A. Waves on a bending beam or a stiff string
We have already met a system whose wave-propagation properties are dispersive. Bending (or flexural) waves on a beam can be described by the Euler-Bernoulli model, and we saw in section 3.2.1 that the equation of motion for free vibration of such a beam is
$$m \dfrac{\partial^2 y}{\partial t^2}+EI \dfrac{\partial^4 y}{\partial x^4}=0 \tag{1}$$
where $y(x,t)$ is the transverse displacement of the beam, $EI$ is the flexural rigidity, and $m$ is the mass per unit length.
We can look for a wave-like solution with frequency $\omega$ and wavenumber $k$: if we substitute
$$y(x,t)=Y e^{i\omega t – ikx} \tag{2}$$
into eq. (1), where $Y$ is a constant, we find the condition
$$EIk^4 = m \omega^2 \tag{3}$$
which is called the dispersion relation for bending waves. For a given (real) value of $\omega$, there are four possible values for $k$
$$k = \pm \alpha \omega ^{1/2},~~~\pm i \alpha \omega^{1/2} \tag{4}$$
where
$$\alpha^4 = \dfrac{m}{EI} . \tag{5}$$
The first pair of these are real, corresponding to travelling waves in the two directions along the beam. The second pair are evanescent waves, decaying exponentially along the beam as we saw in section 3.2.1.
Since $\omega = k^2/\alpha^2$, it is easy to see that the pair of travelling flexural waves are dispersive. From eq. (2) the speed of propagation, called the phase velocity, is
$$c_p = \dfrac{\omega}{k} = \dfrac{k}{\alpha^2} = \dfrac{\omega^{1/2}}{\alpha} \tag{6}$$
so that waves with higher frequency or higher wavenumber (i.e. shorter wavelength) travel faster.
In section 4.4, an example was given of dispersive waves involving bending. However, that example was not for a simple bending beam, it was for a stretched string with some bending stiffness. For such a system, the governing equation in place of eq. (1) is
$$m \dfrac{\partial^2 y}{\partial t^2}+EI \dfrac{\partial^4 y}{\partial x^4} – P\dfrac{\partial^2 y}{\partial x^2} =0 \tag{7}$$
where $P$ is the tension in the string. If we make the substitution from eq. (2) to look for wave-like solutions, we now obtain the dispersion relation
$$EIk^4 + P k^2 = m \omega^2 . \tag{8}$$
This can be rearranged to read
$$\omega = c_0 k \sqrt{1 + \beta k^2} \tag{9}$$
where
$$c_0=\sqrt{P/m} \tag{10}$$
is the (non-dispersive) wave speed on the string in the absence of bending stiffness, and
$$\beta = \dfrac{EI}{P} . \tag{11}$$
The phase velocity is now
$$c_p = c_0 \sqrt{1+ \beta k^2} , \tag{12}$$
so that the waves are indeed still dispersive but with a different pattern from the simple bending beam. For low frequencies, when $k \rightarrow 0$, the phase velocity is approximately equal to $c_0$. But as $k$ increases the effect of bending stiffness increases and eventually for very large values of $k$ it dominates, and the behaviour tends towards that of the beam as given in eq. (6).
B. Surface waves on deep water
The other example system studied in section 4.4 concerns waves on the surface of water. We will give a brief derivation of the dispersion relation for such waves, which will be needed in later subsections. We will treat only the simplest case: small-amplitude surface waves on deep water, with no allowance for surface tension or viscosity. We will take advantage of the vector calculus notation developed in sections 4.1.1 and 4.1.4. We will use the governing equations for fluid flow — if these are not familiar to you, a summary can be found later, in section 11.2.1.
