7.3.1 Coupling of damped systems and the first-order method

So far, we have treated damping in a rather cavalier, way, simply by analysing undamped systems, then finding a suitable damping factor or Q-factor to assign to each individual mode. For most purposes this is all that is required. However, occasionally it is necessary dig a little deeper, and the problem of coupling two strings via a soundboard mode is one of those occasions. The underlying issue is that, taken by themselves, the string modes and the body mode will have very different Q-factors: especially for metal strings, the strings will have far higher Q-factors (or far lower damping) than the body mode. When we want to analyse how things change when we couple these together, this difference of damping levels will play a significant role in the behaviour. We will first develop the basic theory, then apply it to the coupled-string problem.

There is no completely general mathematical model of damping. This is in strong contrast to the behaviour of mass and stiffness for small vibration of any system. As we showed back in section 2.2.5, there are very strong reasons to expect any undamped system to be described by a mass matrix and a stiffness matrix, capturing the dominant behaviour of the potential energy and the kinetic energy of the system, respectively. However, a good indication of the behaviour to be expected from damped systems can be gained from a particular damping model. We showed in section 2.2.5 that undamped vibration of a system with $N$ degrees of freedom can be represented by the equation

$$M \ddot{\mathbf{q}} + K \mathbf{q}= 0, \tag{1}$$

where $\mathbf{q}$ is a vector of generalised coordinates, $M$ is the mass matrix and $K$ is the stiffness matrix. Both matrices have dimensions $N \times N$. Physically, these equations can be interpreted as representing $N$ coupled harmonic oscillators. It seems very natural to extend this to damped systems by adding a new term by analogy with the expression for a damped harmonic oscillator derived in section 2.2.7, to represent $N$ coupled, damped oscillators:

$$M \ddot{\mathbf{q}} + C \dot{\mathbf{q}} + K \mathbf{q}= 0, \tag{2}$$

where $C$ is the damping matrix or dissipation matrix. Damping governed in this way by the generalised velocities $\dot{\mathbf{q}}$ is called viscous damping.

Now recall from section 2.2.5 that the undamped equation (1) can be derived from Lagrange’s equations, and that the mass and stiffness matrices can be assumed symmetric because they arise from the quadratic expressions for the kinetic and potential energy respectively. A similar argument can be used to deduce that the damping matrix can also be assumed symmetric. Multiplying eq. (2) by $\dot{\mathbf{q}}^t$ (where $t$ denotes the transpose of the vector) gives

$$\dot{\mathbf{q}}^t M \ddot{\mathbf{q}} + \dot{\mathbf{q}}^t C \dot{\mathbf{q}} + \dot{\mathbf{q}}^t K \mathbf{q}= 0$$

$$\therefore \dot{\mathbf{q}}^t C \dot{\mathbf{q}} = -\dfrac{d}{dt}\left[ \dfrac{1}{2}\dot{\mathbf{q}}^t M \dot{\mathbf{q}} + \dfrac{1}{2}\mathbf{q}^t C \dot{\mathbf{q}}\right] = -\dfrac{d}{dt}\left[ \mathrm{total~energy} \right] \tag{3}$$

making use of the fact that both $K$ and $M$ are symmetric. This makes it clear that the function

$$F= \dfrac{1}{2} \dot{\mathbf{q}}^t C \dot{\mathbf{q}} \tag{4}$$

known as the Rayleigh dissipation function, is half the rate of dissipation of energy. This is another quadratic expression, in which we can choose the entries in the matrix to be symmetric. Working back via Lagrange’s equations, with a generalised force

$$Q_j=\dfrac{\partial F}{\partial \dot{q_j}} \tag{5}$$

representing the damping force, we arrive back at eq. (2).

