## 2 Analysis of Cable Vibration under Double-End Excitation
### Theoretical Model
During cable vibration, a relationship exists between the force acting on each micro-segment of the cable and its horizontal component force, as in Equation (6). \[\frac{T}{T_{v}}=\frac{ds}{du} \tag{6}\]  
where \(T=EA\varepsilon\) is the axial force of the cable. Then, the dynamic tension \(T_{v}\) can be derived as Equation (7). \[T_{v}\frac{ds}{du}\left(\frac{ds_{0}}{du}\right)^{2}=EA\left\{\frac{dv}{du} \frac{dv_{s}}{du}+\frac{1}{2}\Bigg{[}\left(\frac{dv}{du}\right)^{2}+\left( \frac{dw}{du}\right)^{2}\Bigg{]}\right\} \tag{7}\]  
The approximate average dynamic tension can be obtained by integrating Equation (7) along the \(u\)-axis for the entire cable. \[T_{v}=\frac{EA}{l_{v}}\Bigg{\{}\int_{0}^{l}\Bigg{[}\frac{dv}{du}\frac{dv_{s}} {du}+\frac{1}{2}\Bigg{(}\left(\frac{dv}{du}\right)^{2}+\left(\frac{dw}{du} \right)^{2}\Bigg{)}\Bigg{]}du\Bigg{\}} \tag{8}\]  
where \(l_{v}\) is the in-plane virtual length. \[l_{v}=\int_{0}^{l}\frac{ds}{du}\bigg{(}\frac{ds_{0}}{du}\bigg{)}^{2}du=\int_{ 0}^{l}\sqrt{1+\frac{d^{2}v}{du^{2}}}\cdot\left(1+\frac{d^{2}v_{s}}{du^{2}} \right)du \tag{9}\]  
The out-plane dynamic tension \(T_{w}\) can be derived analogically, and it should be noted that there is no out-plane sag. \[T_{w}=\frac{EA}{l_{w}}\left\{\int_{0}^{l}\frac{1}{2}\Bigg{(}\left(\frac{dv}{ du}\right)^{2}+\left(\frac{dw}{du}\right)^{2}\Bigg{)}du\right\} \tag{10}\]where \(l_{w}\) is the out-plane virtual length. \[l_{w}=\int_{0}^{l}\frac{ds}{du}du=\int_{0}^{l}\sqrt{1+\frac{d^{2}v}{du^{2}}}du \tag{11}\]  
The dynamic tension caused by the variation of the boundary condition is deduced using Equation (12). \[ \begin{split} T_{m}=\frac{EA}{l}\bigg{(}\sqrt{\left(l+\Delta U \right)^{2}+\Delta V^{2}+\Delta W^{2}}-l\bigg{)}\\ \approx\frac{EA}{l}\Delta U(t)+EA\frac{\Delta U^{2}(t)}{2l^{2}} +EA\frac{\Delta V^{2}(t)}{2l^{2}}+EA\frac{\Delta W^{2}(t)}{2l^{2}}\end{split}  \tag{12}\]  
The variation of the boundary condition is the relative displacement of the cable's two ends, which is relatively small compared to the length of the cable, so the high-order terms in Equation (12) can be ignored to simplify the equation as Equation (13). \[T_{m}=\frac{EA}{l}\Delta U(t) \tag{13}\]  
Assuming the force excitation of the cable caused by the end vibration is a linear triangular distribution along the axis of the cable, the force in-plane and out-plane excitation can be deduced as Equations (14) and (15). \[f_{v}(u,t)=m\Big{(}\frac{x}{l}\ddot{V}_{j}+\Big{(}1-\frac{x}{l}\Big{)}\ddot{V }_{i}\Big{)} \tag{14}\]  
\[f_{w}(w,t)=m\Big{(}\frac{x}{l}\ddot{W}_{j}+\Big{(}1-\frac{x}{l}\Big{)}\ddot{W }_{i}\Big{)} \tag{15}\]  
Substituting Equation (3) to Equation (15) into Equation (2), the partial differential equations of the cable's in-plane and out-plane can be rewritten as:  
\[ \begin{split}\bigg{\{}T_{s}+\frac{EA}{l}\Delta U(t)+\frac{EA}{l_{ v}}\bigg{\{}T_{0}+\frac{EA}{l_{v}}\bigg{\{}T_{s}+\frac{EA}{l}\Delta U(t)+\frac{ EA}{l_{v}}\bigg{\{}T_{0}+\frac{EA}{l_{v}}\bigg{\{}T_{0}+\frac{EA}{l_{v}}\bigg{\}}\bigg{\}} \bigg{\}}\frac{d^{2}v}{du^{2}}+\\ \bigg{\{}\frac{EA}{l_{v}}\bigg{\{}T_{s}+\frac{EA}{l}\Delta U(t)+ \frac{EA}{l_{w}}\bigg{\{}T_{0}+\frac{EA}{l_{w}}\bigg{\}}\bigg{\}}\bigg{\}} \frac{d^{2}v}{du^{2}}=m\ddot{w}+c_{w}\ddot{w}+m\Big{(}\frac{x}{l}\ddot{W}_{j}+ \Big{(}1-\frac{x}{l}\Big{)}\ddot{W}_{i}\Big{)}\end{split}  \tag{16}\]  
\[\bigg{\{}T_{s}+\frac{EA}{l}\Delta U(t)+\frac{EA}{l_{w}}\bigg{\{}T_{0}^{l}\frac {1}{2}\bigg{(}\bigg{(}\frac{dv}{du}\bigg{)}^{2}+\bigg{(}\frac{dw}{du}\bigg{)} ^{2}\bigg{)}du\bigg{\}}\bigg{\}}\bigg{\}}\frac{d^{2}v}{du^{2}}=m\ddot{w}+c_{w} \ddot{w}+m\Big{(}\frac{x}{l}\ddot{W}_{j}+\Big{(}1-\frac{x}{l}\Big{)}\ddot{W}_{i }\Big{)} \tag{17}\]  
The Galerkin method [25] is adopted to solve the partial differential equations, considering that the vibration of the cable is the superposition of the first m-order sinusoidal modes.