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2 Masses Connected by a Spring 
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#1
Jun3009, 09:18 PM

P: 1

I was just wondering about this problem (this is not a homework problem or anything like that), and I'm not sure that I know how to solve it:
1. The problem statement, all variables and given/known data 2 masses, M1 and M2, are connected in the middle by a spring (spring constant k). How would I describe the force exerted by the spring on each mass so that the force is zero at some predetermined equilibrium, outward when the spring is compressed, and inward when the spring is stretched? 2. Relevant equations F=kx (Hooke's Law) 3. The attempt at a solution I have thought about putting the center (zero) of my coordinate system at the center of the spring, but I am not sure if this is correct. In other words, the displacement x is defined by the distance of either end of the spring from the center of the spring (at x=0). I set the equilibrium displacement at x=3 and x=3 (3 units to the left of the center and 3 units to the right, respectively). Thus, when the spring is compressed (x>3 on the left side of the center position, and x<3 on the right side), the spring exerts an outward force on each mass (left (negative) on M1, right (positive) on M2), so the Hooke's Law expression for the force on M1 would be F1=k(x+3) and the expression for the force on M2 would be F2=k(x3). Similarly, the spring would exert an inward force when stretched (x<3 on the left side, x>3 on the right). When the ends of the spring are at x=3 on the left and x=3 on the right, the force exerted by the spring on either mass is zero Ultimately, I would like to apply a force to one of the masses (a leftward force to the leftmost mass, M1), add friction to the surface upon which the masses slide, and describe the equations of motion for the system, but I would like to be sure if I'm on the right track. I suppose my biggest hangup involves the assumption that each end of the spring (on either side of x=0) is the same, which seems like an oversimplification, and the fact that the forces exerted by the spring on each mass do not balance. I'm sure it is just something simple that I'm overlooking, but any help that could be offered would be appreciated. 


#2
Jun3009, 11:21 PM

HW Helper
P: 1,982

Hello, welcome to PF.
The centre of mass of the system experiences zero force. It would be convenient to take that as the origin. All the other points on the spring experiences tensional or compressional force. After simplification, you'll end up with something called a reduced mass, where you have to solve a single differential equation, instead of two DEs for two masses. 


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