# 12.2.4 Reflection at a clavichord tangent

In order to obtain a satisfactory model for the modal damping in a clavichord string, we need to allow for a new energy dissipation mechanism in addition to the ones introduced in section 5.4. When the player’s finger holds the tangent in contact with the string, this does not make a rigid boundary condition on the string: small movement of the tangent can occur, which allows some energy to leak past the tangent, to be dissipated by the felt woven into the non-playing lengths of string. We will use a very simple model to investigate this leakage: an ideal flexible string, with the tangent/key/finger combination modelled as a mass $M$ and a damper (or dashpot) with strength $d$, as sketched in Fig. 1.

We imagine an incident wave with frequency $\omega$ and unit amplitude arriving at the tangent from the left-hand side. Interaction with the tangent generates outgoing reflected and transmitted waves as sketched, with (complex) amplitudes $R$ and $S$ respectively. The transmitted wave will be dissipated by the felt, so that no waves arrive back at the tangent from the right-hand side. So for $x \le 0$ the transverse displacement of the string is

$$w(x,t)=e^{i \omega (t-x/c)} + R e^{i \omega (t+x/c)} \tag{1}$$

while for $x \ge 0$

$$w(x,t)=S e^{i \omega (t-x/c)} \tag{2}$$

where $c=\sqrt{T/m}$ is the wave speed on the string, $T$ being the tension and $m$ the mass per unit length.

At $x=0$ we have two conditions, which will give us the two equations we need to solve for $R$ and $S$. First, we have the continuity condition: the string is not broken at $x=0$ so we must get the same answer for $w(0,t)$ from both expressions. This requires

$$1+R=S . \tag{3}$$

Second, we have the equation of motion for the mass $M$:

$$M \dfrac{\partial^2 w}{\partial t^2} = -d \dfrac{\partial w}{\partial t} + T\left[ \left. \dfrac{\partial w}{\partial x} \right|_{x=0+} – \left. \dfrac{\partial w}{\partial x} \right|_{x=0-} \right] \tag{4}$$

so from equations (1) and (2)

$$(-M \omega^2 + i \omega d)(1+R) = \dfrac{i \omega T}{c} (-S -R+1) . \tag{5}$$

Combining with equation (3) and rearranging gives

$$R=-\dfrac{i \omega M + d}{2 Z_0 + i \omega M + d} \tag{6}$$

where $Z_0=T/c = \sqrt{T m}$ is the wave impedance of the string. Now from equation (6) of section 5.1.2 we know that the associated loss factor $\eta_t^{(n)}$ for the $n$th mode of the string is given by

$$\eta_t^{(n)} \approx \dfrac{1 -|R|^2}{2 \pi n} , \tag{7}$$

so substituting equation (6) and simplifying we obtain

$$\eta_t^{(n)} \approx \dfrac{2 Z_0}{\pi n} \left[\dfrac{Z_0 + d}{(2Z_0+d)^2 + \omega^2 M^2} \right] . \tag{8}$$

To obtain a complete model for the damping of the string, we need to combine this with the other loss factors we already know: energy loss at the other end of the string due to coupling with the clavichord soundboard is described by $\eta^{(n)}_{body}$ from section 5.1.2; and energy loss due to viscosity in the surrounding air is described by $\eta^{(n)}_{air}$ from section 5.4.5. Finally, we have energy loss internal to the string itself. For metal strings as on the clavichord, the internal damping of the material is very low, and it will turn out to be good enough to include a crude estimate by simply adding a small, constant background loss factor $\eta_{background} = 10^{-4}$. So the combined modal loss factor $\eta^{(n)}$ is given by

$$\eta^{(n)} = \eta_t^{(n)} + \eta^{(n)}_{body} + \eta^{(n)}_{air} + \eta_{background} . \tag{9}$$

In order to obtain an estimate of $\eta^{(n)}_{body}$, we need the bridge admittance. Figure 2 shows the measured admittance of the clavichord soundboard, close to the position of the note chosen for analysis. It also shows, in a blue dashed curve, the corresponding admittance for a harpsichord that we will come to in a moment. Both soundboards, being plate-like, show a broadly horizontal trend. For our present purpose, it is enough to estimate the mean level of each: the clavichord is around –35 dB, the harpsichord around –45 dB.

Figure 3 shows a plot of the various loss factor contributions for a particular clavichord string, which had length $L$ = 0.99 m and diameter 0.48 mm. The new factor, $\eta_t^{(n)}$, is plotted in a red dash-dot line. This has been calculated using the values $M=10\mathrm{~g}$, $d=10\mathrm{~Ns/m}$, chosen to give a good fit to measurements to be shown in Fig. 4. The other contributions are shown in various types of broken blue line. The sum of all the blue contributions gives the solid blue curve; adding in $\eta_t^{(n)}$ gives the solid red curve. It can be seen that $\eta_t^{(n)}$ dominates at the lowest frequencies, causing a big divergence between red and blue curves. Note that $\eta^{(n)}_{body}$, shown in the dashed blue line, makes only a rather small contribution, which is why we can get away with a crude estimate of the bridge admittance.

The red curve in Fig. 4 shows the Q-factors calculated by inverting the loss factor shown in the red curve in Fig. 3. This can be compared to the set of measured Q-factors, the red stars. The blue circles show corresponding measurements from the same note played on a harpsichord string with length 1.48 m and diameter 0.36 mm. The prediction for this string is shown in the blue curve. This is similar to the inverse of the blue curve in Fig. 3, but not quite the same because it has been calculated with the different length and diameter of the harpsichord string, and also with the different bridge admittance estimate. For both strings, the model captures the trend of the data very well.