In terms of a Cartesian coordinate system $(x,y,z)$, suppose that the surface of the water lies at $z=\zeta(x,y,t)$ as sketched in Fig.1. We will assume that the water is incompressible, with a constant density $\rho$. The velocity field $\underline{u} (\underline{r},t)$ then satisfies
$$\nabla \cdot \underline{u} = 0 . \tag{13}$$
When the free surface of the water is flat and undisturbed by waves, the pressure distribution with depth is hydrostatic:
$$p_0 = p_a – \rho g z \tag{14}$$
where $p_a$ is atmospheric pressure and $\rho$ is the density of water. Note that $z$ is defined to be positive upwards, so pressure increases with depth as expected because $z$ is negative beneath the surface. Disturbance of the free surface by waves will result in an additional pressure
$$p_e = p~-~p_0 . \tag{15}$$
The linearised momentum equation is now
$$ \rho \dfrac{\partial \underline{u}}{\partial t} = -\nabla p_e \tag{16}$$
(ignoring the nonlinear term $\underline{u}\cdot \nabla \underline{u}$ for small-amplitude waves). Taking the curl of this equation give
$$\rho \dfrac{\partial}{\partial t}(\nabla \times \underline{u}) = -\nabla \times \nabla p_e = 0 \tag{17}$$
so that any “rotational” component of the velocity field is independent of time, and hence contributes nothing to $p_e$ by eq. (16). It thus does not contribute to the motion of the free surface, and can be ignored. The interesting part of the flow can be assumed to be irrotational:
$$\nabla \times \underline{u} = 0 \mathrm{~,~~so~~} \underline{u} = \nabla \phi \tag{18}$$
where $\phi$ is the velocity potential. Now by eq. (13)
$$\nabla \cdot \nabla \phi = \nabla^2 \phi = 0 . \tag{19}$$
Thus the flow is governed by Laplace’s equation, which involves no time derivatives. These enter the problem via the boundary conditions for the free surface, at
$$z= \zeta(x,y,t) . \tag{20}$$
We will assume that the restoring force on this free surface is influenced only by the effect of gravity: we neglect the effect of surface tension.
In order that the pressure $p$ takes the value $p_a$ at the surface, we require
$$p_e = \rho g \zeta . \tag{21}$$
Substituting eq. (18) into eq. (16) gives
$$p_e = -\rho \dfrac{\partial \phi}{\partial t} \tag{22}$$
and so
$$\dfrac{\partial \phi}{\partial t} = -g \zeta . \tag{23}$$
Strictly speaking this derivative of $\phi$ should be evaluated at the disturbed surface, $z=\zeta$. However, within a linearised theory for small-amplitude waves it is good enough to evaluate it at $z= 0$ since the difference would only involve a second-order term.
A second boundary condition at the free surface follows from the fact that the rate of change of $\zeta$, following a particle of fluid, is equal to the vertical component of $\nabla \phi$ there. Within the linearised theory, this requires
$$\dfrac{\partial \zeta}{\partial t} = \left[ \dfrac{\partial \phi}{\partial z} \right]_{z=0} . \tag{24}$$
Differentiating eq. (23) and substituting eq. (24) finally yields the condition
$$\dfrac{\partial^2 \phi}{\partial t^2} = -g \dfrac{\partial \phi}{\partial z} \tag{25}$$
at $z=0$.
This is the governing equation we are after, and we can look for wave-like solutions by trying the substitution
$$\phi = f(z) e^{i \omega t – i k x} \tag{26}$$
where $f(z)$ is an as yet unknown function. Laplace’s equation (19) then requires
$$\dfrac{d^2 f}{d z^2} – k^2 f = 0 , \tag{27}$$
which has solutions $e^{kz}$ and $e^{-kz}$. The first of these behaves as we require for deep water: the motion decays to zero with increasing depth ($z$ large and negative). The second of the solutions, on the other hand, grows exponentially with increasing depth, and is physically unacceptable. So for a simple wave on deep water we must take
$$\phi = A e^{kz} e^{i \omega t – i k x} \tag{28}$$
where $A$ is a constant.
Now we substitute into the boundary condition eq. (25), and obtain the simple result
$$\omega^2 = g k . \tag{29}$$
This is the dispersion relation for our water waves. It follows that the phase velocity is
$$c_p = \dfrac{g}{\omega} = \sqrt{\dfrac{g}{k}} . \tag{30}$$
Waves of low frequency travel fast, those of higher frequency progressively slower.