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?

 

[PJC01]

To find the maximum deflection of the spring between the two masses, we can use the concept of equilibrium. At maximum deflection, the spring will be stretched to its maximum length and the system will come to a stop. This means that the net force on each mass will be zero.

To find this point of equilibrium, we can set up equations for the forces acting on each mass. For mass M1, the force from the spring will be k(x1-x2) and the force from the applied force F will be M1a1. Similarly, for mass M2, the force from the spring will be k(x1-x2) and the force from the applied force F will be M2a2.

At equilibrium, the net force on each mass will be zero, so we can set these equations equal to zero and solve for the displacement x1-x2. This will give us the maximum deflection of the spring.

To simplify the equations, we can use the acceleration of the center of mass of the system, a = F/(m1+m2), and substitute it into the equations for a1 and a2. This will give us:

k(x1-x2) = M1(F/(m1+m2))
k(x1-x2) = M2(F/(m1+m2))

Solving for x1-x2, we get:

x1-x2 = (2M1*M2*F)/(k*(m1+m2)^2)

This is the maximum deflection of the spring between the two masses, which matches the answer provided. I hope this explanation helps clarify the process for finding the maximum deflection of the spring in this system.