# Woodwind Reed Susceptance

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Hello musicians and everyone!

About woodwind instruments, I mentioned sometimes the reed susceptance. Here are explanations.

This sketches a clarinet mouthpiece and reed. A double reed works similarly
https://en.wikipedia.org/wiki/Double_reed

The pressure oscillations of the air column move the flexible reed. When the pressure is lower, it closes the reed. When it's higher, the more open reed lets more breath pass. More throughput when the downstream pressure is bigger is the opposite of a usual obstacle or loss: it's a negative conductance that provides power to the oscillation of the air column. That's known.

The reed must be closed when the downstream pressure is low to sustain the oscillation. This needs that mainly its stiffness determines the position. Its inertia would close the reed when the downstream pressure is high, damping the oscillation. That is, the reed's resonance is higher than the instrument's notes. It can be heard as a high pitched hiss if the musician takes the mouthpiece much too far, and during articulation using an imperfect saxophone mouthpiece.

Similarity of acoustics with electricity lets replace Pa by V, m3/s by A, then we use ohm and siemens: S=ohm-1. For sine U and I of given frequency, Y=G+jB where Y is called admittance, G conductance and B susceptance, while Z=R+jX where Z is called impedance, R resistance and X reactance, with j2=-1. We think of sine sounds and linear reeds to understand but none is. Horribly common.

If a clarinettist blows 5L in 25s which the reed modulates by full 200µA peak while the pressure oscillates by 0.1bar peak, the reed's conductance is -20nS=-(50Mohm)-1. The air column of D=14.6mm has a wave impedance of 2.5Mohm and its losses are less than (50Mohm)-1 to oscillate with this reed, resulting from a strong resonance consistent with stiff intonation.

As the pressure oscillation moves the reed, the sweeping area displaces air. Here pressure lets absorb a current (air throughput) when it rises, not when it's low or high, and lets provides a current when it falls. Like air's compressibility does, the reed adds a susceptance representable by a capacitor. I haven't seen this in books nor research papers, and at least one instrument maker ignored it. But bassoonists know it by experience while clarinettists may ignore it, not to mention organists.

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A Bb clarinet mouthpiece can have 18mm facing length and 1.1mm tip opening, which the clarinettist reduces to a variable amount as he presses the reed with his jaw covered by the lip. Let's take 12mm width, remaining 15mm length and parabolic 0.8mm opening: the 38mm3 displaced by 0.05bar make 7.6pF, as compliant as 1.1cm3 of air, equivalent to 6.4mm length of air column; for a low note it's 1.2% of the length or 0.2 half-note. This is consistent with how much a clarinettist can pull the pitch with his embouchure.

A bassoon's double reed can be 3mm open for low notes, mean 10mm wide and move over 10mm length. 0.02bar moving most of the 67mm3 make 33pF like 4.7cm3 of air. At the bassoon's narrow D/L=0.02 cone, it's as much as 0.36m from the apex or 12% of the length. Consistently, a bassoonist can pull the pitch by half a tone with his embouchure, and tunes the instrument by cutting the reed.

When sounding a double reed alone, or a single reed on the naked mouthpiece, the reed's capacitance resonates with the outlet's inductance. Where the bassoon's bocal fits, L=20mm D=3mm make 3.5kH that resonate at 470Hz, not too bad estimate, and much lower than the cane's flexural resonance.

At a double reed, the embouchure varies the tip opening and the mobile width, but the mobile length little. At a single reed, it varies the mobile length and tip opening but not the width. The equivalent capacitance can diminish much more with a double reed, possibly because the opening can reduce to nothing; one can pull the resonance of a naked double reed much more than of a single reed.

The embouchure influences also the resonance mode of the air column - the register. By the reed's conductance or the susceptance, which are about as big? I'm not quite sure. The reed and mouthpiece must fit an instrument to sound its full range. At a bassoon, the better controlled reed lets play all the range without the lone register key, which wouldn't suffice for the range, is built little efficient and renamed "whisper key". As opposed, a clarinet would be unplayable without its speaker key.

Marc Schaefer, aka Enthalpy

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To estimate reed susceptances, I took a pressure and a displacement a priori. Both are imprecise, the estimation is worse.

Measuring the pressure oscillation and the reed's movement under the same conditions would be better.

More simply, we can load the reed with a known inductance and measure the oscillation frequency. If the volume of the chosen inductor is small, much capacitance comes from the reed, whose susceptance is accurately measured.

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For a bassoon reed, I estimated here above a capacitance of 33pF at low notes, equivalent to 4.7cm3 air. The naked reed has a near-cylindric part where it fits on the bocal, whose approximately L=20mm ID=4mm contain only 0.25cm3. The wider tip of the reed contains some 0.15cm3.

The reed's capacitance is the major contribution and remains important with a stiff embouchure that produces higher notes (more than an octave higher on a naked bassoon reed). A naked reed lets measure the susceptance on the whole range, including at some points to help analyze or design an instrument: play the reed on the instrument and alone with the same embouchure.

The inductance must be modelled to deduce the capacitance from the frequency. The reed's cone is deduced from the reamer. Or a short narrow tube of accurate machined dimensions can fit the reed. L=0.1m ID=4mm would make 10kH that resonate the capacitance at 300Hz to 800Hz, in the instrument's range.

About 1/3 of the tube's 1.3cm3 add to the reed's capacitance - or model that better or use a shorter narrower tube at 600Hz, since 100mm are 0.23*lambda there. Flow and thermal losses make only Q~10 at 300Hz. A few tubes let check that the model is sensible.

I have nothing here to make the experiment, alas.

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An oboe reed is said to be W=5mm h~1.3mm and vibrate over 10mm, swinging by +-7.4mm3 at low notes from +-5kPa (all vague estimates). The resulting 1.5pF act as 0.21cm3 more.

An added tube of L=30mm ID=1.2mm would bring 32kH to resonate at 0.7 to some 2.2kHz. It would give only Q~5 at low notes but is already 0.19*lambda long at high notes, so several tubes would improve.

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A reed for soprano clarinet is 13mm wide, a known mouthpiece opens by 1.15mm over 20mm length, but the musician reduces that: I take 15mm and 0.65mm for low notes. If peak 5kPa close the reed, the reed's capacitance is 8.5pF, as much as 1.2cm3 air more.

The mouthpiece contains over 10cm3 air, hampering a measure as is. Shortening holes wouldn't bring enough.

Instead, we can put a tube of ID=3mm L=80mm in the mouthpiece and fill the rest with modelling clay. A pyramidal void in clay, W=10mm L=20mm h=5mm over the reed around the tube's end, adds 0.3cm3 which must be measured. Then, 14kH by the tube let resonate at 420 to maybe 800Hz, in the instrument's range. Q~9 at 420Hz, L~0.18*lambda at 800Hz, so the tube isn't just an inductor. Two tubes would improve.

Bass clarinets and low saxophones make it easier.

Marc Schaefer, aka Enthalpy

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