A normal violin bridge has two important in-plane resonances that can be modelled in idealised form by the two mass-spring systems shown in Figs. 1 and 2. Both models have a rigid, massless base with two feet spaced a distance $d$ apart. In both cases, forces $F_1$ and $F_2$ act on the violin body through the two feet of the base. The displacements of these two feet are $w_1$ and $w_2$ respectively. Both forces and displacements are shown here as vertical: this cannot, of course, be literally true, an issue to which we return in subsection C below.

Figure 1 shows the model of the rocking mode, in which a rigid weighted bar is pivoted about the mid-point of the base. This bar has moment of inertia $I$ about the pivot point, and a torsional spring of stiffness $K$ connects it to the base. A transverse force $F_3$ is applied to the top of the bar, a distance $L$ from the pivot point, and the displacement at the forced point is $w_3$. Figure 2 shows the corresponding model for the bouncing mode of the bridge. A mass $m$ is attached to the mid-point of the base through a linear spring of stiffness $k$. A vertical force $F_3$ is applied to this mass, causing displacement $w_3$.

For both models, the dynamics of the violin body is described by a $2 \times 2$ matrix of admittances, so that

$$i \omega \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22} \end{bmatrix} \begin{bmatrix} F_1 \\ F_2 \end{bmatrix} \tag{1}$$

where reciprocity requires that $Y_{12}=Y_{21}$.

It should be emphasised that both these models ignore a number of effects in order to get a first approximation to what happens. We will relax a lot of those approximations in subsection C, when a more complete model will be developed. For example, Fig. 1 does not include the lateral forces on the violin body at the bridge feet.

*A. The rocking mode*

Both bridge models are symmetrical, and the applied force respects the symmetry: it is antisymmetric or symmetric in the two cases sketched. However, the violin body is not symmetrical because of the bass bar and soundpost. We can obtain useful approximate models by ignoring one manifestation of this effect: we will come back to it later, to check how well the approximation stands up. For the rocking mode, it is very convenient to ignore the small vertical acceleration of the bridge mass induced by the body asymmetry. In that case, force balance in the vertical direction for the whole bridge model requires that

$$F_1 + F_2 = 0 . \tag{2}$$

Equation (1) then tells us that

$$i \omega w_1 = (Y_{11}-Y_{12})F_1 \tag{3}$$

and

$$i \omega w_2 = (Y_{21}-Y_{22})F_1 . \tag{4}$$

For the next step, it is useful to look at the separated version of the bridge model sketched in Fig. 3. This shows the forces and moments on the bar and the base. The foot forces $F_1$ and $F_2$ have reversed in direction, because these are forces acting *on* the bridge, rather than on the violin body. There are equal and opposite moments acting on the two subsystems, due to the torsion spring. This moment has magnitude $K \theta$ where $\theta$ is the relative rotation angle between bar and base:

$$\theta \approx \dfrac{w_3}{L} + \dfrac{w_1\ -\ w_2}{d}. \tag{5}$$

Making use of eqs. (3) and (4), this can be written

$$\theta \approx \dfrac{w_3}{L} + \dfrac{d F_1 R}{i \omega} \tag{6}$$

where

$$R(\omega) = \dfrac{Y_{11}\ -\ Y_{12}\ -\ Y_{21} + Y_{22}}{d^2} \tag{7}$$

is the *rotational admittance* of the violin top [1], the ratio of the angular velocity of the bridge base to the moment applied to the violin by the bridge feet.

Ignoring, temporarily, a small contribution to moment from the horizontal forces at the feet and the pivot point, moment balance for the base (which we are assuming to be massless) gives

$$(F_2\ -\ F_1)d/2=K \theta \tag{8}$$

so

$$F_1 d = -K \theta = -K\left[ \dfrac{w_3}{L} + \dfrac{d F_1 R}{i \omega} \right] \tag{9}$$

from eq. (6). Rearranging gives

$$F_1 d \left[ 1 + \dfrac{KR}{i \omega} \right] = -\dfrac{K w_3}{L} \tag{10}$$

so that

$$F_1=-\dfrac{i \omega K w_3}{Ld[i \omega + KR]}. \tag{11}$$

The equation of rotational motion of the bar is now

$$- I \omega^2 \dfrac{w_3}{L} = F_3 L -K \theta = F_3 L + dF_1$$

$$=F_3 L -\dfrac{i \omega K w_3}{L[i \omega + KR]}. \tag{12}$$

It follows that the bridge admittance for this rocking motion, $Y_r(\omega)$, is given by

$$Y_r=\dfrac{i \omega w_3}{F_3} = \dfrac{L^2 (i \omega + KR)}{K – \omega^2 I + i \omega I K R} . \tag{13}$$

