## 4 Discussion
By solving Equation (11), the response equation of rock particles under regular coordinates can be obtained, as follows:  
\[\eta_{i}(t)=\frac{F_{0}\sin\omega t}{\omega_{i}^{2}-\omega^{2}}\sqrt{\frac{2}{ m}}\sin\frac{(2i-1)\pi}{2}+\frac{F_{s}}{\omega_{i}^{2}}\sqrt{\frac{2}{m}}\sin \frac{(2i-1)\pi}{2} \tag{18}\]  
The response of rock to the combination of static load and ultrasonic vibration load is as follows:  
\[u(x,t)=\frac{2}{m}\sum_{1}^{\infty}\left(\frac{F_{0}\sin\omega t}{\omega_{i} ^{2}-\omega^{2}}+\frac{F_{s}}{\omega_{i}^{2}}\right)\sin\frac{(2i-1)\pi}{2} \sin\frac{(2i-1)\pi}{2h}x \tag{19}\]  
When the rock is assumed to be a single-degree-of-freedom model, Equation (15) can be converted as follows:  
\[u(x,t)=\frac{2}{m}\left[\left(\frac{F_{0}\sin\omega t}{\omega_{0}^{2}-\omega ^{2}}+\frac{F_{s}}{\omega_{0}^{2}}\right)\sin\frac{\pi}{2h}x\right] \tag{20}\]  
Taking \(m=1\) kg, \(\omega/\omega_{0}=1/2\), \(F_{s}=40\) N, \(F_{0}=8\) kN, The variation curves of particle displacement at various section positions in rock with respect to excitation time under the influence of ultrasonic high-frequency vibration are presented in \[U_{i}(x)=D_{i}\sin\frac{(2i-1)\pi}{2h}x,i=1,2,\ldots \tag{12}\]0 The figure indicates that the amplitude of particle displacement within the rock progressively decreases as the distance between the rock layer and the excitation surface increases. The vibration amplitude of particles at the excitation surface (height \(h\)) reaches 5.2 \(\upmu\)m. When the layer height varies from \(h\) to 3/4 \(h\), \(h/2\), and \(h/4\), the displacement amplitudes of rock particles in the corresponding horizons decrease by 0.076, 0.293, and 0.617, respectively. Furthermore, a larger distance from the excitation surface corresponds to a greater reduction in the maximum displacement amplitude. \[U_{i}(x)=D_{i}\sin\frac{(2i-1)\pi}{2h}x,i=1,2,\ldots \tag{12}\]1 The displacement of particles in different positions of rocks varies. In the vibrational regime, the force on the unit particle in different layers of rock is expressed as follows:  
\[U_{i}(x)=D_{i}\sin\frac{(2i-1)\pi}{2h}x,i=1,2,\ldots \tag{12}\]2  
where \(m_{p}\) stands for particle mass; \(a(x,t)\) represents particle acceleration. The force experienced by particles within the rock is directly proportional to their displacement, as illustrated in Equation (17). Consequently, the closer the individual particle is to the excitation surface, the greater the force acting upon it, and the more likely it is to reach the rock's fracture strength, thereby causing damage. Conversely, as particles move further away from the excited surface, the amplitude of the stress decay within the rock increases, making it more challenging for the rock to sustain damage. When the distance from the excitation surface surpasses a specific value, the rock in this region remains undamaged and undergoes only elastic deformation due to the vibrational load. This phenomenon elucidates why localized rock fractures occur under the influence of an ultrasonic vibration load. Presently, the majority of mechanical models developed in pertinent studies consider the load force acting on the rock's end face as a single harmonic force uniformly distributed along the rock's axial direction, neglecting the static load effect. This holistic model fails to investigate the deformation and displacement distribution characteristics of the rock itself. This paper presents an optimized model. Firstly, the load force on the rock is considered as a concentrated force acting on the end face, which provides an explanation for the variation in the damage degree at different rock layers under vibratory load to some extent. However, this model is not without limitations. It assumes the rock to be an isotropic ideal viscoelastic body; thus, the model can only describe the elastic deformation stage of the rock during vibration excitation.