## 4 Special Case of Low Frequency and Low Viscosity  
When the frequency of the incident surface wave is low and the ratio of the crack length to wavelength of the incident surface wave is sufficiently small, the normal and shear tractions generated by the incident surface wave can be considered to be uniformly distributed on the crack surface. In addition, for low-frequency waves, wave dispersion can be neglected when the viscosity is small, that is, the phase velocity of sound waves remains unchanged during propagation. Thus, for the simplicity of verification, the crack length was set to be sufficiently smaller than the wavelength of the incident surface wave. Thus, the tractions on the crack surface can be expressed by the stresses at the center of the crack, which are as follows:  
\[\overline{P}=\overline{\tau}_{xx}=A_{in}T_{xx}(h)e^{-ikx_{1}}\]  
\[\overline{T}=\overline{\tau}_{xz}=A_{in}T_{xz}(h)e^{-ikx_{1}} \tag{35}\]  
where \(x_{1}\) is the distance between the origin of the incident surface wave and the surface crack. In the special case, \(x_{1}=5\) mm. For the special case of low frequency and low viscosity, a numerical calculation was carried out to verify the analytical results. The numerical simulation was solved by the commercial software ABAQUS. In the simulations, the material was rubber, of which the density and Poisson's ratio were 1140 kg/m\({}^{3}\) and 0.49, respectively. When the frequency of the surface wave is 5 kHz, the _Lame_ parameters and Young's modulus can be determined by Equations (6) and (7). Then, the corresponding velocities of the longitudinal wave, shear wave and surface wave can be determined based on Equations (4)-(7), which were 810 m/s, 100 m/s and 95 m/s, respectively. Thus, the corresponding wavelength (\(\lambda\)) was 5.8 mm. In order to make the wavelength of the incident surface wave sufficiently larger than the crack length, the crack lengths (\(L\)) were between 100 \(\upmu\)m and 2000 \(\upmu\)m with a step of 400 \(\upmu\)m. It should be noted that the crack shape was rectangular in the numerical model, and the width of the crack was set to be 10 \(\upmu\)m. The ratio of the crack length to width was greater than 10, which meant the rectangular-shaped defect could be used to simulate crack. For the plane strain condition, the numerical simulation can be modeled in a two-dimensional plane, which would save a lot of calculation time of the model. In order to minimize the influence of the reflected waves from model boundaries, and to reduce the size and running time of the simulation model, absorbing boundaries were applied to the left and right sides and the bottom of the model. The free surface of the model adopted free boundary conditions, and the initial displacement field was zero. In order to accurately solve the wave propagation problems and track the generation process of the scattering surface wave from the crack, a small time step and element size were needed. In general, to solve wave propagation problems, the element size should be smaller than 1/20 of the wavelength to obtain better spatial resolution of the wave propagation, and the time step should be smaller than 1/20 of the reciprocal of the wave frequency, namely:  
\[\Delta t<\frac{1}{20f} \tag{36}\]  
\[\Delta x<\frac{\lambda}{20} \tag{37}\]  
where \(\Delta t\) is the time step of the numerical calculation, \(\Delta x\) is the element size, and \(f\) and \(\lambda\) are the frequency and wavelength ultrasonic wave, respectively. It is worth noting that in the simulation model, the element size in the region around the crack should be equal to the crack width to accurately simulate the wave propagation in the region around the crack. Thus, the element size was 10 \(\upmu\)m near the crack, and gradually increased to 200 \(\upmu\)m in the far field, and the time step of calculation was 2 ns. The amplitude of the incident surface wave was set as 10 nm, which was \(A_{in}=10\) nm. The central depth of the surface crack was \(h=L/2\).