8.5: Damping

It is instructive, at this point, to consider waves in systems with frictional forces. We have postponed this until now because it will be easier to understand what is happening in systems with damping now that we have discussed traveling waves.

The key observation is that in a translation invariant system, even in the presence of damping, the normal modes of the infinite system are exactly the same as they were without damping, because they are still determined by translation invariance. The normal modes are still of the form, $e^{\pm i k x}$, characterized by the angular wave number $k$. Only the dispersion relation is different. To see how this goes in detail, let us recapitulate the arguments of chapter 5.

The dispersion relation for a system without damping is determined by the solution to the eigenvalue equation

$$\left[-\omega^{2}+M^{-1} K\right] A^{k}=0$$

where $A^{k}$ is the normal mode with wave number $k$,

$$A_{j}^{k} \propto e^{i j k a} ,$$

with time dependence $e^{-i \omega t}$.⁵ We already know that $A^{k}$ is a normal mode, because of translation invariance. This implies that it is an eigenvector of $M^{- 1} K$. The eigenvalue is some function of $k$. We will call it $\omega_{0}^{2}(k)$, so that

$$M^{-1} K A^{k}=\omega_{0}^{2}(k) A^{k}$$

This function $\omega_{0}^{2}(k)$ determines the dispersion relation for the system without damping, because the eigenvalue equation, (8.75) now implies

$$\omega^{2}=\omega_{0}^{2}(k) .$$

We can now modify the discussion above to include damping in the infinite translation invariant system. In the presence of damping, the equation of motion looks like

$$M \frac{d^{2}}{d t^{2}} \psi(t)=-M \Gamma \frac{d}{d t} \psi(t)-K \psi(t) ,$$

where $M \Gamma$ is the matrix that describes the velocity dependent damping. Then for a normal mode,

$$\psi(t)=A^{k} e^{-i \omega t} ,$$

the eigenvalue equation now looks like

$$\left[-\omega^{2}-i \Gamma \omega+M^{-1} K\right] A^{k}=0 .$$

Now, just as in (8.77) above, because of translation invariance, we know that $A^{k}$ is an eigenvector of both $M^{- 1}K$ and $\Gamma$,

$$M^{-1} K A^{k}=\omega_{0}^{2}(k) A^{k}, \quad \Gamma A^{k}=\gamma(k) A^{k} .$$

Then, as above, the eigenvalue equation becomes the dispersion relation

$$\omega^{2}=\omega_{0}^{2}(k)-i \gamma(k) \omega .$$

For all $k$, $\gamma (k) \geq 0$, because as we will see in (8.84) below, the force is a frictional force. If $\gamma(k)$ were negative for any $k$, then the “frictional” force would be feeding energy into the system instead of damping it. Note also that if $\Gamma = \gamma I$, then $\gamma(k) = \gamma$, independent of $k$. However, in general, the damping will depend on $k$. Modes with different $k$ may get damped differently.

In (8.83), we see the new feature of translation invariant systems with damping. The only difference is that the dispersion relation becomes complex. Both $\omega_{0}^{2}(k)$ and $\gamma(k)$ are real for real $k$. Because of the explicit $i$ in (8.83), either $\omega$ or $k$ (or both) must be complex to satisfy the equation of motion.

Free Oscillations

For free oscillations, the angular wave numbers, $k$, of the allowed modes are determined by the boundary conditions. Typically, the allowed $k$ values are real and $\omega_{0}^{2}(k)$ is positive (corresponding to a stable equilibrium in the absence of damping). Then the modes of free oscillation are analogous to the free oscillations of a damped oscillator discussed in chapter 2. In fact, if we substitute $\alpha \rightarrow-i \omega$ and $\Gamma \rightarrow \gamma(k)$ in (2.5), we get precisely (8.83). Thus we can take over the solution from (2.6),

$$-i \omega=-\frac{\gamma(k)}{2} \pm \sqrt{\frac{\gamma(k)^{2}}{4}-\omega_{0}^{2}(k)} .$$

This describes a solution that dies out exponentially in time. Whether it oscillates or dies out smoothly depends on the ratio of $\gamma(k)$ to $\omega_{0}(k)$, as discussed in chapter 2.

