Skip to content

Dispersion curve of Lamb wave on complex plane and its properties.

Our mode are like this:

Lamb wave is a kind of elastic guide wave, its properties and dispersion curve equation can be found in any solid acoustic book. I’ll introduce the equation briefly, then I’ll discusss about the properties of complex dispersion curve and what is Lamb mode.

1. Dispersion curve of Lamb wave The dispersion curve equation can be wrote as:
\begin{equation}\label{eq 1}
\frac{\tan ({k_t}h/2)}{\tan ({k_l}h/2)} = – {\left[ {\frac{4{k^2}{k_l}{k_s}}{\left(k_s^2 – {k^2}\right)^2}} \right]^{ \pm 1}}

‘$+$’ and ‘$-$’ correspond to symmetry mode and antisymmetry mode, respectively. $k_l$ is the longitudinal wave velocity, $k_t$ is the transverse wave velocity, $ y = \pm \frac{h}{2} $ is the free boundary, i.e. $h$ is the thickness of plate, and
\begin{equation}\label{eq 2}
k_l^2 + k^2 = \left( \frac{\omega}{c_l} \right)^2,\ \ k_t^2 + k^2 = \left( \frac{\omega}{c_t} \right)^2

In order to simplify eq.\eqref{eq 1}, we set $k’ = kh$, $k’_l = k_lh$, $k’_t = k_th$, $k’ = kh$, $\omega’ = \omega d$, and equations become:
\begin{equation}\label{eq 3}
\frac{\tan ({k’_t}/2)}{\tan ({k’_l}/2)} = – {\left[ {\frac{4{k’}^2{k’_l}{k’_s}}{\left({k’_s}^2 – {k’}^2\right)^2}} \right]^{ \pm 1}}

\begin{equation}\label{eq 4}
{k’_l}^2 + {k’}^2 = \left( \frac{\omega’}{c_l} \right)^2,\ \ {k’_t}^2 + {k’}^2 = \left( \frac{\omega’}{c_t} \right)^2

Now the thickness $h$ disappears. $\omega’ $ is called the frequency-thickness product.
We can solve eq.\eqref{eq 3} numerically, and plot the graph of $\omega’ $ w.r.t. $k’$:

Since phase velocity $c_p$ equates ${\omega’}/{k’}$, we can get another common graph:

The graphs are consist of many lines, we call each line a Lamb mode. Symmetry modes are $S_0, S_1, \cdots$, while antisymmetry modes are $A_0, A_1, \cdots$.

2. What’s the use of Lamb mode? Lamb modes have explicit physical meaning. Given a frequency-thickness product $\omega’_0$, each Lamb modes are supposed to correspond a possible displacement field $U_n(x,y)$ and $V_n(x,y)$. Here $U_n$ and $V_n$ are complete orthogonal basis, i.e. any displacement field $(u,v)$ can be composed by the basis.

Since Lamb modes can be the basis, each displacement field in plate should be composed by Lamb modes, even when the thickness of plate is varying.

balabala… skip this just for now.

In short, each Lamb mode should be single-valued and continuous w.r.t. frequency-thickness product.

3. The properties of Lamb modes

I choose a $\omega’_0$ to show the Lamb modes. As you can see below, $\omega’_0$ corresponds to 3 $k’$. 3 points must be 3 Lamb modes, nevertheless, it seems like 2 points belong to $s_1$ mode. There must be something wrong.

The key is that $k’$ can be a complex value. So please allow me to replot the graph:

The graph only contains symmetry Lamb modes, or it will be a little confusing. We can see the original left part of $S_1$ is $S_2$ actually. The two modes coalesce at the point where the group velocity is zero, then with the decreasing of $\omega’$, the point forks and forms two complex conjugate Lamb modes.

The conditions of coalescence are more common on imaginary axis. When we decompose a guided wave into Lamb modes, the coalescence of modes will be the problem mathematically. Fortunately, we find a possible solution: when we meet a coalescence at given $\omega’_0$, we set an imaginary part $\epsilon = 10^{-2}$ of $\omega’_0$, and the Lamb modes will not coalesce:

The imaginary part of frequency-thickness product doesn’t have any physical meaning. It’s a purely mathematical process, and calculations show that the process won’t influence the results.

No comments yet.

Leave a Reply

Your email address will not be published. Required fields are marked *

SidebarComments (0)