The excitation mechanism for a clavichord string is unusual: it consists of moving one end of the string with a specified displacement $W(t)$, representing the movement of the tangent as it strikes the string and is then brought to rest. Following Thwaites and Fletcher [1], we will assume an exponential form with velocity

$$\dot{W}(t)= V_0 e^{-\alpha t} \tag{1}$$

where $V_0$ is the initial striking speed and $\alpha$ is the exponential deceleration rate. It follows that the displacement is

$$W(t)=\dfrac{V_0}{\alpha}\left(1- e^{-\alpha t} \right) \tag{2}$$

on the assumption that $W(0)=0$. The asymptotic displacement is thus given by

$$W(t) \rightarrow V_0/\alpha \mathrm{~~~ as~~~} t \rightarrow\infty . \tag{3}$$

We will choose a fixed displacement $d$ for this asymptotic value, which thus requires

$$\alpha = V_0/d . \tag{4}$$

We now consider a string of length $L$ with transverse displacement $w(x,t)$ satisfying

$$w(0,t) = W(t) \mathrm{~~ and~~} w(L,t)=0 . \tag{5}$$

We can express this in the form

$$w(x,t)=\sum_n{a_n(t) \sin \dfrac{n \pi x}{L}} + W(t) \left(1-\dfrac{x}{L} \right) \tag{6}$$

where $\sin n \pi x/L$ is the $n$th mode shape and $\left(1-\dfrac{x}{L} \right)$ is called a “constraint mode”, which describes the static response to moving the end of the string at $x=0$. We will analyse this system initially assuming an ideal, undamped string, but once we have expressed the response in modal terms we will be able to add in the effects of bending stiffness and damping.

The governing equation of an ideal string is

$$m \dfrac{\partial^2 w}{\partial t^2} – T \dfrac{\partial^2 w}{\partial x^2} = 0 \tag{7}$$

where $m$ is the mass per unit length and $T$ is the tension. Substituting,

$$m \left[ \sum_n{\ddot{a}_n \sin \dfrac{n \pi x}{L}} + \ddot{W} \left(1-\dfrac{x}{L} \right) \right] + T \sum_n{a_n \left(\dfrac{n \pi}{L}\right)^2 \sin \dfrac{n \pi x}{L}} = 0 . $$

$$ \tag{8}$$

Now we use the usual Fourier series trick: multiply through by $\sin j \pi x/L$ and integrate over the range $0-L$. The orthogonality of the $\sin$ functions over this range means that only the term $j=n$ survives from each of the summation. The result is

$$\ddot{a}_n + \omega_n^2 a_n = -\dfrac{2}{L} \ddot{W} \int_0^L{\left(1-\dfrac{x}{L} \right) \sin \dfrac{n \pi x}{L} dx} \tag{9}$$

where

$$\omega_n= \sqrt{\dfrac{T}{m}}\left( \dfrac{n \pi}{L}\right) \tag{10}$$

is the $n$th natural frequency of the string. The integral on the right-hand side of equation (9) can be evaluated straightforwardly by integration by parts. It has the value $\dfrac{L}{n \pi}$ so that equation (9) becomes

$$\ddot{a}_n + \omega_n^2 a_n = -\dfrac{2}{n \pi} \ddot{W} \tag{11}$$

where from equation (1) we know that

$$\ddot{W}=V_0 \delta(t) – \alpha V_0 e^{-\alpha t} \tag{12}$$

in terms of the Dirac delta function $\delta(t)$.

We can solve equation (11) by expressing the solution as the convolution of the right-hand side with the impulse response $g_n(t)$, which is

$$g_n(t)=\dfrac{\sin \omega_n t}{\omega_n} . \tag{13}$$

So

$$a_n=-\dfrac{2}{n \pi \omega_n} \int_0^t{\ddot{W}(\tau) \sin \omega_n (t-\tau) d \tau}$$

$$=-\dfrac{2 V_0}{n \pi \omega_n} \left[ \sin \omega_n(t) -\alpha \int_0^t{e^{- \alpha \tau} \sin \omega_n (t-\tau) d \tau} \right] . \tag{14}$$

The integral on the right-hand side can be expressed as the imaginary part of a complex quantity:

$$\mathrm{Im} \int_0^t{e^{- \alpha \tau} e^{i\omega_n (t-\tau)} d \tau} \tag{15}$$

$$=\mathrm{Im} \left[ \dfrac{e^{-\alpha \tau} e^{i \omega_n (t-\tau)}}{-\alpha -i\omega_n} \right]_0^t = \mathrm{Im} \left[\dfrac{e^{-\alpha t} – e^{i \omega_n t}}{-\alpha -i\omega_n} \right]$$

$$=\dfrac{\alpha \sin \omega_n t + \omega_n\left(e^{-\alpha t} – \cos \omega_n t \right)}{\alpha^2 + \omega_n^2} . \tag{16}$$

Substituting back into equation (14)gives our final result

$$a_n=-\dfrac{2 V_0}{n \pi \omega_n} \left[ \sin \omega_n(t) -\dfrac{\alpha^2 \sin \omega_n t + \alpha \omega_n\left(e^{-\alpha t} – \cos \omega_n t \right)}{\alpha^2 + \omega_n^2} \right] . $$

$$ \tag{17}$$

At this stage we can augment the model to be more realistic: we can replace the value of $\omega_n$ with the one appropriate for a stiff string with inharmonicity, and we can allow for damping with a modal Q-factor $Q_n$ by multiplying the terms $\sin \omega_n(t)$ and $\cos \omega_n(t)$ by the decay factor $e^{-\omega_n t/2 Q_n}$. Equation (6) then gives an explicit equation for the response of the string.

To calculate the bridge force $f_b(t)$ we can note that

$$f_b(t)= -T\left.\dfrac{\partial w}{\partial x} \right|_{=L} = \dfrac{T}{L} \left[W- \sum_n{n \pi (-1)^n a_n} \right] . \tag{18}$$

The parameter values for the simulations were as follows: string length $L$ = 0.99 m; string diameter 0.48 mm; tuned frequency 73.4 Hz; material properties $E$ = 100 GPa, density $\rho = 8200 \mathrm{~kg/m}^3$; sampling frequency 44.1 kHz.

[1] Suszanne Thwaites and N. H. Fletcher, “Some notes on the clavichord”; *Journal of the Acoustical Society of America* **69**, 1476–1483 (1981).