# 5.3.1 The bridge hill: a resonator near the driving point

To see the effect of driving a system through a resonant mass-spring system, we need to analyse the idealised version shown in Fig. 1. A force $Fe^{i \omega t}$ is applied to a mass $m$. This is connected via a spring of stiffness $k$ to the original system, which has admittance $Y_v(\omega)$ (v for ‘violin’). The aim is to calculate the new admittance $Y_b(\omega)$ (b for ‘bridge’) at the forcing position. We can denote the displacement of the mass by $x_b e^{i \omega t}$ and the displacement of the underlying system by $x_v e^{i \omega t}$. For the rest of the analysis, we will drop the $e^{i \omega t}$ factors.

The force from the spring is $k(x_b-x_v)$: upwards at the top and downwards at the bottom. So Newton’ s law for the mass states that

$$-\omega^2 m x_b = F -k(x_b -x_v) \tag{1}$$

while the force and velocity on the ‘violin body’ must satisfy

$$i \omega x_v = k (x_b – x_v) Y_v(\omega) . \tag{2}$$

We don’t really want $x_v$, so we eliminate it between these two equations. From eq. (2),

$$x_v(i \omega +k Y_v)=kY_v x_b .\tag{3}$$

Substituting in eq. (1) then gives

$$-\omega^2 m x_b = F -k x_b + \dfrac{k^2 Y_v x_b}{i \omega + kY_v} .\tag{4}$$

We are trying to calculate

$$Y_b(\omega) = \dfrac{i \omega x_b}{F} \tag{5}$$

and so using eq. (4),

$$Y_b(\omega) = \dfrac{i \omega}{k – \dfrac{k^2 Y_v}{i \omega +k Y_v} – \omega^2 m}$$

so that after rearranging,

$$Y_b(\omega) = \dfrac{i \omega + kY_v}{k-\omega^2 m + i \omega k Y_v} .\tag{6}$$