# Two rigid links with central linear spring support

### Potential energy function

Begin by considering the deflected shape

Note that the two displacement variables, $$u$$ and $$v$$ are related by the equation

\begin{align} u + \sqrt{L^{2} - v^{2}} & = \frac{u}{2} + L \\ \frac{u}{2} & = L - \sqrt{L^{2} - v^{2}} \\ \frac{u}{2L} & = 1 - \sqrt{1 - \frac{v^{2}}{L^{2}}} \\ & \approx \frac{1}{2} \frac{v^{2}}{L^{2}} \\ \frac{u}{2} & = \frac{1}{2} v^{2} \frac{1}{L} \\ \end{align}

Now use the substitution $$k_{g} = \frac{2}{L}$$

$u = \frac{1}{2} k_{g} v^{2}$

Work done through moving $$P$$ a distance $$u$$:

\begin{align} W & = P u \\ & = P \frac{1}{2} k_{g} v^{2} \end{align}

Energy stored in spring:

$V = \frac{1}{2} k_{m} v^{2}$

Total energy in system

\begin{align} \Pi & = V - W \\ & = \frac{1}{2}k_{m} v^{2} - P \frac{1}{2} k_{g} w^{2} \end{align}