Apr 13, 2018

$\left\{\begin{matrix}m {\ddot{x}}_{1} + k \left({x}_{1} - {x}_{2}\right) = - F \\ 2 m {\ddot{x}}_{2} + k \left({x}_{2} - {x}_{1}\right) = 2 F\end{matrix}\right.$

Applying the Laplace transform

{((s^2+k/m)X_1-k/mX_2=-1/mF+dot x_1(0)+s x_1(0)), (-k/(2m)X_1+(s^2+k/(2m))X_2 = 1/(2m)F+dot x_2(0)+s x_2(0)):}

Now assuming

${\dot{x}}_{1} \left(0\right) = - u$
${\dot{x}}_{2} \left(0\right) = 2 u$

${x}_{1} \left(0\right) = {x}_{2} \left(0\right) = 0$

$\left(\begin{matrix}{s}^{2} + \frac{k}{m} & - \frac{k}{m} \\ - \frac{k}{2 m} & {s}^{2} + \frac{k}{2 m}\end{matrix}\right) \left(\begin{matrix}{X}_{1} \\ {X}_{2}\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{m} F - u \\ \frac{1}{2 m} F + 2 u\end{matrix}\right)$

and after the anti transformation gives

$\left\{\begin{matrix}{x}_{1} = u t - \sqrt{\frac{2}{3 k m}} \left(F + 2 m u\right) \sin \left(\sqrt{\frac{3 k}{2 m}} t\right) \\ {x}_{2} = u t + \sqrt{\frac{1}{6 k m}} \left(F + 2 m u\right) \sin \left(\sqrt{\frac{3 k}{2 m}} t\right)\end{matrix}\right.$

hence

${x}_{2} - {x}_{1} = \left(\sqrt{\frac{1}{6 k m}} + \sqrt{\frac{2}{3 k m}}\right) \left(F + 2 m u\right) \sin \left(\sqrt{\frac{3 k}{2 m}} t\right)$

The conclusions are left to the reader.

Apr 13, 2018

${\xi}_{\max} = \sqrt{\frac{16 {F}^{2}}{9 {k}^{2}} + \frac{6 m {u}^{2}}{k}} + \frac{4 F}{3 k}$

#### Explanation:

For spring extension is $\xi = {x}_{B} - {x}_{A}$, Newton's 2nd Law applied to blocks is:

Block A: $q \quad k \xi - F = m {\ddot{x}}_{A}$

Block B: $q \quad 2 F - k \xi = 2 m {\ddot{x}}_{B}$

Newton's Law for the system (ignoring spring as it is in the system):

$2 F - F = \left({m}_{A} + {m}_{B}\right) {\ddot{x}}_{c m} \implies F = 3 m {\ddot{x}}_{c m} q \quad \triangle$

Centre of mass of system:

${x}_{c m} = \frac{{m}_{A} {x}_{A} + {m}_{B} {x}_{B}}{{m}_{A} + {m}_{B}} = \frac{1}{3} {x}_{A} + \frac{2}{3} {x}_{B}$, so:

....so:

$\left(\begin{matrix}\xi \\ {x}_{c m}\end{matrix}\right) = \left(\begin{matrix}- 1 & 1 \\ \frac{1}{3} & \frac{2}{3}\end{matrix}\right) \left(\begin{matrix}{x}_{A} \\ {x}_{B}\end{matrix}\right)$

Inverting this:

$\left(\begin{matrix}{x}_{A} \\ {x}_{B}\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{3} & 1 \\ \frac{1}{3} & 1\end{matrix}\right) \left(\begin{matrix}\xi \\ {x}_{c m}\end{matrix}\right)$

and

$\left(\begin{matrix}{\ddot{x}}_{A} \\ {\ddot{x}}_{B}\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{3} & 1 \\ \frac{1}{3} & 1\end{matrix}\right) \left(\begin{matrix}\ddot{\xi} \\ {\ddot{x}}_{c m}\end{matrix}\right)$

Our equations for A& B become:

Block A: $q \quad k \xi - F = m {\ddot{x}}_{c m} - \frac{2}{3} m \ddot{\xi}$

Block B: $q \quad 2 F - k \xi = 2 m {\ddot{x}}_{c m} + \frac{2}{3} m \ddot{\xi}$

We can eliminate ${x}_{c m}$ using $\triangle$:

Block A: $q \quad k \xi - F = \frac{F}{3} - \frac{2}{3} m \ddot{\xi}$

$\implies \ddot{\xi} + \frac{3 k}{2 m} \xi = 2 \frac{F}{m}$

This solves as:

$\xi \left(t\right) = \alpha \sin \left(\omega t\right) + \beta \cos \left(\omega t\right) + \frac{4 F}{3 k}$ with $\omega = \sqrt{\frac{3 k}{2 m}}$

$\xi \left(0\right) = 0 \implies \beta = - \frac{4 F}{3 k}$

$\xi ' = \alpha \omega \cos \left(\omega t\right) - \beta \omega \sin \left(\omega t\right)$

$\xi ' \left(0\right) = 3 u \implies \alpha \omega = 3 u$

$\implies \xi \left(t\right) = \frac{3 u}{\omega} \sin \left(\omega t\right) - \frac{4 F}{3 k} \cos \left(\omega t\right) + \frac{4 F}{3 k}$

$= \Gamma \left(\frac{\frac{3 u}{\omega}}{\Gamma} \sin \left(\omega t\right) - \frac{\frac{4 F}{3 k}}{\Gamma} \cos \left(\omega t\right)\right) + \frac{4 F}{3 k}$

where Gamma = sqrt(((3u)/omega)^2 + ((4 F)/(3 k) )^2) = sqrt(16 F^2 ω^2 + 81 k^2 u^2)/(3k omega)

=sqrt(16 F^2 ω^2 + 81 k^2 u^2)/(3k omega) sin(omega t - phi) + (4 F)/(3 k)

EDIT: Now ${\omega}^{2} = \frac{3 k}{2 m}$ so this simplifies further to:

$= \sqrt{\frac{16 {F}^{2}}{9 {k}^{2}} + \frac{6 m {u}^{2}}{k}} \sin \left(\omega t - \phi\right) + \frac{4 F}{3 k}$

Max Extension is therefore:

${\xi}_{\max} = \sqrt{\frac{16 {F}^{2}}{9 {k}^{2}} + \frac{6 m {u}^{2}}{k}} + \frac{4 F}{3 k}$