Seesaw

To the deflection of the mass and the lengths of springs arise geometrically based on the following diagram:

The position and orientation of the seesaw can be described by the height $h$ of its center and the angle of inclination $\alpha$ . The vertical deflection then gives $\delta = (h_0 - h) + l \sin \alpha \approx (h_0 - h) + l \alpha + \mathcal{O}(\alpha^3)$ The lengths $s_1$ and $s_2$ of the springs results acccordingly $\begin{aligned} s_{1,2} &= \sqrt{a^2 (1 - \cos \alpha)^2 + (h \pm a \sin \alpha)^2 } \\ &= \sqrt{h^2 + 2 a^2 (1 - \cos \alpha) \pm 2 a h \sin \alpha} \end{aligned}$ Now we make an Taylor expansion of length $s_1$ and $s_2$ as a function of the variables $h$ and $\alpha$ . Therefore, we need the partial derivatives up the the second order $\begin{aligned} \frac{\partial s_i}{\partial h} &= \frac{h \pm a \sin \alpha}{s_i} \\ \frac{\partial s_i}{\partial \alpha} &= \frac{a^2 \sin \alpha \pm a h \cos \alpha}{s_i} \\ \frac{\partial^2 s_i}{\partial h^2} &= \frac{1}{s_i} - \frac{(h + a \sin \alpha)^2}{s_i^3} \\ \frac{\partial^2 s_i}{\partial \alpha^2} &= \frac{a^2 \cos \alpha \mp a h \sin \alpha}{s_i} - \frac{(a^2 \sin \alpha \pm a h \cos \alpha)^2}{s_i^3} \\ \frac{\partial^2 s_i}{\partial \alpha \partial h} = \frac{\partial^2 s_i}{\partial h \partial \alpha} &= \pm \frac{a \cos \alpha}{s_i} - \frac{(h \pm a \sin \alpha)(a^2 \sin \alpha \pm a h \cos \alpha)}{s_i^3} \end{aligned}$ The second order Taylor expansion results to $\begin{aligned} s_i &\approx \left. s_i \right|_{h_0,0} + \left. \frac{\partial s_i}{\partial h} \right|_{h_0,0} (h - h_0) + \left. \frac{\partial s_i}{\partial \alpha} \right|_{h_0,0} \alpha \\ & \quad + \left. \frac{1}{2}\frac{\partial^2 s_i}{\partial h^2} \right|_{h_0,0} (h - h_0)^2 + \left. \frac{1}{2}\frac{\partial^2 s_i}{\partial \alpha^2} \right|_{h_0,0} \alpha^2 + \left. \frac{\partial^2 s_i}{\partial h \partial \alpha} \right|_{h_0,0} (h_0 - h) \alpha \\ &= h \pm a \alpha \end{aligned}$ The total energy of the system is the sum of the elastic energies of the springs and the potential energy of the mass: $\begin{aligned} E(h, \alpha) &= \frac{k}{2} (s_1 - h_0)^2 + \frac{k}{2} (s_2 - h_0)^2 - m g \delta \\ &\approx \frac{k}{2} (h - h_0 + a \alpha)^2 + \frac{k}{2} (h - h_0 - a \alpha)^2 - m g \left(h_0 - h + l \alpha \right) \\ &\approx k (h - h_0)^2 + k a^2 \alpha^2 - m g \left(h_0 - h + l \alpha \right) \end{aligned}$ In equilibrium, the energy als function of $h$ and $\alpha$ is minimal. Therefore, both partial derivatives must be zero $\begin{aligned} 0 \stackrel{!}{=} \frac{\partial E}{\partial h} &= 2 k (h - h_0) + m g \\ \Rightarrow \qquad h &\approx h_0 - \frac{m g}{2 k} \\ 0 \stackrel{!}{=} \frac{\partial E}{\partial \alpha} &= 2 k a^2 \alpha - m g l \\ \Rightarrow \qquad \alpha &\approx \frac{m g l}{2 k a^2} \end{aligned}$ Thus, we get the final result $\begin{aligned} \delta &\approx h_0 - h + l \alpha \\ &\approx \frac{m g}{2 k} \left(1 + \frac{l^2}{a^2} \right) \\ &\approx \frac{1 \cdot 10}{2 \cdot 1000} \left(1 + 9 \right) \,\text{m} \\ &= 5 \,\text{cm} \end{aligned}$