is inherently unsteady. I can’t explain the details on how that works. It may have something to do with turbulence and the shedding of eddies and vortices under the tongue. It seems I have difficulty convincing you of the difference between unsteady flow and steady flow, and in your explanations, you don’t make distinctions. When anyone invokes the so-called “Bernoulli” explanation on how a free reed works, they implicitly adopt a quasi-steady assumption, because the Bernoulli equation referred to is the one for steady (and incompressible) flow. As I wrote before, with this assumption, the gap under the tongue during starting will cause a jet of air, and the pressure all along the bottom surface of the tongue is P2. This is true in part because the pressure inside a subsonic jet must equal the pressure immediately outside the jet. In this case, the jet velocity is the maximum velocity that can be obtained by the pressure difference (P1 – P2), and the pressure difference (P1 – P2) is the maximum for the system. Let me try another way. Bernoulli can be written: P1 + 0 = p + 0.5*rho*v^2 The zero is because P1 represents a quiescent reservoir. We can write this as P1 – p = 0.5*rho*v^2 I’m saying that p = P2, which is the lowest pressure in the system, and thus, (P1 – P2) is the maximum pressure difference that the air could experience – when unsteady effects are not important. Thus, when you adopt a quasi-steady approach and say that there must be a pressure force greater than (P1 – P2), you are equivalently saying that p is negative, which is impossible. I’m in favor of discarding the quasi-steady assumption for the very beginning of the start transient, and to accept that the flow field is inherently unsteady, and I am at a loss to explain the details of that flow field. The question I asked in my previous post was in the context of the quasi-steady assumption, and when I make that assumption, I cannot explain completely the start mechanism. And I’m sorry, I don’t understand your explanation. Perhaps this means that the quasi-steady assumption is inadequate to explain the start transient, simple because this assumption is wrong, and one needs to invoke some principles of unsteady flow in order to understand it. The “pre-vibration” you describe suggests to me that the quasi-steady assumption is inadequate. Most other musical instruments contain sound sources that involve unsteady effects, such as the flute, trumpet, and reed instruments, thus, why shouldn’t the free reed? However, for the air motion during periodic motion (speaking), as I mentioned, the quasi-steady assumption may be accurate for the lower frequency reeds. Indeed theoretical papers have been written that make this assumption. Best regards, Tom Quote Link to comment Share on other sites More sharing options...
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