Mechanics |18-09-2020|

Let the X coordinates of the lower, middle and upper bars be $x_1$ , $x_2$ , and $x_3$ respectively. Then applying Newton's second law to each of the masses gives:

$m\ddot{x}_1 = - k\dot{x}_1 + k(\dot{x}_2 - \dot{x}_1)$ $m\ddot{x}_2 = k(\dot{x}_3 - \dot{x}_2) - k(\dot{x}_2 - \dot{x}_1)$ $m\ddot{x}_3 = -k(\dot{x}_3 - \dot{x}_2)$

$x_1(0) = x_2(0) = x_3(0) = \dot{x}_1(0) = \dot{x}_2(0) = 0 \ ; \ \dot{x}_3(0) = u$

Integrating all the above equations, appying initial conditions, equating the velocities to zero, and simplifying gives:

$-2kx_1 + kx_2 = 0$ $kx_1 - 2kx_2 + kx_3 =0$ $-kx_2 + kx_3 = mu$

Solving and adding gives:

$x_1 + x_2 + x_3 = \frac{6mu}{k}$

The answer is $\boxed{\alpha = 6}$