Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of force constant $k$ . Block of mass $m_{1}$ is pulled by a constant force $F_{1}$ and other block is pulled by a constant force of $F_{2}$ . Find the maximum elongation that the spring will suffer.

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$x_{m a x} = \frac{1}{2 k} (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$

$x_{m a x} = \frac{2}{k} (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$

$x_{m a x} = \frac{3}{k} (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$

$x_{m a x} = \frac{1}{3 k} (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$

answer is B.

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Detailed Solution

Let us take the two blocks plus spring as the system. The centre of mass of system moves with an acceleration

$a_{c} = \frac{(F_{1} - F_{2})}{m_{1} + m_{2}}$

Supposing $F 1 > F 2$ . As this frame is accelerated with respect to ground, we have to apply pseudo force on the blocks.

Therefore net external force on $m_{1}$

$F_{1}^{'} = F_{1} - m_{1} a_{c} = F_{1} - m_{1} (\frac{F_{1} - F_{2}}{m_{1} + m_{2}})$

$= (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$ towards right.

and on $m_{2}$ $F_{2}^{'} = F_{2} - m_{2} a_{c} = F_{2} - m_{2} (\frac{F_{1} - F_{2}}{m_{1} + m_{2}})$

$= (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$ towards left

As the centre mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The spring will have maximum extension at this instant. Suppose the right block displaces distance $x_{1}$ and left displaces a distance $x_{2}$ from their initial positions.

Therefore work done by external force = Increase in P.E. of the spring

$F_{1}^{'} x_{1} + F_{2}^{'} x_{2} = \frac{1}{2} k {(x_{1} + x_{2})}^{2}$

$F_{1}^{'} (x_{1} + x_{2}) = \frac{1}{2} k {(x_{1} + x_{2})}^{2} (x_{1} + x_{2}) = \frac{2 F_{1}^{'}}{k} x_{m a x} = \frac{2}{k} (\frac{F_{1} m_{2} + F_{2} m_{1}}{m_{1} + m_{2}})$