For the undamped system (1) we would find modes by calculating the eigenvalues and eigenvectors from

$$K \mathbf{u}^{(n)}=\omega_n^2 M \mathbf{u}^{(n)} . \tag{6}$$

When we change to modal variables, or “normal coordinates”, both $M$ and $K$ are diagonalised. It is not possible to do this for the system of eq. (2) — it is not in general possible to find a change of variables which diagonalises three matrices simultaneously. There are special cases for which it is possible, for example if

$$C= \alpha M + \beta K \tag{7}$$

where $\alpha, \beta$ are constants (so-called “Rayleigh damping”). This special form is often assumed for convenience, e.g. in Finite Element packages, but there is usually no physical justification whatever for this assumption. If the damping in a system is modelled by a physically sensible matrix $C$, standard modal analysis will usually not work. We need a different mathematical formalism to find what has become of the modes.

A useful approach, which is also widely used to analyse systems in control theory, is to reformulate the governing equations as a set of $2N$ first-order equations, rather than the $N$ second-order equations we have at present. This change is simply an algebraic trick, with no physical significance. Define a new vector

$$\mathbf{y} = \begin{bmatrix}\mathbf{q}\\ \dot{\mathbf{q}}\end{bmatrix},~~~\dot{\mathbf{y}} = \begin{bmatrix}\dot{\mathbf{q}}\\\ddot{\mathbf{q}}\end{bmatrix} . \tag{8}$$

Then

$$\dot{\mathbf{y}} = \begin{bmatrix}0 & I\\ -M^{-1}K & -M^{-1}C\end{bmatrix} \begin{bmatrix}\mathbf{q}\\ \dot{\mathbf{q}}\end{bmatrix} = A \mathbf{y} \tag{9}$$

where $0$ denotes the $N \times N$ matrix of zeros, $I$ denotes the $N \times N$ identity matrix, and the final matrix $A$ is a $2N \times 2N$ matrix.

The top half of this set of equations simply says $\dot{\mathbf{q}}=\dot{\mathbf{q}}$, while the bottom half says

$$\ddot{\mathbf{q}} = -M^{-1} K \mathbf{q} – M^{-1} C \dot{\mathbf{q}} \tag{10}$$

which is a rearranged version of eq. (2).

We can now look for “modal” solutions of eq. (9): try

$$\mathbf{y} = \mathbf{u} e^{\lambda t} . \tag{11}$$

Then we require

$$A \mathbf{u} = \lambda \mathbf{u} \tag{12}$$

which is a standard matrix eigenvalue-eigenvector problem. It is difficult to solve by hand for problems of any realistic size, but it is very easy to compute answers.

Note that the matrix $A$ is not symmetric, so the eigenvalues $\lambda$ will in general be complex. From eq. (11), the imaginary part of $\lambda$ will give the frequency. The real part will be negative, equal to minus the decay rate. Also, we no longer expect the useful result that eigenvectors are orthogonal. There are, however, equivalent (but more complicated) relations between the eigenvectors, which allow things like step response and impulse response to be calculated in a similar way that we did earlier. We will not go into the full details here, we will simply use the results: see for example Chapter 8 of Newland [1].

We can now apply the approach to the coupled-strings problem. We have two strings of length $L$, with tension $T_j$, mass per unit length $m_j$ and Q-factor $Q_j$, where $j=1,2$. At position $x=0$ both strings are rigidly anchored, while at $x=L$ they are both attached to a mass $m$, which is itself attached to a fixed base through a parallel combination of a spring of stiffness $k$ and a dashpot of strength $c$. We will only consider the first resonance of each string, considered in isolation. So suppose that the displacement of string $j$ is given approximately by

$$w_j(x,t) = b_j(t) \sin \dfrac{\pi x}{L} + a(t)\dfrac{x}{L} \tag{13}$$

where $a(t)$ is the displacement of the oscillator representing the body mode.