In the limiting case when the feet of the bridge are rigidly clamped, $R \rightarrow 0$ so that

$$Y_r \rightarrow \dfrac{i \omega L^2}{K – \omega^2 I } . \tag{14}$$

The clamped-foot resonance occurs at $\omega=\sqrt{K/I}$. Another interesting limiting case is when the bridge is made rigid by letting $K \rightarrow \infty$, in which case

$$Y_r \rightarrow \dfrac{L^2 R}{1+i \omega I R} \tag{15}$$

so that the admittance represents a balance between the moment of inertia of the bridge and the rotational admittance of the body.

We can deduce some other interesting transfer functions from this analysis. From eq. (10), the dimensionless transfer functions linking the applied force $F_3$ to the bridge-foot forces $F_1$ and $F_2$ are given by

$$\dfrac{F_1}{F_3} = -\dfrac{F_2}{F_3} =-\dfrac{K Y_r}{Ld(i \omega +KR)}. \tag{16}$$

The motion at the bridge feet can then be written in terms of two transfer admittances

$$Y_{t1}=\dfrac{i \omega w_1}{F_3} = -\dfrac{(Y_{11} – Y_{12})Y_r K}{Ld(i \omega +KR)} \tag{17}$$

and

$$Y_{t2}=\dfrac{i \omega w_2}{F_3} = -\dfrac{(Y_{21} – Y_{22})Y_r K}{Ld(i \omega +KR)} \tag{18}$$

using eqs. (3) and (4).

*B. The bouncing mode*

Analysing the bouncing mode from Fig. 2 is slightly easier. We start with an approximation, similar to the one used for the rocking mode: we will ignore for the moment the small rotational motion of the base that may be induced by the asymmetry of the body response matrix. In that case, moment balance for the base requires that

$$F_1\ -\ F_2=0 . \tag{19}$$

From eq. (1) we can then deduce that

$$i \omega w_1 = (Y_{11} +Y_{12})F_1 \tag{20}$$

and

$$i \omega w_2 = (Y_{21} + Y_{22})F_1 . \tag{21}$$

Vertical force balance for the base (assuming for the moment that it is massless) requires

$$2F_1=k \left( w_3\ -\ \dfrac{w_1 + w_2}{2} \right)$$

$$=k w_3\ -\ \dfrac{2kF_1}{i \omega} \bar{Y} \tag{22}$$

from eqs. (20) and (21), where

$$\bar{Y}=\dfrac{Y_{11} +Y_{12}+Y_{21} +Y_{22}}{4} . \tag{23}$$

Rearranging eq. (22) gives

$$F_1=\dfrac{i \omega k w_3}{2(i \omega + k \bar{Y})} . \tag{24}$$

Newton’s law for the mass requires

$$-m \omega^2 w_3 = F_3\ -\ k \left( w_3\ -\ \dfrac{w_1 + w_2}{2} \right)$$

$$ = F_3\ -\ 2F_1 = F_3 -\dfrac{i \omega k w_3}{i \omega + k \bar{Y}} . \tag{25}$$

Now the bridge admittance for this vertical bouncing motion is defined by

$$Y_b=\dfrac{i \omega w_3}{F_3}, \tag{26}$$

so

$$i \omega m F_3 Y_b = F_3\ -\ \dfrac{k F_3 Y_b}{i \omega + k \bar{Y}} \tag{27}$$

and rearranging gives

$$Y_b=\dfrac{i \omega + k \bar{Y}}{k\ -\ m \omega^2 + i \omega k m \bar{Y}} , \tag{28}$$

a result with obvious similarities to eq. (13) for the rocking motion.

We can derive limiting cases, just we did for the rocking case. If the bridge had clamped feet so that $\bar{Y} \rightarrow 0$, then

$$Y_b \rightarrow \dfrac{i \omega}{k\ -\ m \omega^2} , \tag{29}$$

analogous to eq. (14), so that the rigid-foot bouncing frequency is given by $\sqrt{k/m}$. If the bridge is turned into a rigid body by letting $k \rightarrow \infty$ then

$$Y_b \rightarrow \dfrac{\bar{Y}}{1+ i \omega m \bar{Y}} , \tag{30}$$

analogous to eq. (15).

