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Slinky springs and eq of motion.

  1. Jul 13, 2006 #1
    A mass is resting on a horizontal surface without friction. It is connected on both sides by slinky springs k1 and k2 (that is they have different spring constants, and the equilibrium position will not be in the center of the system). First I had to find the equation of motion after setting up the forces. I've seen an example of a mass connected on either side by a spring and k=k1+k2. This is because whichever side of equilibrium you place the mass, the forces act in the same direction, one pulls and one pushes, basically. In this however, the slinky springs naturally are of length L0, or they are close to zero before stretched. Therefore both springs are pulling away from the center no matter where you set the displacement. It is just that one is pulling more than the other, being displaced further. (This is my reasoning so far.) So if I displace to the right, I say sum of forces=ma=-k1x+k2x. That is the force of k1 pulls to the left while k2 to the right. So I have mx"+(k1-k2)x=0 or x"+((k1-k2)/m)x=0. Is this right? Or should I have done (K1+k2)?

    Thanks, Josh
  2. jcsd
  3. Jul 13, 2006 #2

    Doc Al

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    Staff: Mentor


    Are you saying that the springs are unstretched when the mass is at equilibrium? In any case, the only thing that matters is the change in the force that each spring exerts when the mass is displaced from equilibrium.

    When the mass is at the equilibrium position, the net force is always zero. The springs are either both pushing or both pulling or both exerting no force. When the mass is displaced from equilibrium, the change in force exerted by each spring is always in the same direction: towards the equilibrium position. Thus both springs contribute positively to the net restoring force.
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