# Analysis of Dynamic Characteristics of Attached High Rise Risers
## 3 Dynamic Analysis of VNA System under Combined Excitation
### Vibration Response of the VNA System under First Order Frequency Locking
\begin{split} O(\varepsilon^{1})&:D_{0}^{2}\overline{ \eta}_{11}+\Omega_{g}{}^{2}\overline{\eta}_{11}=0\\ & D_{0}^{2}\overline{\eta}_{11}+\Omega_{g}{}^{2}\overline{\eta}_{ 11}=0\end{split}  \tag{34}\]  
\[ \begin{split} O(\varepsilon^{3})&:D_{0}^{2}\overline{ \eta}_{13}+\Omega_{g}{}^{2}\overline{\eta}_{13}=-\left(\begin{array}{c} \widehat{\Omega}_{1}{}^{2}b_{5}\overline{\eta}_{11}+\widehat{b}_{1}D_{0} \overline{\eta}_{11}+2D_{0}D_{2}\overline{\eta}_{11}-\sigma_{1}\overline{\eta}_ {11}+\sigma_{2}\overline{\eta}_{11}\\ +b_{5}\sigma_{2}\overline{\eta}_{11}+b_{2}\overline{\eta}_{11}D_{0} \overline{\eta}_{11}+\widehat{b}_{3}\overline{\eta}_{11}+\widehat{b}_{4}D_{0} \overline{\eta}_{11}+b_{6}\overline{\eta}_{11}^{3}\end{array}\right)\end{split}  \tag{35}\]The general solution form of Equation (34) can be expressed as  
\[ \begin{array}{l}\overline{\eta}_{11}(T_{0},T_{2})=A_{11}(T_{2})\mathrm{e}^{ \mathrm{i}\Omega_{g^{\prime}}T_{0}}+\overline{A_{11}(T_{2})}\mathrm{e}^{- \mathrm{i}\Omega_{g^{\prime}}T_{0}}\\ \overline{q}_{11}(T_{0},T_{2})=B_{1}(T_{2})\mathrm{e}^{\mathrm{i}\Omega_{g^{ \prime}}T_{0}}+\overline{B_{1}(T_{2})}\mathrm{e}^{-\mathrm{i}\Omega_{g^{ \prime}}T_{0}}\end{array}  \tag{36}\]  
where  
\[ \begin{array}{l}A_{11}(T_{2})=\frac{1}{2}u_{1}(T_{2})\mathrm{e}^{\mathrm{i} \theta_{1}(T_{2})}\\ B_{1}(T_{2})=\frac{1}{2}u_{2}(T_{2})\mathrm{e}^{\mathrm{i}\theta_{2}(T_{2})} \end{array}  \tag{37}\]  
Here, \(u_{1}\) and \(u_{2}\) respectively represent the time term amplitude of the VNA system and the time term amplitude of the vortex oscillator. \(\theta_{1}\) and \(\theta_{2}\) respectively represent the initial phase of the VNA system and the vortex oscillator. Substitute Equations (36) and (37) into Equation (25), eliminate the perpetual term, and collate to obtain the average equation for the amplitude and phase of the system as follows:  
\[ \begin{array}{l}{u_{1}}^{\prime}=\frac{\widehat{d_{4}}\sin(\theta_{1})- \widehat{d_{2}}\Omega_{g}u_{1}+\sin(\phi)\left(\widehat{\Omega_{1}}^{-2}{}_{ 2}s_{12}+s_{5}\sigma_{2}u_{2}\right)}{2\Omega_{g}}\\ {u_{2}}^{\prime}=-\frac{1}{8\Omega_{g}}\Big{(}4\cos(\phi)\widehat{b_{4}} \Omega_{g}u_{1}+\sin(\phi)\Big{(}4\widehat{b_{3}}u_{1}+3b_{6}{u_{1}}^{3} \Big{)}+4\widehat{b_{1}}\Omega_{g}u_{2}+b_{2}\Omega_{g}{u_{2}}^{3}\Big{)}\\ {\theta_{1}}^{\prime}=\frac{1}{8\Omega_{g^{\prime}}u_{1}}\Big{(}4\cos(\theta_ {1})\widehat{a_{4}}+4\widehat{a_{1}}u_{1}-4\widehat{\Omega_{1}}^{-2}{}_{1}u_{ 1}-4\sigma_{1}u_{1}+3a_{3}{u_{1}}^{3}+\cos(\phi)\Big{(}4\widehat{\Omega_{1}}^ {-2}{}_{2}s_{12}+4a_{5}\sigma_{2}u_{2}\Big{)}\Big{)}\\ {\phi}^{\prime}=\frac{1}{8\Omega_{g^{\prime}}u_{1}}\Big{(}4\cos(\theta_{1}) \widehat{a_{4}}+4\widehat{a_{1}}u_{1}-4\widehat{\Omega_{1}}^{-2}{}_{1}-4 \sigma_{1}u_{1}+3a_{3}{u_{1}}^{3}+\cos(\phi)\Big{(}4\widehat{\Omega_{1}}^{-2}{ }_{2}s_{12}+4a_{5}\sigma_{2}u_{2}\Big{)}\Big{)}\\ {-\frac{1}{8\Omega_{g}{u_{2}}}\Big{(}{-4\widehat{b_{4}}\sin(\phi)\Omega_{g}u_ {1}+\cos(\phi)\Big{(}4\widehat{b_{3}}u_{1}+3b_{6}{u_{1}}^{3}\Big{)}+4\widehat{ \Omega_{1}}^{-2}{}_{2}b_{5}u_{2}-4\sigma_{1}u_{2}+4\sigma_{2}u_{2}+4b_{5} \sigma_{2}u_{2}\Big{)}}\end{array}  \tag{38}\]  
where \(\phi=\theta_{1}-\theta_{2}\). Let \({u_{1}}^{\prime}=0\), \({u_{2}}^{\prime}=0\), \({\theta_{1}}^{\prime}=0\), \(\phi^{\prime}=0\), then the equation for determining the steady solution can be obtained  
\[