We can also deduce an expression for the non-dimensional transfer function from the applied force to the foot forces, analogous to eq. (16): from eq. (24) we immediately obtain

$$\dfrac{F_1}{F_3} = \dfrac{F_2}{F_3} = \dfrac{k Y_b}{2(i \omega + k \bar{Y})} . \tag{31}$$

The motion at the bridge feet can then be written in terms of two transfer admittances analogous to eqs. (17) and (18):

$$Y_{t1}=\dfrac{i \omega w_1}{F_3} = -\dfrac{(Y_{11} + Y_{12}) k Y_b}{2(i \omega + k \bar{Y})} \tag{32}$$

and

$$Y_{t2}=\dfrac{i \omega w_2}{F_3} = -\dfrac{(Y_{21} + Y_{22}) k Y_b}{2(i \omega + k \bar{Y})} \tag{33}$$

using eqs. (20) and (21).

*C. A combined and extended model*

The two simple models presented above both involve approximations and missing ingredients. In order to assess how important these might be, we need to compare the predictions with a more complete model. This model needs to include the asymmetry of the violin body, which breaks the symmetry of the bridge models so that the two types of motion (rocking and bouncing) are in fact coupled together. In addition, we can allow for the mass of the base, and for the effect of the horizontal force at the bridge feet. We can also allow for the fact that in measurements, the force is not necessarily applied at the centre of the bridge: in particular, for a rocking measurement (i.e. the usual bridge admittance) the hammer normally strikes on the corner of the bridge in some direction.

To allow for a more general force, we can use the geometry sketched in Fig. 4. The weighted bar representing the rocking deformation of the bridge can include a rigid body, sketched to include the top curve of a bridge. The force $F_3$ can then be applied at a chosen position with offsets $x$ and $y$ from the top centre of the bridge, and at an angle $\alpha$ to the horizontal, as indicated in the figure. Later, we will need to know the perpendicular distance from the pivot point to the line of action of this force. A little trigonometry shows that this distance is

$$L_1=x \sin \alpha + (L-y) \cos \alpha . \tag{34}$$

Note that to avoid clutter, Fig. 4 does not try to represent every feature of the combined model: it shows the rotational components but not the vertical stiffness and mass, which are exactly as in Fig. 2. Some new features are included. An equal and opposite pair of horizontal forces $P$ act on the base and bridge top at the pivot point, and horizontal forces $Q_1$ and $Q_2$ act at the bridge feet. The effective mass associated with rotation of the bridge top is now denoted $M$, at a distance $a$ from the pivot, so that the moment of inertia used earlier is given by $I=Ma^2$. The height of the bridge base is $h$. Although it seems counter-intuitive in Fig. 4, we should note that for a real violin bridge $h$ might possibly turn out to be negative: the effective centre of rotation of the top part of the bridge could be lower than the feet.

The output displacements can be taken at the top centre of the bridge as before: we will denote the horizontal displacement of that point by $w_3$ and the downward vertical displacement by $w_4$. We can define corresponding admittances by

$$Y_3(\omega)=i \omega w_3/F_3 \tag{35}$$

and

$$Y_4(\omega)=i \omega w_4/F_3 . \tag{36}$$

In addition to the parameters of the two previous models, we will allow the base to have a mass $m_b$ below the linear spring of Fig. 2, and to have a moment of inertia $I_b$ about the pivot point.

In this combined model, the compressive force in the linear spring is

$$k\left(w_4\ -\ \dfrac{w_1+w_2}{2} \right) \tag{37}$$

and the moment generated by the torsional spring is $K \theta$ where the relative rotation angle $\theta$ is given by

$$\theta = \dfrac{w_3}{L} + \dfrac{w_1-w_2}{d} . \tag{38}$$

All of this is essentially the same as in the separate models presented earlier.

