## 1 Introduction
### The Analytical Solution
\tag{16}\]  
where \(d_{1,\cdots,8}\
eq 0\), \(\mu_{1,2}\
eq 0\) and \(\mu_{1}<\mu_{2}\). The root configuration of the polynomial \(f(\xi)\) can be written as  
\[f(\xi)=\left\{ \begin{array}{ll}[\xi-(ib_{1}+b_{2})][\xi-(ib_{1}-b_{2})](\xi -ic_{1})(\xi-ic_{2}),&g_{\mu_{0}}<0,\\ [\xi-(ib_{3}+b_{4})][\xi-(ib_{3}-b_{4})][\xi-ic_{0})^{2},&g_{\mu_{0}}=0,\\ [\xi-(ib_{5}+b_{6})][\xi-(ib_{5}+b_{6})][\xi-(ib_{7}+b_{8})][\xi-(-ib_{7}-b_{8} )],&g_{\mu_{0}}>0,\end{array} \right. \tag{17}\]  
where \(b_{i}=\sqrt{\beta_{1}}d_{i}\), \(i=1,2,\cdots,8\); \(c_{j}=\sqrt{\beta_{1}}\mu_{j}\), \(j=0,1,2\). To compute the inverse Fourier transform of Equation (7) using the residue theorem, the signs of \(b_{1,3,5,7}\) and \(c_{1,2}\) must be determined. Based on the relationship between roots and coefficients in the first equation of Equation (16), the signs of \(d_{1,3,5,7}\) and \(c_{1,2}\) can be presented as  
\[\left\{ \begin{array}{ll}b_{1}>0,0>c_{1}>c_{2},&g_{\mu_{0}}<0,\\ b_{3}>0,c_{0}<0,&g_{\mu_{0}}=0,\\ b_{5}>0,b_{7}<0,&g_{\mu_{0}}>0.\end{array} \right. \tag{18}\]  
Substituting Equation (13) into Equation (15) yields  
\[g_{\mu_{0}}(\gamma_{1})=\frac{1}{72}(e_{1}(\gamma_{1})+e_{2}(\gamma_{1}))^{2} -\frac{1}{8}\gamma_{1}(e_{1}(\gamma_{1})+e_{2}(\gamma_{1}))+\gamma_{2}. \tag{19}\]  
Taking the derivative of \(g_{\mu_{0}}(\gamma_{1})\) with respect to \(\gamma_{1}\) results in  
\[g_{\mu_{0}}^{\prime}(\gamma_{1})=\frac{1}{4}(\frac{e_{1}(\gamma_{1})+e_{2}( \gamma_{1})}{9}-\frac{\gamma_{1}}{2})\frac{d(e_{1}(\gamma_{1})+e_{2}(\gamma_{ 1}))}{d\gamma_{1}}-\frac{\gamma_{1}}{8}(e_{1}(\gamma_{1})+e_{2}(\gamma_{1})). \tag{20}\]  
It can be easily verified that \(g_{\mu_{0}}^{\prime}(\gamma_{1})<0\) holds for any \(\gamma_{1}>0\), indicating that \(g_{\mu_{0}}\) decreases as \(\gamma_{1}\) increases. Considering the definition of \(\gamma_{1}\) in Equation (10), there must exist a unique velocity \(v=v_{m}\) (where \(v\) is the velocity of the moving load) such that \(g_{\mu_{0}}(v_{m})=0\) for given mechanical and geometric properties of the beam and the moving load (with fixed values of \(c\), \(k\), \(m\), and \(P\)). Moreover, \(g_{\mu_{0}}(v)>0\) when \(v>v_{m}\), and \(g_{\mu_{0}}(v)<0\) when \(v<v_{m}\). Therefore, Equations (17) and (18) can be rewritten as  
\[f(\xi)=\left\{ \begin{array}{ll}[\xi-(ib_{1}+b_{2})][\xi-(ib_{1}-b_{2})](\xi -ic_{1})(\xi-ic_{2}),&v>v_{m},\\ [\xi-(ib_{3}+b_{4})][\xi-(ib_{3}-b_{4})](\xi-ic_{0})^{2},&v=v_{m},\\ [\xi-(ib_{5}+b_{6})][\xi-(ib_{5}-b_{6})][\xi-(ib_{7}+b_{8})][\xi-(ib_{7}-b_{8 })],&v<v_{m},\end{array} \right.\]  
\[\text{and},\quad\left\{ \begin{array}{ll}b_{1}>0,0>c_{1}>c_{2},&v>v_{m},\\ b_{3}>0,c_{0}<0,&v=v_{m},\\ b_{5}>0,b_{7}<0,&v<v_{m},\end{array} \right. \tag{21}\]where \(v=v_{m}\) is the unique positive real root of the equation \(g_{n_{0}}(v)=0\). Since the case \(v=v_{m}\) is less meaningful, the following analysis focuses on \(v>v_{m}\) and \(v<v_{m}\). Applying the inverse Fourier transform to Equation (7) and using the residue theorem yields  
\[w=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{\beta_{4}e^{i\xi y}}{\xi^{4}- \beta_{1}\xi^{2}+i\beta_{2}\xi+\beta_{3}}d\xi=\left\{ \begin{array}{ll}W_{b},& \quad y<0,\\ W_{f},&\quad y\geq 0,\end{array} \right. \tag{22}\]  
where  
\[W_{b}=\left\{ \begin{array}{ll}q_{1}e^{-c_{1}y}+q_{2}e^{-c_{2}y},&\quad v>v_{ m},\\ (q_{3}e^{-ib_{3}y}+q_{4}e^{ib_{3}y})e^{-b_{7}y},&\quad v<v_{m},\end{array} \right. \tag{23}\]  
\[W_{f}=\left\{ \begin{array}{ll}(q_{5}e^{-ib_{2}y}+q_{6}e^{ib_{2}y})e^{-b_{1}y },&\quad v>v_{m},\\ (q_{7}e^{-ib_{6}y}+q_{8}e^{ib_{6}y})e^{-b_{5}y},&\quad v<v_{m},\end{array} \right. \tag{24}\]  
\[q_{1}=-\beta_{4}i[f_{d}(i_{c})]^{-1},\quad q_{2}=-\beta_{4}i[f_{d}(i_{c}z)]^{ -1},\quad q_{3}=-\beta_{4}i[f_{d}(-b_{8}+ib_{7})]^{-1},\]  
\[q_{4}=-\beta_{4}i[f_{d}(b_{8}+ib_{7})]^{-1},\quad q_{5}=\beta_{4}i[f_{d}(-b_{ 2}+ib_{1})]^{-1},\quad q_{6}=\beta_{4}i[f_{d}(b_{2}+ib_{1})]^{-1},\]  
\(q_{7}=\beta_{4}i[f_{d}(-b_{6}+ib_{5})]^{-1},\quad q_{8}=\beta_{4}i[f_{d}(b_{6} +ib_{5})]^{-1},\)  
\(f_{d}(\cdot)\) is given by Equation (8).