# Two block-spring system - finding max extension in the spring

1. Oct 8, 2011

### Ashu2912

1. The problem statement, all variables and given/known data
2 blocks, mass 'M1' & 'M2' are connected by an ideal spring of force constant 'k' and placed on a frictionless surface. Force 'F' is applied on the 'M2' block. We have to find the maximum extension in the spring.
2. The attempt at a solution
(1) The conservation of energy law won't work on the blocks-spring because I don't know the work done by the force F.
(2) I also tried to apply Newton's Second Law on the blocks individually but it also won't work.
Thanks.....

2. Oct 8, 2011

### Kishor Bhat

Well, the work done by the force is equal to (-k*x^2)/2, where x is the max. deformation of the spring. This can be found by integration. But I'm not sure how the setup looks and how you're applying the force.

3. Oct 8, 2011

### ehild

Show the set-up and the free-body diagram, please.

ehild

4. Oct 10, 2011

### Ashu2912

See the diagram and the FBD in the attachment please....

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• ###### untitled.bmp
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5. Oct 10, 2011

### ehild

This is not a simple problem. The centre of mass of the system will accelerate with aCM=F/(m1+m2) and the two blocks oscillate with respect to the CM. Have you learned about two-body problems?

You can set-up the equation of motion for both blocks, and express their position coordinate with the position of the CM and the length of the spring. If x1 is the coordinate of block1 and x2 is the coordinate of block2, then the the position of CM is Xcm=(m1x1+m2x2)/(m1+m2), and the length of the spring is L=x2-x1.
The CM moves with acceleration acm=F/(m1+m2)

Write out Newton's second law for both x1 and x2. Express x2 and x1 in terms of Xcm and L. You get a second-order differential equation for L, similar the one when a mass is hanged on a spring. The mass will oscillate around an equilibrium position, determined by gravity and the spring constant. At this equilibrium position the net force is zero. If the spring was unstretched initially and then released, the amplitude of the vibration is equal to the difference between the equilibrium length and the unstretched length. So the length is maximum when it is stretched by two amplitudes.

The situation is similar here: the constant force plays the role of gravity. Find the length L of the spring when the acceleration of the two blocks are equal. Calculate L-L0, and add to L: that will be the maximum length of the spring.

ehild

6. Oct 10, 2011

### Staff: Mentor

Sometimes, when analyzing physical systems, it's advantageous to consider using an analogous electrical model. The forms of the equations that govern mechanical and electrical systems are essentially identical, and the solutions to the resulting differential equations are thus also identical in form. Being able to bring the well-established methods of circuit analysis to bear on a mechanical problem can at times be a real boon.

In this problem one can use the established analogs (you can find tables of such analogs for various system types on the web if you persevere):

Mass <---> Capacitance; M --> C
Spring <---> Inductance; 1/k --> L
Current <---> Force; F --> I

In the present case the mechanical system and its electrical equivalent are straightforward:

Solve for the current through the inductor and you are essentially solving for the spring force as well. Laplace Transforms will take care of this problem in a few lines

#### Attached Files:

• ###### Fig1.gif
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7. Oct 10, 2011

### Ashu2912

Thanks a lot gneill and ehild. Actually, I have still not studied electromagnetic induction or current electricity. My teacher says that the extension in the spring will be max. when the velocity of both the blocks will be same. Is it true?? Then how can we visualize or better, mathematically prove this?? Thanks...
Ashu2912

8. Oct 10, 2011

### Staff: Mentor

Since a constant force F is being applied to the system, the center of mass of the system must be accelerating uniformly according to F = M*A, where here M is the total mass of the system.

If you were sitting at and moving with the center of mass and watching the springs go through their motions relative to you, then they would be oscillating back and forth (or in and out) from your point of view. They would alternately extend to their maximum extent, then rebound to their minimum extent. Their individual masses would determine the proportional lengths of those excursions for each mass since the center of mass must be fixed in your reference frame, and momentum should be conserved.

When the masses are at minimum or maximum extension, they will have zero velocity with respect to the center of mass (they are just turning around), and so zero velocity with respect to each other as well. Their velocity from an outside observer then should be equal to the then current velocity of the center of mass.