Two bars connected by a weightless spring of stiffness `ϰ` and length (in the non-deformed state) `l_0` rest on a horizontal plane. A constant horizontal force F starts acting on one of the bars as shown in figure. Find the maximum and minimum distances between the bars during the subsequent motion of the system, if the masses of the bars are:
(a) equal:
(b) equal to `m_1` and `m_2`, and the force F is applied to the bar of mass `m_2`.

Explanation

Let us consider both blocks and spring as the physical system. The centre of mass of the system moves with acceleration `a=(F)/(m_1+m_2)` towards right. Let us work in the frame of centre of mass. As this frame is a non-inertial frame (accelerated with respect to the ground) we have to apply a pseudo force `m_1a` towards left on the block `m_1` and `m_2a` towards left on the block `m_2`
As the centre of mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The elongation of the spring will be maximum or minimum at this instant. Assume that the block `m_1` is displaced by the distance `x_1` and the block `m_2` through a distance `x_2` from the initial positions.
From the energy equation in the frame of C.M.
`Deltaoverset~T+U=A_(ext)`,
(where `A_(ext)` also includes the work done by the pseudo forces)
Here,
`Deltaoverset~T=0`, `U=1/2k(x_1+x_2)^2` and
`W_(ext)=((F-m_2F)/(m_1+m_2))x_2+(m_1F)/(m_1+m_2)x_1=(m_1F(x_1+x_2))/(m_1+m_2)`
or, `1/2k(x_1+x_2)^2=(m_1(x_1+x_2)F)/(m_1+m_2)`
So, `x_1+x_2=0` or, `x_1+x_2=(2m_1F)/(k(m_1+m_2))`
Hence the maximum separation between the blocks equals: `l_0+(2m_1F)/(k(m_1+m_2))`
Obviously the minimum sepation corresponds to zero elongation and is equal to `l_0`