A simple model for a transverse flute, to allow us to investigate the effect of the cork position, follows straightforwardly from earlier calculations we have seen in sections 11.1.1 and 11.1.3. The geometry of interest is sketched in Fig. 1. Three cylindrical tubes meet at the point O, which we will use as our origin of coordinates: an open-ended tube of length $L$ and cross-sectional area $S$, a closed-ended tube of length $b$ and cross-sectional area $S$, and a short open-ended tube of length $h$ and cross-sectional area $C$ (representing the embouchure hole, including its end corrections).

From section 11.1.1 we already know general expressions for the acoustical pressure and volume flow rate at a frequency $\omega$, in any cylindrical pipe. The total pressure $p$ in the pipe can be written

$$p=Ae^{i(\omega t – kx)}+Be^{i(\omega t + kx)} , \tag{1}$$

and the corresponding total volume flow rate is then

$$v=\dfrac{AS}{Z_0}e^{i(\omega t – kx)}-\dfrac{BS}{Z_0}e^{i(\omega t + kx)} \tag{2}$$

where $S$ is the cross-sectional area, $Z_0$ is the characteristic impedance for sound waves in air, and the wavenumber $k=\omega/c$ where $c$ is the speed of sound. Distance along the pipe is measured by $x$, and $t$ is time as usual.

The coefficients $A$ and $B$ describe (complex) amplitudes of waves travelling in the positive and negative $x$ directions, respectively. We can use corresponding expressions for our three tubes. We can measure $x$ outwards from the origin O along each of the three tubes, and we can use coefficients $(A_1,B_1)$, $(A_2,B_2)$ and $(A_3,B_3)$ respectively, as labelled in Fig. 1. For the embouchure tube we replace $S$ in equation (2) by $C$.

We now need 6 equations for these 6 unknown wave amplitudes. At the open end of the main tube we require $p=0$ at $x=L$, so

$$A_1 e^{-ikL} + B_1 e^{ikl} =0. \tag{3}$$

At the closed end of the shorter tube we require $v=0$ at $x=b$, so

$$A_2 e^{-ikb} – B_2 e^{ikb} =0. \tag{4}$$

At the junction, we require several things. The pressures in all three tubes must be the same at $x=0$, so that

$$A_1+B_1 = A_2 + B_2 = A_3 + B_3 . \tag{5}$$

Also, the combined volume outflow rate from O must be zero, so that

$$\dfrac{C}{Z_0}(A_3 – B_3) + \dfrac{S}{Z_0}(A_1 – B_1 + A_2 – B_2) = 0 . \tag{6}$$

Finally, our aim is to calculate the input admittance at the open end of the short tube representing the embouchure hole. We can impose unit pressure at this point by setting $p=1$ at $x=h$ so that

$$A_3 e^{-ikh} + B_3 e^{ikh} =1, \tag{7}$$

and then the admittance $Y$ is given by the value of volume flow rate at this point:

$$Y=\dfrac{C}{Z_0}\left( -A_3 e^{-ikh} + B_3 e^{ikh} \right) \tag{8}$$

where a negative sign has been introduced because for input admittance the relevant volume flow rate is *into* the tube, not out of it.

After some straightforward but tedious algebra, the linear simultaneous equations (3)–(7) can be solved to give the admittance in the form

$$Y = \dfrac{S (\cot kL – \tan kb ) – C \tan kh}{i Z_0 \left[ \tan kh (\cot kL – \tan kb ) +1 \right]} . \tag{9}$$

In order to plot the admittance shown in Figs. 7, 8 and 9 of section 11.8, some parameter values were estimated to give a reasonable match to the measured admittance. The length $L$ was 0.6 m, the cork position was chosen so that $b$ was 18 mm, the main tube was assumed to be circular with internal radius 10 mm, the embouchure hole was taken to be circular with radius 5 mm, and the effective length $h$ was 12 mm. In addition, the values of $k$ used to compute the admittance were complex, allowing for wall damping in the tube using the model due to Fletcher as described in section 11.1.1.

In order to plot the pressure distributions shown in Fig. 10 of section 11.8, the analysis here was used in a slightly different way. At each chosen frequency, the simultaneous equations (3)–(7) were solved to obtain values for the coefficients $A_1, B_1, A_2$ and $B_2$. These were substituted in the relevant versions of equation (1) to give the pressure distributions in the long tube and the closed tube. These pressures are, of course, complex. For simplicity, the plots show the imaginary part of this complex pressure, which contains the dominant component.