Now we can write down the set of governing equations. For vertical motion of the base we require

$$F_1+F_2\ -\ k\left(w_4\ -\ \dfrac{w_1+w_2}{2} \right) = m_b \omega^2 \left( \dfrac{w_1+w_2}{2} \right) . \tag{39}$$

For vertical motion of the top we require

$$F_3 \sin \alpha + k\left(w_4\ -\ \dfrac{w_1+w_2}{2} \right) = m \omega^2 w_4 . \tag{40}$$

Lateral force balance for the base simply gives

$$P=Q_1+Q_2 \tag{41}$$

because no horizontal motion is allowed at the feet — if we had a measurement of the *horizontal* admittances of the violin body at the foot positions, we could enhance the model by factoring those in at this stage, but for the present we are assuming that these admittances are zero. (If we were modelling a cello bridge rather than a violin bridge, we would have to include another effect here: the long legs can deform in shear, so that there is some horizontal motion of the top of the base, even if there is none at the feet.)

With these assumptions, Newton’s law in the horizontal direction for the bridge top requires

$$F_3 \cos \alpha – P = -M\omega^2 \left(\dfrac{a w_3}{L} \right) . \tag{42}$$

For rotational motion of the base we require

$$(F_2 – F_1) \dfrac{d}{2} -K\theta -Ph = -I_b \omega^2 \left(\dfrac{w_1 – w_2}{d} \right) . \tag{43}$$

Finally, for rotational motion of the top we require

$$F_3 L_1\ -\ K \theta = -I \omega^2 \dfrac{w_3}{L} \tag{44}$$

where the distance $L_1$ is given by eq. (34).

From eq. (1), we know that

$$w_1+w_2=\dfrac{1}{i \omega} \left[ (Y_{11} + Y_{21}) F_1 + (Y_{12} +Y_{22}) F_2 \right] \tag{45}$$

and

$$w_1\ -\ w_2=\dfrac{1}{i \omega} \left[ (Y_{11}\ -\ Y_{21}) F_1 + (Y_{12}\ -\ Y_{22}) F_2 \right] . \tag{46}$$

Substituting into eq. (43) and rearranging gives

$$\dfrac{K w_3}{L} + \dfrac{haM \omega^2}{L} w_3 + h F_3 \cos \alpha = $$

$$-F_1 \left[ \dfrac{d}{2} + (Y_{11} – Y_{21})\left(\dfrac{K-\omega^2 I_b}{i \omega d} \right) \right]$$

$$ + F_2 \left[ \dfrac{d}{2} – (Y_{12} – Y_{22})\left(\dfrac{K-\omega^2 I_b}{i \omega d} \right) \right] . \tag{47}$$

Using this result to get rid of $w_3$ from eq. (44) leads to

$$\left[ \dfrac{K+haM \omega^2}{K – I \omega^2} L_1 + h \cos \alpha \right] F_3 = $$

$$F_1 \left\lbrace -\dfrac{d}{2} + \dfrac{(Y_{11} – Y_{21})}{i \omega d} \left[ I_b\omega^2 -K + \dfrac{K(K+haM \omega^2)}{K-I\omega^2} \right] \right\rbrace$$

$$+ F_2 \left\lbrace \dfrac{d}{2} + \dfrac{(Y_{12} – Y_{22})}{i \omega d} \left[ I_b\omega^2 -K + \dfrac{K(K+haM \omega^2)}{K-I\omega^2} \right] \right\rbrace$$

$$=A F_1 + B F_2 \tag{48}$$

say.

In a similar way, substituting into eq. (39) and rearranging gives

$$2 i \omega k w_4 = F_1 \left[ 2i \omega + (k-\omega^2 m_b)(Y_{11}+Y_{21}) \right]$$

$$ + F_2 \left[ 2i \omega + (k-\omega^2 m_b)(Y_{12}+Y_{22}) \right] . \tag{49}$$

Using this result to get rid of $w_4$ from eq. (40) leads to

$$F_3 \sin \alpha = $$

$$F_1 \left\lbrace \dfrac{k}{2i\omega}(Y_{11}+Y_{21}) – \left( \dfrac{k- \omega^2 m}{k}\right) \left[ 1+(Y_{11}+Y_{21}) \left(\dfrac{k- \omega^2 m_b}{2i \omega}\right) \right] \right\rbrace$$

$$+F_2 \left\lbrace \dfrac{k}{2i\omega}(Y_{12}+Y_{22}) – \left( \dfrac{k- \omega^2 m}{k}\right) \left[ 1+(Y_{12}+Y_{22}) \left(\dfrac{k- \omega^2 m_b}{2i \omega}\right) \right] \right\rbrace$$

$$=C F_1 + D F_2 \tag{50}$$

say.

