# 2.2.5 Vibration frequency response

It is worth showing a little detail on linear vibration theory, because this leads to a formula which will repeatedly be useful as we look at the behaviour of musical instruments. For simplicity, we will show the results for a discrete vibrating system: one that has only a finite number of variables (or degrees of freedom). This is not an important restriction: continuous systems like stretched strings or wooden plates can always be approximately described by discrete systems, provided we allow enough variables. For example, we might do the equivalent of sampling the string motion at regularly spaced points. If those points are close enough together then we won’t be missing anything important in the gaps between the points. Another way to approximate a continuous system by a discrete system is to build a finite element model of it, perhaps using a commercial software package. The details of that method do not matter here, the important thing is that the bit of theory we are about to look at, for general discrete systems, would apply automatically to any of these approximate treatments of continuous systems.

The rest of this discussion will be quite technical, but the end result is very important. If the detail is too much, then skip to eq. (11) and its discussion. All you really need to know is that $u_j^{(n)}$ represents the amplitude of the $n$th vibration mode, at a position labelled by $j$, and that $\omega_n$ is the natural frequency of that mode. But here is the fuller discussion. Suppose we have a system with $N$ degrees of freedom, described by parameters (called “generalised coordinates”) $q_1,q_2,\dots,q_N$. The first step is to calculate approximate expressions for the potential energy $V$ and the kinetic energy $T$ in terms of the $q$s and their time derivatives, $\dot{q}$s, assuming that the motion is small as outlined in section 2.2.3. The result for the potential energy can be written

$$V=\dfrac{1}{2}\sum_j \sum_k K_{jk} q_j q_k, \tag{1}$$

a quadratic expression involving a matrix $K$ whose terms are $K_{jk}$, called the stiffness matrix. (The notation $\sum_j(\dots)$ means “add up the terms $(\dots)$ for all the values of $j$.”) We can always choose the values so as to make $K$ symmetric. The total kinetic energy $T$ is the sum of terms like “$\frac{1}{2}mv^2$” for each bit of mass making up the system. It follows that it will be described by a quadratic expression

$$T=\dfrac{1}{2}\sum_j \sum_k M_{jk} \dot{q}_j \dot{q}_k . \tag{2}$$

Again, we can always choose to make the matrix $M$, whose terms are $M_{jk}$, symmetric. It is called the mass matrix.

Now we can use Lagrange’s equations: for free motion of the system,

$$\dfrac{d}{dt}\left( \dfrac{\partial T}{\partial \dot{q}_j}\right)+\dfrac{\partial V}{\partial q_j}=0 \tag{3}$$

so that

$$\sum_k M_{jk} \ddot{q}_k + \sum_k K_{jk} q_k =0. \tag{4}$$

These are linear equations, representing a set of coupled mass-spring oscillators. Finding the vibration modes allows us to uncouple them. A vibration mode is simply defined to be a solution with sinusoidal time dependence:

$$q_j(t)=u_j e^{i\omega t}. \tag{5}$$

So we require

$$-\omega^2 \sum_k M_{jk} u_k + \sum_k K_{jk} u_k =0, \tag{6}$$

or in matrix-vector form,

$$K \mathbf{u}=\omega^2 M \mathbf{u}, \tag{7}$$

which is a standard mathematical equation called a “generalised eigenvalue problem”. For the purposes of hand calculation, the usual solution procedure is:

(i) solve

$$\det[K -\omega^2 M ]=0 \tag{8}$$

for the natural frequencies $\omega=\omega_n$;

(ii) for each allowed $\omega_n$, solve the simultaneous equations

$$K \mathbf{u}^{(n)}=\omega_n^2 M \mathbf{u}^{(n)}, \tag{9}$$

for the mode shape $\mathbf{u}^{(n)}$.

Because $M$ and $K$ are both symmetric, there are standard results from linear algebra which show that

(i) the values of $\omega_n^2$ are all real, as are the eigenvectors;

(ii) the eigenvectors are orthogonal with respect to both the mass and stiffness matrices. This means that

$$\mathbf{u}^{(m)T} M \mathbf{u}^{(n)} = 0 \mathrm{~~~~if~} n \ne m \tag{10}$$

and

$$\mathbf{u}^{(m)T} K \mathbf{u}^{(n)} = 0 \mathrm{~~~~if~} n \ne m \tag{11}$$

where $.^T$ denotes the transpose.

Mode shapes have no built-in amplitude, they can be scaled by any factor. It is convenient to fix the scale factor using “mass normalisation”: scale each mode so that

$$\mathbf{u}^{(n)T} M \mathbf{u}^{(n)}=1 . \tag{12}$$

The archetypal vibration measurement, on a musical instrument or any other structure, is to apply a sinusoidal force to one point on a structure and observe the response at another point. Suppose we apply a sinusoidal force $F$ at frequency $\omega$ to the $j$th generalised coordinate. Now the frequency response function we want is the response at “point” $k$ to this forcing at “point” $j$. It can be expressed rather simply in terms of the mode shapes and frequencies:

$$G(j,k,\omega)=\dfrac{q_k}{F}=\sum_n \dfrac{u_j^{(n)}u_k^{(n)}}{\omega_n^2-\omega^2}. \tag{13}$$

The word “point” was deliberately written in quotes in the sentence above, because eq. (13) applies to a much wider range of problems than is immediately apparent. We will return to that question after some qualitative comments about what eq. (13) tells us.

The substance of eq. (13) can be described in words. First, the frequency response function for any chosen driving point $j$ and measuring point $k$ can be found once we know the mode shapes and natural frequencies: no other information is needed about the system. Modal information tells you everything. Second, the amplitude of a given modal response is given by the product of the (mass-normalised) mode shape at the driving point with the same mode at the observation point. (Those could be the same point if $j = k$, in which case it would be called a “driving-point response”.) For the drum problem, if the measuring point is kept fixed while the drummer hits at different positions, the “amount” of each mode in the mixture will change in a way which is easily visualised: it simply follows the modal amplitude at the tapping point. Third, each of the separate terms inside the summation in eq. (13) looks exactly like the response of a simple mass-spring oscillator as shown in eq. (12) of section 2.2.2. Comparing the two expressions, we see that the $n$th mode has a natural frequency $\omega_n$ and an effective mass which is the inverse of the mode-shape product $u_j^{(n)}u_k^{(n)}$.

This equivalence extends to the free motion of the system, as well. The most general free motion involves a mixture of the vibration modes, with each mode behaving just like a mass-spring oscillator, bouncing away at its own frequency quite independent of what any other modes might be doing. The total response is simply the sum of these separate modal responses.

The components of the vector $\mathbf{u}$ are not necessarily displacements at points on the structure: they might be rotation angles if we are looking at a problem involving torsional motion, or they might be less obvious things like the coefficients in a Finite-Element model of a structure. The transfer function (13) applies equally well to all these cases, provided the correct definition is used for the “force” $F$. For internal consistency, this has to be the correct “generalised force” to appear on the right-hand side of Lagrange’s equation. The definition of the generalised force $Q_j$ associated with a generalised coordinate $q_j$ is as follows. Imagine a small movement of the system in which $q_j$ increases by a small amount $\delta q_j$ while all the other generalised coordinates remain unchanged. Suppose the work done on the system by external forces during this displacement is $\delta W$: then define $Q_j$ so that

$$\delta W=Q_j \delta q_j. \tag{14}$$

If $q_j$ is actually the displacement of a particular point on the system, then $Q_j$ is indeed the force applied at that point, as implied by the description during the derivation of the transfer function (13) above. If $q_j$ is the rotation angle of a system component, then $Q_j$ is the external moment applied to that component.