Coupled Horizontal Spring Mass Oscillator

Coupled Oscillator Spring Mass Oscillator: State Space

Problem Statement

Two 4 Kg Weights are suspended between two walls. They are connected by a spring between them with a spring constant k2. They are connected to the walls by two springs k1 and k3 with k1=k3. m1 is a distance x1 form m2 and m2 is x2 from the wall.

Solution

Things we know

$m_{1}=5kg{\frac {}{}}$

$m_{2}=5kg{\frac {}{}}$

$k_{1}=50Nm{\frac {}{}}$

$k_{2}=100Nm{\frac {}{}}$

$k_{3}=50Nm{\frac {}{}}$

${\text{So now that we have are problem we need to start setting up the equations we need to solve it.}}\,$

${\dot {x_{1}}}={\dot {x_{1}}}$

${\ddot {x_{1}}}+{\frac {k_{1}+k_{2}}{m_{1}}}{x_{1}}-{\frac {k_{2}}{m_{1}}}{x_{2}}=0$

${\dot {x_{2}}}={\dot {x_{2}}}$

${\ddot {x_{2}}}+{\frac {k_{3}+k_{2}}{m_{2}}}{x_{2}}-{\frac {k_{2}}{m_{2}}}{x_{1}}=0$

${\text{Now we take these equations and put them in a state space model.}}\,$

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}\,$ = ${\begin{bmatrix}0&1&0&0\\{\frac {(k_{1}+k_{2})}{m_{1}}}&0&{\frac {-k_{1}}{m_{1}}}&0\\0&0&0&1\\{\frac {-k_{1}}{m_{2}}}&0&{\frac {(k_{1}+k_{2})}{m_{2}}}&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}+{\begin{bmatrix}0\end{bmatrix}}$

${\text{Now we make the appropriate numerical substitutions.}}\,$

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}\,$ = ${\begin{bmatrix}0&1&0&0\\{\frac {150}{5}}&0&{\frac {-50}{5}}&0\\0&0&0&1\\{\frac {-50}{5}}&0&{\frac {150}{5}}&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}+{\begin{bmatrix}0\end{bmatrix}}$

${\begin{bmatrix}{\dot {x_{1}}}\\{\ddot {x_{1}}}\\{\dot {x_{2}}}\\{\ddot {x_{2}}}\end{bmatrix}}\,={\begin{bmatrix}0&1&0&0\\30&0&-10&0\\0&0&0&1\\-10&0&30&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}+{\begin{bmatrix}0\end{bmatrix}}$

${\text{So using Maple I was able to obtain the eigenvalues and eigenvectors.}}\,$

${\text{Eigenvalues.}}\,$

$\lambda _{1}=2{\sqrt {10}}\,$ $\lambda _{2}=-2{\sqrt {10}}\,$ $\lambda _{3}=2{\sqrt {5}}\,$ $\lambda _{4}=-2{\sqrt {5}}\,$

${\text{Eigenvectors.}}\,$

$\ {K_{1}=}\,$${\begin{bmatrix}-1\\-2{\sqrt {(}}10)\\1\\2{\sqrt {(}}10)\end{bmatrix}}\,$,$\ {K_{2}=}\,$${\begin{bmatrix}-1\\2{\sqrt {(}}10)\\1\\-2{\sqrt {(}}10)\end{bmatrix}}\,$,$\ {K_{3}=}\,$${\begin{bmatrix}1\\2{\sqrt {(}}5)\\1\\2{\sqrt {(}}5)\end{bmatrix}}\,$,$\ {K_{4}=}\,$${\begin{bmatrix}1\\-2{\sqrt {(}}5)\\1\\-2{\sqrt {(}}5)\end{bmatrix}}\,$