Forced Oscillation

8-3 – 8-5
Now consider a forced oscillation, in which we drive one end of a translation invariant system with angular frequency $\omega$. After the free oscillations have died away, we are left with oscillation at the single, real angular frequency $\omega$. As always, in forced oscillation problems, we think of the real displacement of the end of the system as the real part of a complex displacement, proportional to $e^{-i \omega t}$. Then the dispersion relation, (8.83), applies. Now the dispersion relation determines $k$, and $k$ must be complex.

You may have noticed that none of the dispersion relations that we have studied so far depend on the sign of $k$. This is not an accident. The reason is that all the systems that we have studied have the property of reflection symmetry. We could change $x \rightarrow-x$ without affecting the physics. In fact, a translation invariant system that did not have this symmetry would be a little peculiar. As long as the system is invariant under reflections, $x \rightarrow-x$, the dispersion relation cannot depend on the sign of $k$. The reason is that when $x \rightarrow-x$, the mode $e^{i k x}$ goes to $e^{-i k x}$. If $x \rightarrow-x$ is a symmetry, these two modes with angular wave numbers $k$ and $- k$ must be physically equivalent, and therefore must have the same frequency. Thus the two solutions for fixed $\omega$ must have the form:

$$k=\pm\left(k_{r}+i k_{i}\right)$$

Because of the $\pm$ sign, we can choose $k_{r} > 0$ in (8.85).

In systems with frictional forces, we always find

$$k_{i} \geq 0 \text { for } k_{r}>0 .$$

The reason for this is easy to see if you consider the traveling waves, which have the form

$$e^{-i \omega t} e^{\pm i\left(k_{r}+i k_{i}\right) x}$$

or

$$e^{i\left(\pm k_{r} x-\omega t\right)} e^{\mp k_{i} x} .$$

From (8.88), it should be obvious what is going on. When the $\pm$ is $+$, the wave is going in the $+ x$ direction, so the sign of the real exponential is such that the amplitude of the wave decreases as $x$ increases. The wave peters out as it travels! This is what must happen with a frictional force. The other sign would require a source of energy in the medium, so that the wave amplitude would grow exponentially as the wave travels. A part of an infinite damped traveling wave is animated in program 8-3.

The form, (8.88) has some interesting consequences for forced oscillation problems in the presence of damping. In damped, discrete systems, even in a normal mode, the parts of the system do not all oscillate in phase. In damped, continuous systems, the distinction between traveling and standing waves gets blurred.

Consider a forced oscillation problem for the transverse oscillation of a string with one end, at $x = 0$ fixed, and the other end, $x = L$ driven at frequency $\omega$. It will not matter until the end of our analysis whether the string is continuous, or has beads with separation a such that $n a = L$ for integer $n$. The boundary conditions are

$$\psi(L, t)=A \cos \omega t, \quad \psi(0, t)=0 .$$

As usual, we regard $\psi(x, t)$ as the real part of a complex displacement, $\tilde{\psi}(x, t)$, satisfying

$$\tilde{\psi}(L, t)=A e^{-i \omega t}, \quad \tilde{\psi}(0, t)=0 .$$

If $k$, for the given angular frequency $\omega$, is given by (8.85), then the relevant modes of the infinite system are those in (8.87), and we must find a linear combination of these two that satisfies (8.89). The answer is

$$\tilde{\psi}(x, t)=A\left[\left(\frac{e^{i\left(k_{r}+i k_{i}\right) x}-e^{-i\left(k_{r}+i k_{i}\right) x}}{e^{i\left(k_{r}+i k_{i}\right) L}-e^{-i\left(k_{r}+i k_{i}\right) L}}\right) e^{-i \omega t}\right] .$$

The factor in parentheses is constructed to vanish at $x = 0$ and to equal 1 at $x = L$.

For a continuous string, the solution, (8.91), is animated in program 8-4. The interesting thing to notice about this is that near the $x = L$ end, the solution looks like a traveling wave. The reason is that here, the real exponential factors in (8.91) enhance the left-moving wave and suppress the right-moving wave, so that the solution is very nearly a traveling wave moving to the left. On the other hand, near $x = 0$, the real exponential factors are comparable, and the solution is very nearly a standing wave. We will discuss the more complicated behavior in the middle in the next chapter.

The same solution works for a beaded string (although the dispersion relation will be different). An example is shown in the animation in program 8-5. Here you can see very clearly that the parts of the system are not all in phase.

______________________
⁵In the presence of damping, the sign of i matters. The relations below would look different if we had used eiωt, and we could not use cos ωt or sin ωt.