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In the solution of above question, the authors of the book I'm reading solved it by changing the reference frame from the ground frame to the frame of the center of mass of the masses. But we know that, to move from one frame to another, we have to subtract the mass times acceleration product. But in the above case, the pseudo-force is applied in two opposite direction.

Is their any other approach to solve it from ground frame?

  • $\begingroup$ Sure, just set up Newton's 2nd law on the block M and then on the block m, and then remember that their accelerations are equal. $\endgroup$
    – Steeven
    Commented Jul 12, 2018 at 20:36
  • 1
    $\begingroup$ FYI, your title doesn't really make sense. The pseudo force does not act in the opposite direction when in the ground frame (inertial frame) compared to in the centre-of-mass frame (non-inertial frame), because there is no pseudo force in the ground frame. That pseudo force is an addition for non-inertial frames only, because Newton's laws otherwise wouldn't hold true in non-inertial frames. $\endgroup$
    – Steeven
    Commented Jul 12, 2018 at 20:39
  • $\begingroup$ There is no need to consider conceptually changing frame 0f references to solve this, you can just try to calculate the difference between F and f when m reaches the same speed as M. Although actually it's the same calculation. $\endgroup$
    – JMLCarter
    Commented Jul 12, 2018 at 20:46
  • $\begingroup$ You may find this answer to a related question useful. $\endgroup$
    – kbakshi314
    Commented Mar 17, 2021 at 2:31

2 Answers 2


From the ground frame, let $X$ represent the displacement of the mass M to the right (from its original location), and let $x$ represent the displacement of the mass m to the right (from its original location). Then, from free body diagrams on each of the two masses, the force balances are: $$M\frac{d^2X}{dt^2}=F-k(X-x)$$and$$m\frac{d^2x}{dt^2}=+k(X-x)$$

  • $\begingroup$ And how do I find acceleration of particular body? $\endgroup$
    – The ReBel
    Commented Jul 13, 2018 at 4:45
  • $\begingroup$ Divide both the equation by the mass term on the left and subtract them you would get the differential equation of SHM $\endgroup$
    – LM2357
    Commented Nov 13, 2019 at 7:53

In the horizontal direction, the center of mass will accelerate to the right at: A = F/(M+m). In that frame, each mass will experience a pseudo-force to the left, (mA and MA), and the two masses will undergo harmonic motion relative to the center of mass. At maximum stretch of the spring, they both have a zero velocity relative to the CM.


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