${\text{So then the answer is...}}\,$

$\ x=c_{1}$${\begin{bmatrix}-1\\-2{\sqrt {(}}10)\\1\\2{\sqrt {(}}10)\end{bmatrix}}\,$$e^{2{\sqrt {10}}}+c_{2}$${\begin{bmatrix}-1\\2{\sqrt {(}}10)\\1\\-2{\sqrt {(}}10)\end{bmatrix}}\,$$e^{2*-2{\sqrt {10}}}+c_{3}$${\begin{bmatrix}1\\2{\sqrt {(}}5)\\1\\2{\sqrt {(}}5)\end{bmatrix}}\,$$e^{3*2{\sqrt {5}}}+c_{4}$${\begin{bmatrix}1\\-2{\sqrt {(}}5)\\1\\-2{\sqrt {(}}5)\end{bmatrix}}\,$$e^{4*-2{\sqrt {5}}}\,$

Solve with the Matrix exponential

${\text{So first we need to know what the matrix exponential equation looks like.}}\,$

${\text{it is...}}\,$

${\tilde {x}}=e^{{\tilde {A}}t}{\tilde {x(0)}}\,$

${\text{Where A is a matrix}}\,$

${\text{Also }}\,$

${\tilde {z}}={\tilde {T}}{\tilde {x}}\,$

${\tilde {x}}={\tilde {T}}^{-1}{\tilde {z}}\,$

${\text{Where }}\,$

${\tilde {T}}^{-1}=\,$${\begin{bmatrix}-1&-1&1&1\\-2{\sqrt {(}}10)&2{\sqrt {(}}10)&2{\sqrt {(}}5)&-2{\sqrt {(}}5)\\1&1&1&1\\2{\sqrt {(}}10)&-2{\sqrt {(}}10)&2{\sqrt {(}}5)&-2{\sqrt {(}}5)\end{bmatrix}}\,$

${\text{I converted the T matrix to decimal form for make it easier to write up on here }}\,$

${\tilde {T}}=\,$</math>${\begin{bmatrix}-.25&-.039528&.25&.039528\\-.25&.039528&.25&-.039528\\.25&.055902&.25&.055902\\.25&-.055902&.25&-.055902\end{bmatrix}}\,$

${\text{and}}\,$

${\hat {A}}=\,$${\begin{bmatrix}e^{\lambda _{1}t}&0&0&0\\0&e^{\lambda _{2}t}&0&0\\0&0&e^{\lambda _{3}t}&0\\0&0&0&e^{\lambda _{4}t}\end{bmatrix}}\,$

$e^{{\hat {A}}t}=\,$${\begin{bmatrix}e^{2{\sqrt {10}}t}&0&0&0\\0&e^{-2{\sqrt {10}}t}&0&0\\0&0&e^{2{\sqrt {5}}t}&0\\0&0&0&e^{-2{\sqrt {5}}t}\end{bmatrix}}\,$

${\text{Then the next step is}}\,$

${\tilde {z}}=e^{{\hat {A}}t}{\tilde {z}}(0)\,$

${\text{So that implies}}\,$

${\tilde {x}}={\tilde {T}}^{-1}e^{{\hat {A}}t}{\tilde {z}}(0)\,$$={\tilde {T}}^{-1}e^{{\hat {A}}t}{\tilde {T}}^{-1}{\tilde {x}}(0)\,$

${\text{Now Simply substitute back in and we have the answer. }}\,$

${\tilde {x}}=\,$${\begin{bmatrix}-1&-1&1&1\\-2{\sqrt {(}}10)&2{\sqrt {(}}10)&2{\sqrt {(}}5)&-2{\sqrt {(}}5)\\1&1&1&1\\2{\sqrt {(}}10)&-2{\sqrt {(}}10)&2{\sqrt {(}}5)&-2{\sqrt {(}}5)\end{bmatrix}}{\begin{bmatrix}e^{2{\sqrt {10}}t}&0&0&0\\0&e^{-2{\sqrt {10}}t}&0&0\\0&0&e^{2{\sqrt {5}}t}&0\\0&0&0&e^{-2{\sqrt {5}}t}\end{bmatrix}}$${\begin{bmatrix}-.25&-.039528&.25&.039528\\-.25&.039528&.25&-.039528\\.25&.055902&.25&.055902\\.25&-.055902&.25&-.055902\end{bmatrix}}{\tilde {x}}(0)\,$

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Created By: Mark Bernet