How to find max deflection of spring between 2 masses ?

bksree

Hi
I came across a problem as below :
2 masses M1 and M2 are connected by a spring and kept on a frictionless table horizontally. A force F is applied to M2. What is the maximum displacement of the spring ?

The acc of the COM of the system is a = F/(m1 + m2).
However, each mass will move with different acceleration because of the spring between them. If x1 and x2 are the displacements at M1 and M2, the force exerted by the spring on each mass will be k(x1-x2).
FBD of M1 gives
k(x1-x2) = M1 a1
FBD of M2 gives
F-k(x1-x2) = M2 a2

I'm not clear how to proceed from here. The ans is
2M1 * F / (M1 + M2)

Cany anypone explain ?

TIA

mfb

I would split the system in the center of mass movement and the movement of the individual objects. The whole system accelerates with a=F/(m1+m2). In this system, an effective force of F-a*m1 accelerates m1 inwards and a*m2 accelerates m2 inwards. As a(m1+m2)=F, both have the same magnitude - this is just a cross-check.

Let x be the distance relative to the springs at rest (x=0), with positive x as larger distance. The spring potential is now V(x)=1/2*k*x^2 and the external force adds a potential of -2*a*m2*x. The system will oscillate in this parabola, with 0 potential as endpoints. One is x=0, can you find the other one?

bksree

Thanks for the reply. But why should a1 be equal to a2 ? The acceleration of COM is a = F/(m1 + m2)
Hence a = (m1a1 + m2a2) / (m1+m2).
Since x1, m1 and F1 are not equal to x2, m2 and F2, how can we conclude that a1 = a2 ?

TIA

mfb

Where did I say a1=a2?