## 3 Solution Method
To obtain the global dynamic equilibrium equations of the nanoplate, a global nodal displacement vector is defined as:  
\[\overline{\delta}(X,T)=\Big{[} \begin{array}{cccccccc}W_{1}&\Theta_{1}&W_{2} &\Theta_{2}&\cdots&W_{NE+1}&\Theta_{NE+1}\end{array} \Big{]}^{\mathsf{T}}. \tag{23}\]  
Like the finite element method, the contributions of each strip element are added together and the global dynamic equilibrium equations are obtained as:  
\[\overline{\mathbf{K}}^{\overline{\partial}^{4}\overline{\delta}} \overline{\delta X^{4}}+\overline{\mathbf{K}}^{2}\frac{\partial^{2}\overline{ \delta}}{\partial X^{2}}+\overline{\mathbf{K}}^{0}\overline{\delta}+ \overline{\mathbf{K}}_{1}^{1}\frac{\partial^{2}\overline{\delta}}{\partial X \partial T}+\overline{\mathbf{M}}\frac{\partial^{2}\overline{\delta}}{ \partial T^{2}}+\overline{\mathbf{K}}_{2}^{2}\frac{\partial^{4}\overline{ \delta}}{\partial X^{2}\partial T^{2}}+\overline{\mathbf{K}}_{1}^{3}\frac{ \partial^{4}\overline{\delta}}{\partial X^{3}\partial T}=\mathbf{0}, \tag{24}\]  
where \(\overline{\mathbf{K}}^{(4)},\overline{\mathbf{K}}^{(2)},\overline{\mathbf{K}}^ {(0)},\overline{\mathbf{K}}_{1}^{(1)},\overline{\mathbf{K}}_{2}^{(2)}, \overline{\mathbf{K}}_{1}^{(3)}\) are global stiffness matrices, and \(\overline{\mathbf{M}}\) is a global mass matrix. Further, dealing with the displacement boundary conditions imposed on nodal lines, the global dynamic equilibrium equations are rewritten as:  
\[\tilde{\mathbf{K}}^{4}\frac{\partial^{4}\overline{\delta}^{\prime}_{w}}{ \partial X^{4}}+\tilde{\mathbf{K}}^{2}\frac{\partial^{2}\overline{\delta}^{ \prime}_{w}}{\partial X^{2}}+\tilde{\mathbf{K}}^{0}\overline{\delta}^{\prime }_{w}+\tilde{\mathbf{K}}_{1}^{1}\frac{\partial^{2}\overline{\delta}^{\prime }_{w}}{\partial X\partial T}+\tilde{\mathbf{M}}\frac{\partial^{2}\overline{ \delta}^{\prime}_{w}}{\partial T^{2}}+\tilde{\mathbf{K}}_{2}^{2}\frac{ \partial^{4}\overline{\delta}^{\prime}_{w}}{\partial X^{2}\partial T^{2}}+ \tilde{\mathbf{K}}_{1}^{3}\frac{\partial^{4}\overline{\delta}^{\prime}_{w}}{ \partial X^{3}\partial T}=0, \tag{25}\]  
where  
\[ \begin{array}{c}\tilde{\mathbf{K}}^{4}=\mathbf{T}_{1}^{\mathsf{T}}\overline{ \mathbf{K}}^{4}\mathbf{T}_{1},\widetilde{\mathbf{K}}^{2}=\mathbf{T}_{1}^{ \mathsf{T}}\overline{\mathbf{K}}^{2}\mathbf{T}_{1},\tilde{\mathbf{K}}^{0}= \mathbf{T}_{1}^{\mathsf{T}}\overline{\mathbf{K}}^{0}\mathbf{T}_{1},\\ \tilde{\mathbf{K}}_{1}^{1}=\mathbf{T}_{1}^{\mathsf{T}}\overline{\mathbf{K}}_{1 }^{1}\mathbf{T}_{1},\tilde{\mathbf{K}}_{2}^{2}=\mathbf{T}_{1}^{\mathsf{T}} \overline{\mathbf{K}}_{2}^{2}\mathbf{T}_{1},\tilde{\mathbf{K}}_{1}^{3}= \mathbf{T}_{1}^{\mathsf{T}}\overline{\mathbf{K}}_{1}^{3}\mathbf{T}_{1}, \tilde{\mathbf{M}}=\mathbf{T}_{1}^{\mathsf{T}}\overline{\mathbf{M}}\mathbf{T} _{1},\end{array}  \tag{26}\]  
where \(\mathbf{T}_{1}\) are row transformation matrices containing \(0\) and \(1\), which is referred to literature [39]. Next, the Fourier transform of Equation (25) with respect to time are given by:  
\[\widetilde{\mathbf{K}}^{4}\frac{\partial^{4}\hat{\delta}^{\prime}_{w}}{ \partial X^{4}}+i\omega\tilde{\mathbf{K}}_{1}^{3}\frac{\partial^{3}\hat{ \delta}^{\prime}_{w}}{\partial X^{3}}+\Big{(}\widetilde{\mathbf{K}}^{2}+i \omega\widetilde{\mathbf{K}}_{1}^{2}+(i\omega)^{2}\widetilde{\mathbf{K}}_{2} ^{2}\Big{)}\frac{\partial^{2}\hat{\delta}^{\prime}_{w}}{\partial X^{2}}+i \omega\widetilde{\mathbf{K}}_{1}^{1}\frac{\partial\hat{\delta}^{\prime}_{w}}{ \partial X}+\Big{(}\widetilde{\mathbf{K}}^{0}+(i\omega)^{2}\widetilde{\mathbf{M }}\Big{)}\hat{\delta}^{\prime}_{w}=\mathbf{0}, \tag{27}\]  
where \(\hat{\delta}^{\prime}_{w}\) is the Fourier transformation form of \(\overline{\delta}^{\prime}_{w}\), \(i\omega\) is the Fourier transform parameter, \(\omega\) is the angular frequency and \(i=\sqrt{-1}\). If denoting a status vector as:  
\[\mathbf{\eta}(X,i\omega)=\Big{[}