The transfer functions relating $F_3$ to the bridge-foot forces are then found by solving the two simultaneous equations (48) and (50):

$$\begin{bmatrix} F_1/F_3 \\ F_2/F_3 \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix}^{-1} \begin{bmatrix} \dfrac{K+haM \omega^2}{K – I \omega^2} L_1 + h \cos \alpha \\ \sin \alpha \end{bmatrix} . \tag{51}$$

After that, the admittances $Y_3(\omega)$ and $Y_4(\omega)$ are found by substituting back into equations (47) and (49) respectively:

$$Y_3 = \dfrac{i \omega L}{K+haM \omega^2}\left\lbrace -h \cos \alpha -\dfrac{F_1}{F_3} \left[ \dfrac{d}{2} + (Y_{11} – Y_{21})\left(\dfrac{K-\omega^2 I_b}{i \omega d} \right) \right]\right.$$

$$\left. + \dfrac{F_2}{F_3} \left[ \dfrac{d}{2} – (Y_{12} – Y_{22})\left(\dfrac{K-\omega^2 I_b}{i \omega d} \right) \right]\right\rbrace \tag{52}$$

and

$$Y_4 = \dfrac{1}{2k}\left\lbrace \dfrac{F_1}{F_3} \left[ 2i \omega + (k-\omega^2 m_b)(Y_{11}+Y_{21}) \right]\right.$$

$$\left. + \dfrac{F_2}{F_3} \left[ 2i \omega + (k-\omega^2 m_b)(Y_{12}+Y_{22}) \right]\right\rbrace . \tag{53}$$

We can use this extended model to check the significance of various approximations made in the derivation of the simplified models: we ignored several effects, including the fact that the body response matrix violates the symmetry of the simple bridge models. We will illustrate with rocking motion. A plausible set of parameter values were calculated by best-fitting the extended model to the response of a particular violin and bridge, assuming that the excitation force was as shown in Fig. 1: horizontal, at the bridge centre. In terms of the variables defined in Fig. 4, this case has $\alpha=x=y=0$. Other model parameter values for this example were: $L=20\mathrm{~mm}$, $h=13\mathrm{~mm}$, $a=16\mathrm{~mm}$, $d=30\mathrm{~mm}$. The estimate for $L$ made use of the fit quality to the measured admittance (see Fig. 6 from section 7.5), while the other values were estimated from the geometry of a typical violin bridge.

Figure 5 shows the result. The red curve is the original measured rocking admittance, and the blue curve shows the best-fitted response. If the same parameter values are used in the simplified model from subsection A, the result is the dashed green curve. Over most of the range, the two curves are quite similar. The systematic difference at low frequency is mainly caused by the non-zero fitted value of the base height $h$: if this value is set to zero, the two curves become almost identical at low frequency. We can conclude that as far as this admittance is concerned the various approximations may not matter a great deal, and the simplified model gives a very good guide. The same is found when the test is repeated with the admittance to a vertical force as in Fig. 2.

However, this does not mean that the extended model is useless. It gives three advantages, all of which will be useful. First, there is a non-trivial consequence of the fact that the body behaviour is not symmetric. As eqs. (52) and (53) show, the extended model predicts non-zero response in the vertical direction to a horizontal force, and vice versa. The magenta curve in Fig. 5 shows this predicted vertical motion at the bridge centre, corresponding to the horizontal motion shown in the blue curve. The simplified model ignores this vertical motion entirely.

The second advantage of the extended model is that it allows forces to be applied to the bridge at positions other than the centre, and at any angle. This will allow us to compare the response to forcing at the four string notches, each with its particular bowing direction.

The third advantage comes when we want to consider sound radiated by the violin, rather than bridge admittance. It was argued in section 7.5 that the influence of the bridge on radiated sound is captured by the transfer functions relating the applied force to the forces at the bridge feet. In both simplified models, the magnitudes of those two force are always equal, shown by eqs. (2) and (19). But the extended model shows that this will not in general be the case: the two separate transfer functions are given by eq. (51). Figure 6 shows the result, for the simulation that gave the blue curve in Fig. 5. Force at the bass foot is shown in red, force at the treble foot in blue. At the lowest frequencies the two are equal, but after that they diverge and at the highest frequencies there is a small but systematic level difference between the two.