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NOT a duplicate of Maximum length stretch of vertical spring with a mass?, I am asking about a system with two connected springs, as shown in this diagram

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For a single spring, you can simply equate the forces in the vertical direction, i.e. $$ d_{\text{max}} = \frac{mg}{k} $$ How do you do this for a double spring pendulum system as shown above with spring constants $k_1$ and $k_2$, masses $m_1$ and $m_2$, and equilibrium lengths $L_1$ and $L_2$. I think the maximum extensions should occur when both springs/pendula are poiting directly downwards, i.e. $\theta_1 = \theta_2 = \frac{\pi}{2}$, as then gravity pulls the pendula parallell to the spring extensions.

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2 Answers 2

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If the masses are stationary and $\theta_1 = \theta_2 = \pi/2$, then you can draw out force diagrams for each of the masses which will give you two equations for two unknowns ($r_1$ and $r_2$). However, if these masses are swinging, the situation becomes a whole lot more complicated. For one, you need to now consider not just the spring and gravitational forces, but also centripetal forces on the two masses ($mv^2/r$) due to their pendulum-like motion, and their velocities are not constant at all. As an aside, even without springs, double pendulum systems exhibit chaotic motion and you can find some very cool demonstrations of this (https://en.wikipedia.org/wiki/Double_pendulum).

I would suggest to consider a bit of a different approach. When the springs are released, you can calculate the total energy (including both gravitational potential energy and spring potential energy). At a later time, the masses will be in different locations, but the total energy must be the same as the initial energy in the absence of friction or other non-conservative forces. Then you can try to see what values of $\theta$ maximize the spring lengths.

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Intuitive I will also say that you obtain the maximum of the springs deflections at $~\theta_1=\frac{\pi}{2}~,~\theta_2=\frac{\pi}{2}~$

mathematical proof

\begin{align*} &\textbf{the generalized coordinates are}\\ &\boldsymbol q=\left[ \begin {array}{c} r_{{1}}\\ r_{{2}} \\ \theta_{{1}}\\ \theta_{{2}} \end {array} \right]\\ &\textbf{postion vectors}\\ &\boldsymbol R_1=\left[ \begin {array}{c} \left( L_{{1}}+r_{{1}} \right) \cos \left( \theta_{{1}} \right) \\ \left( L_{{1}}+r_{{1}} \right) \sin \left( \theta_{{1}} \right) \end {array} \right] \\ &\boldsymbol R_2=\boldsymbol R_1+\left[ \begin {array}{c} \left( L_{{2}}+r_{{2}} \right) \cos \left( \theta_{{1}} \right) \\ \left( L_{{2}}+r_{{2}} \right) \sin \left( \theta_{{1}} \right) \end {array} \right] \end{align*} according to d'Alembert principal you obtain the static equilibrium from this equation

\begin{align*} &\boldsymbol J^T\,\boldsymbol F=0\qquad\qquad (1)\\\\ &\text{with}\\ &\boldsymbol J=\frac{\partial \boldsymbol R}{\partial \boldsymbol q} \qquad,\boldsymbol R=\begin{bmatrix} \boldsymbol R_1 \\ \boldsymbol R_2 \\ \end{bmatrix}\qquad, \boldsymbol F=\begin{bmatrix} \boldsymbol F_1 \\ \boldsymbol F_2 \\ \end{bmatrix}\\\\ &\text{and the force components of mass 1 and mass 2 are}\\\\ &\boldsymbol F_1= -m_1\,\boldsymbol g-k_1\,(L_1+r_1)\,\boldsymbol e_1-k_2\,(L_2+r_2)\boldsymbol e_2\\ &\boldsymbol F_2= -m_2\,\boldsymbol g+k_2\,(L_2+r_2)\boldsymbol e_2\\\\ &\text{and}\\\\ &\boldsymbol e_1=\left[ \begin {array}{c} \cos \left( \theta_{{1}} \right) \\ \sin \left( \theta_{{1}} \right) \end {array} \right]\qquad, \boldsymbol e_2=-\left[ \begin {array}{c} \cos \left( \theta_{{2}} \right) \\ \sin \left( \theta_{{2}} \right) \end {array} \right] \end{align*}

hence equation (1)

\begin{align*} & \boldsymbol J^T\,\boldsymbol F= \left[ \begin {array}{c} \left( -\sin \left( \theta _{{1}} \right) m_{{1}}-\sin \left( \theta _{{1}} \right) m_{{2}} \right) g-k_{{1}}L_ {{1}}-k_{{1}}r_{{1}}\\ -k_{{2}}L_{{2}}-k_{{2}}r_{{2} }-\sin \left( \theta _{{2}} \right) m_{{2}}g\\ - \left( L_{{1}}+r_{{1}} \right) \cos \left( \theta _{{1}} \right) g \left( m_{{2}}+m_{{1}} \right) \\ - \left( L_{{2}}+ r_{{2}} \right) \cos \left( \theta _{{2}} \right) m_{{2}}g \end {array} \right]=0\\ &\Rightarrow\\\\ &\text{from the first two equations you obtain}\\\\ &r_1(\theta_1)=-{\frac {k_{{1}}L_{{1}}+\sin \left( \theta_{{1}} \right) m_{{1}}g+\sin \left( \theta_{{1}} \right) m_{{2}}g}{k_{{1}}}}\\ &r_2(\theta_2)=-{\frac {k_{{2}}L_{{2}}+\sin \left( \theta_{{2}} \right) m_{{2}}g}{k_{ {2}}}}\\\\ &\text{hence the maxima}\\\\ &\frac{dr_1}{d\theta_1}=0\qquad\Rightarrow~\theta_1=\frac{\pi}{2}\\ &\frac{dr_2}{d\theta_2}=0\qquad\Rightarrow~\theta_2=\frac{\pi}{2}\\ &\Rightarrow\\ &r_1(\pi/2)=-{\frac { \left( m_{{2}}+m_{{1}} \right) g}{k_{{1}}}}-L_{{1}}\\ &r_2(\pi/2)=-{\frac {m_{{2}}g}{k_{{2}}}}-L_{{2}}\\\\\\ &d_{1\text{max}}=r_1+L_1=\bigg|\frac{(m_1+m_2)\,g}{k_1}\bigg|\\ &d_{2\text{max}}=r_2+L_2=\bigg|\frac{m_2\,g}{k_2}\bigg| \end{align*}

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