Question

Derive the system of differential equations describing the straight-line vertical motion of the coupled springs shown in Figure 7.6.6. Use the Laplace transform to solve the system

$\begin{array}{l}\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{n}{\mathbf{k}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{k}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{k}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{a}\mathbf{n}\mathbf{d}{\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ {\mathbf{x}}_{\mathbf{1}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\end{array}$

Short answer

$x_1=-\frac{\sin \sqrt{3}t}{\sqrt{3}},x_2=\frac{\sin \sqrt{3}t}{\sqrt{3}}$

Step-by-step solution

Given Information.

The system is

$$\begin{gathered} \begin{array}{l}{F}_{m_1}=-k_1{x}_1+k_2\left(x_2-x_1\right)\\ F_{m_2}=-k_2\left(x_2-x_1\right)-k_3{x}_2\\ m_1{x}_1^{\text{'}\text{'}}=F_{m_1}\\ m_1{x}_1^{\text{'}\text{'}}=-k_1{x}_1+k_2\left(x_2-x_1\right)\\ m_2{x}_2^{\text{'}\text{'}}=-k_2\left(x_2-x_1\right)-k_3{x}_2\end{array} \\ \begin{array}{l}{k}_1=k_2=k_3=1\\ m_1=m_2=1\end{array} \end{gathered}$$

 Step 2: Apply Laplace Transform

\(\begin{aligned}{*{20}{c}}{x_1^{\prime \prime } = - {x_1} + \left( {{x_2} - {x_1}} \right)}\\{ = - 2{x_1} + {x_2}}\end{aligned}\)

\(x_1^{\prime \prime } + 2{x_1} - {x_2} = 0\)

\(x_2^{\prime \prime } = - \left( {{x_2} - {x_1}} \right) - {x_2}\)

\( = {x_1} - 2{x_2}\)

\(x_2^{\prime \prime } - {x_1} + 2{x_2} = 0\)

\(\begin{aligned}{*{20}{c}}{x_1^{\prime \prime } + 2{x_1} - {x_2} = 0}\\{x_2^{\prime \prime } - {x_1} + 2{x_2} = 0}\end{aligned}\)

\({x_1}(0) = 0,x_1^\prime (0) = - 1,{x_2}(0) = 0,{x_2}(0) = 1\)

\(L\left\{ {{f^n}(t)} \right\} = {s^n}F(s) - {s^{n - 1}}f(0) - {s^{n - 2}}f'(0) - \ldots - {f^{n - 1}}(0)\)

\(L\left\{ {x_1^{\prime \prime }} \right\} + 2L\left\{ {{x_1}} \right\} - L\left\{ {{x_2}} \right\} = L\{ 0\} \)

\(\left( {{s^2}{X_1}(s) - s{x_1}(0) - x_1^\prime (0)} \right) + 2{X_1}(s) - {X_2}(s) = 0\)

\({s^2}{X_1}(s) - s \cdot (0) - ( - 1) + 2{X_1}(s) - {X_2}(s) = 0\)

\(\left( {{s^2} + 2} \right){X_1}(s) - {X_2}(s) = - 1\)

\(L\left\{ {x_2^{\prime \prime }} \right\} - L\left\{ {{x_1}} \right\} + 2L\left\{ {{x_2}} \right\} = L\{ 0\} \)

\(\left( {{s^2}{X_2}(s) - s{x_2}(0) - x_2^\prime (0)} \right) - {X_1}(s) + 2{X_2}(s) = 0\)

\({s^2}{X_2}(s) - s \cdot (0) - (1) - {X_1}(s) + 2{X_2}(s) = 0\)

\(\left( {{s^2} + 2} \right){X_2}(s) - {X_1}(s) = 1\)

\({X_1}(s) - \left( {{s^2} + 2} \right){X_2}(s) = - 1\)

\({X_1}(s) = - \frac{1}{{{s^2} + 3}}\)

\({X_2}(s) = \frac{1}{{{s^2} + 3}}\)

Apply Inverse Laplace Transform

\({x_1} = {L^{ - 1}}\left\{ {{X_1}(s)} \right\}\)

\( = {L^{ - 1}}\left\{ { - \frac{1}{{{s^2} + 3}}} \right\}\)

\( = - \frac{{\sin \sqrt 3 t}}{{\sqrt 3 }}\)

\({x_2} = {L^{ - 1}}\left\{ {{X_2}(s)} \right\}\)

\( = {L^{ - 1}}\left\{ {\frac{1}{{{s^2} + 3}}} \right\}\)

\( = \frac{{\sin \sqrt 3 t}}{{\sqrt 3 }}\)

\({x_2} = \frac{{\sin \sqrt 3 t}}{{\sqrt 3 }}\)

\({x_1} = - \frac{{\sin \sqrt 3 t}}{{\sqrt 3 }},\;\;\;{x_2} = \frac{{\sin \sqrt 3 t}}{{\sqrt 3 }}\)