# 5.1.2 Coupling a string to the instrument body

The result of connecting a stretched string to a non-rigid instrument body is that some energy will be transferred from the string into body vibration. To analyse this, we will suppose the string has tension $T$ and mass per unit length $m$, so that it has a wave speed $c=\sqrt{T/m}$ and a wave impedance $Z_0=\sqrt{Tm}$. One end of the string, at $x=0$, is connected to the instrument body, with bridge admittance $Y(\omega)$. The other end, at $x=L$, is assumed to be rigidly fixed.

It is easiest to work in terms of travelling waves on the string. Suppose that a sine wave with frequency $\omega$ and unit amplitude travels in from $x>0$, and impinges on the bridge. A reflected wave will be generated, with the same frequency but travelling in the opposite direction. It will have a reduced amplitude, and perhaps a phase shift. We can represent both effects through a complex reflection coefficient $R(\omega)$. The incident wave can be written as $e^{i \omega(t+x/c)}$, while the reflected wave is $R e^{i \omega (t-x/c)}$: the situation is sketched in Fig. 1.

The total deflection of the string is given by the sum of the two waves:

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

(remembering that, as usual, the physical solution will be given by the real part of this complex expression). We can now write down expressions for the motion of the string at the bridge, and also for the force exerted on the bridge at that point. These two things must be related via the admittance $Y$, and the resulting equation will allow us to solve for the reflection coefficient $R$.

The string displacement at $x=0$ is given directly from eq. (1) as $(1+R)e^{i \omega t}$. The force on the bridge is given by the resolved component of the tension in the direction normal to the soundboard, which is approximately equal to

$$T \left[ \dfrac{\partial w}{\partial x} \right]_{x=0} = T \left[ \dfrac{i \omega}{c} – R \dfrac{i \omega}{c} \right] e^{i \omega t} = i \omega Z_0 (1-R) e^{i \omega t}\tag{2}$$

since $Z_0=T/c$. So the condition for the admittance $Y$ is that

$$i \omega (1+R) = Y(\omega) i \omega Z_0 (1-R) .\tag{3}$$

The factor of $i \omega$ on the left-hand side is needed to convert the string’s displacement into its velocity. This equation can easily be solved, to give

$$R=\dfrac{Y(\omega) Z_0 – 1}{Y(\omega) Z_0 + 1} .\tag{4}$$

For any realistic stringed instrument, the body motion at the bridge is always small compared to the string motion. This translates into the statement that $|Y(\omega) Z_0| \ll 1$. We can then make use of the binomial theorem to obtain a simpler approximate expression for $R$:

$$R \approx [Y(\omega) Z_0 – 1][1-Y(\omega) Z_0] \approx -1 +2Y(\omega) Z_0 .\tag{5}$$

This makes physical sense: a rigid termination would give an inverted reflection with $R=-1$, and this is slightly modified by the non-rigid bridge.

The final step is to think about the rate of energy loss from the string into the body, and find an expression for the resulting loss factor of a string mode, when any other sources of energy dissipation are ignored. A mode of the string consists of a standing wave, which is the resultant of the forward-travelling and backward-travelling waves in Fig. 1. The wave comes to the bridge, is reflected as we have just analysed, then travels to the other end of the string where it is reflected with no further loss of energy because we have assumed a rigid termination there. It then completes the circuit by arriving back at the bridge.

During one round trip, the amplitude of the wave has been reduced by a factor $|R|$. The energy carried by a wave is proportional to the square of its amplitude, so the energy has been reduced by a factor $|R|^2$ in that time. The proportional loss of energy is thus $(1-|R|^2)$. The definition of loss factor is that it describes the proportional energy loss per radian of vibration, so we need to divide by the number of radians in a round trip. For the $n$th mode of the string, with approximate mode shape $\sin (n \pi x/L)$, there are $n$ cycles per round trip, or $2 \pi n$ radians. So the loss factor due to energy loss through the bridge is

$$\eta^{(n)}_{body} \approx \dfrac{1-|R|^2}{2 \pi n} . \tag{6}$$

Making use of the approximation in eq. (5), the magnitude of $R$ can be calculated via

$$|R|^2=R R^* \approx [-1+2Y(\omega) Z_0] [-1+2Y^*(\omega) Z_0]$$

$$\approx 1 – 2 Z_0 (Y + Y^*) \approx 1 – 4 Z_0 \Re(Y)\tag{7}$$

where $.^*$ denotes the complex conjugate, and $\Re(.)$ denotes the real part of a complex quantity. So finally,

$$\eta^{(n)}_{body} \approx \dfrac{2 Z_0 \Re(Y)}{ \pi n} . \tag{8}$$