# Analysis of Dynamic Behavior of Spray Boom under Step Excitation
## 2 Materials and Methods
### Vibration Model of Spray Boom
\begin{array}{l}\delta W(x,t)=\sum_{i}X_{i}(x)\delta q_{i}.\end{array}  \tag{35}\]  
The virtual work done on the virtual displacement of unit length load on the unilateral spray boom is:  
\[ \begin{array}{l}\delta W=\int_{0}^{L}f(x,t)\left[\sum_{i}X_{i}(x)\delta q_{ i}\right]dx\\ =\sum_{i}\delta q_{i}\int_{0}^{L}f(x,t)X_{i}(x)dx\\ =\sum_{i}Q_{i}\delta q_{i}.\end{array}  \tag{36}\]  
The generalized force \(F_{i}\) is:  
\[F_{i}=\int_{0}^{L}f(x,t)X_{i}(x)dx. \tag{37}\]  
The kinetic energy, potential energy, and generalized force are substituted into the Lagrange equation:  
\[\frac{d}{dt}\left(\frac{\partial E_{K}}{\partial\dot{q}_{i}}-\frac{\partial E _{P}}{\partial\dot{q}_{i}}\right)-\left(\frac{\partial E_{K}}{\partial\dot{q }_{i}}-\frac{\partial E_{P}}{\partial\dot{q}_{i}}\right)=F_{i}. \tag{38}\]  
By simplifying Equation (38), the motion differential equation of the spray boom in generalized coordinates is obtained:  
\[ \begin{array}{l}\ddot{q}_{i}(t)+\omega_{3i}^{2}q_{i}(t)=F_{i}(t)/M_{3i}\ \ \ i=1, \ 2,\ 3.\end{array}  \tag{39}\]Given the damping of the spray boom and let the \(i\)th damping ratio be \(\zeta_{3i}\), then Equation (39) is changed to  
\[\ddot{q}_{i}(t)+2\zeta_{3i}\omega_{3i}\dot{q}_{i}(t)+\omega_{3i}^{2}q_{i}(t)=F_{ i}(t)/M_{3i}\;\;i=1,\;2,\;3. \tag{40}\]  
The inertial force acting on the spray boom when it vibrates with the fixed support end can be expressed as:  
\[f(0,t)=\rho A\frac{d^{2}Y_{2}(t)}{dt^{2}}. \tag{41}\]  
Substituting Equation (41) into Equation (37), the generalized force on the spray boom is  
\[F_{i}=\int_{0}^{L}\rho A\frac{d^{2}Y_{2}(t)}{dt^{2}}X_{i}(x)dx=\rho A\frac{d^{ 2}Y_{2}(t)}{dt^{2}}\int_{0}^{L}X_{i}(x)dx. \tag{42}\]  
By substituting this generalized force into Equation (40), the vibration differential equation of the spray boom with a flexible foundation is  
\[ \begin{split}\ddot{q}_{i}(t)+2\zeta_{3i}\omega_{3i}\dot{q}_{i}(t )+\omega_{3i}^{2}q_{i}(t)&=\rho A\frac{d^{2}Y_{2}(t)}{dt^{2}} \int_{0}^{L}X_{i}(x)dx/\int_{0}^{L}\rho AX_{i}^{2}(x)dx\\ &=\eta_{n}\dot{Y}_{2}\;\;i=1,2,3,\ldots\end{split}  \tag{43}\]  
where \(\eta_{n}=\int_{0}^{L}X_{i}(x)dx/\int_{0}^{L}X_{i}^{2}(x)dx\). The Laplace transform of Equation (43) is:  
\[s^{2}Q_{i}(s)+2\zeta_{3i}\omega_{3i}sQ_{i}(s)+\omega_{3i}^{2}Q_{i}(s)=\eta_{n }\dot{Y}_{2}(s). \tag{44}\]  
Then, the response function between \(q_{i}(t)\) and \(\dot{Y}_{2}(t)\) is:  
\[H_{i}(s)=\frac{\eta_{n}}{s^{2}+2\zeta_{3i}\omega_{3i}s+\omega_{3i}^{2}}. \tag{45}\]  
Since the vibration response of the spray boom does not exceed the first-order mode when it is working, we only calculate the response under the first-order mode. Then, the generalized coordinates of the first order at the extremity of the spray boom are:  
\[q_{1}(t)=\frac{\eta_{1}}{s^{2}-2\zeta_{31}\omega_{31}s+\omega_{31}^{2}}\ddot {Y}_{2}(s). \tag{46}\]  
Then, the differential dynamics model of the extremity of the spray boom in the first-order mode is  
\[W_{1}(x,t)=X_{1}(x)\frac{\eta_{1}}{s^{2}-2\zeta_{31}\omega_{31}s+\omega_{31}^ {2}}\ddot{Y}_{2}(s). \tag{47}\]  
Using the inverse Laplace transform, Equation (47) can be rewritten as [14]:  
\[W_{1}(x,t)=X_{1}(x)\eta_{1}\frac{1}{w_{3d}}e^{-\zeta_{31}w_{3d}t}\sin(w_{3d}t) \cdot\ddot{Y}_{2}. \tag{48}\]