We now calculate the potential and kinetic energies of the system, and deduce the mass and stiffness matrices for the three degree-of-freedom system parameterised by the vector $\mathbf{q} = [a~b_1~b_2]^t$. The potential energy is given by

$$V = \frac{1}{2} k a^2 +\sum_j{\dfrac{1}{2}T_j \int_0^L{\left( \dfrac{\partial w_j}{\partial x} \right)^2 dx}}$$

$$= \frac{1}{2} k a^2 +\sum_j{\dfrac{1}{2}T_j \int_0^L{\left( \dfrac{b_j \pi}{L} \cos \dfrac{\pi x}{L}\right)^2 dx}}$$

$$= \frac{1}{2} k a^2 +\sum_j{\dfrac{\pi^2 T_j b_j^2}{4L}} \tag{14}$$

so that the stiffness matrix is

$$K =\begin{bmatrix}k & 0 & 0\\ 0 & \frac{\pi^2 T_1}{2L} & 0 \\ 0 & 0 & \frac{\pi^2 T_12}{2L}\end{bmatrix} . \tag{15}$$

The kinetic energy is given in a similar way, by

$$T = \frac{1}{2} m \dot{a}^2 +\sum_j{\dfrac{1}{2}m_j \int_0^L{\dot{w}_j^2 dx}}$$

$$= \frac{1}{2} m \dot{a}^2 +\sum_j{\dfrac{1}{2}m_j \int_0^L{\left[\dot{b}_j \sin \dfrac{\pi x}{L} + \dfrac{\dot{a}x}{L}\right]^2 dx}}$$

$$= \frac{1}{2} m \dot{a}^2 +\sum_j{\dfrac{1}{2}m_j \left\lbrace \dot{b}_j^2 \dfrac{L}{2} + \dfrac{\dot{a}^2}{L^2} \dfrac{L^3}{3} + \dfrac{2 \dot{a} \dot{b}_j}{L} \int_0^L{x \sin \dfrac{\pi x}{L} dx}\right\rbrace }$$

$$= \frac{1}{2} m \dot{a}^2 +\sum_j{\dfrac{1}{2}m_j \left\lbrace \dfrac{L}{2} \dot{b}_j^2 + \dfrac{L}{3} \dot{a}^2 + \dfrac{2 \dot{a} \dot{b}_j}{L} \dfrac{L^2}{\pi}\right\rbrace } \tag{16}$$

after using integration by parts on the final term. It follows that the mass matrix is

$$M =\begin{bmatrix}m+L(m_1+m_2)/3 & m_1 L /\pi & m_2 L/\pi\\ m_1 L /\pi & m_1 L /2 & 0 \\ m_2 L /\pi & 0 & m_2 L /2\end{bmatrix} . \tag{17}$$

Finally, we need the damping matrix. The rate of energy dissipation in the body dashpot is $c \dot{a}^2$, so

$$C = \begin{bmatrix}c & 0 & 0 \\ 0 & c_1 & 0 \\ 0 & 0 & c_2 \end{bmatrix} \tag{18}$$

where the equivalent dashpot strengths of the two string modes are given by

$$c_j = \dfrac{\pi \sqrt{T_j m_j}}{2 Q_j} . \tag{19}$$

We can then assemble the $6 \times 6$ matrix $A$ from eq. (9), and compute its eigenvalues and eigenvectors. There, will of course, be 6 of them: but surely we are only expecting to see 3 modes for this system? The answer to that puzzle is simple. Each mode appears twice, related to each other by being complex conjugates. The matrix $A$ is real, so we can deduce from eq. (12) that if $\mathbf{u}$ is an eigenvector with eigenvalue $\lambda$ then $\mathbf{u}^*$ must also be an eigenvector with eigenvalue $\lambda^*$.

Finally, we can calculate the “pluck response” using the formula from Newland [1]. For the una corda case where only one string is initially excited, we apply a step function of generalised force to $b_1$ and calculate the transient response $a(t)$. To obtain the case where both strings are excited equally, we first calculate the corresponding response when a step function of generalised force is applied to $b_2$, then add the result to the previous case. We are still dealing with a linear system, so we can use superposition to obtain the combined response.

[1] David E. Newland; “Mechanical Vibration Analysis and Computation”, Longman Scientific and Technical, Harlow (1989).