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Assume an ideal spring of spring constant $k$ is connected with two blocks of masses $m_1,m_2$ at both of its end. And this system is kept on the horizontal smooth table. And a force $F$ is applied to $m_1$ in direction away from spring. And the task is to find maximum extension during journey.

So, I gone in the frame of $m_1$ assume acceleration of $m_i$ is $a_i$ and $a$ is acceleration of center of mass of the system. Then wrt $m_1$ acceleration of $m_2$ will be $a_2-a_1$. For the equilibrium position $a_1=a_2=a$

$$F-kx=m_1a$$$$ kx=m_2a$$

(In ground frame)

Since this extension ($x$) is at mean position and the motion is SHM. So, maximum extension $X=2x$

Which on solving gives $$X=\frac{2m_2F}{(m_1+m_2)k}$$

Which I also verified by conservation of energy wrt Center of mass.

But I want to know why always it appears to be double the extension on mean position in case where constant force is applied?

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