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Multiple springs in harmonic motion

  1. Dec 2, 2008 #1
    Consider the four block-and-spring systems shown. Each block moves on a horizontal, frictionless table. The blocks all have the same mass and all of the springs are identical and ideal with spring constant k. At the instant shown, each block is released from rest a distance A to the right of its equilibrium position.

    Rank the cases according to:
    a) the magnitude of the new force on the block at the instant shown
    b) the time it takes the bock to return to its equilibrium position
    c) the total potential energy at the instant shown

    Case A is a single spring connected to a wall on the left side and the mass on the right.
    Case B is two springs, one connected on either side of the mass, and the other ends connected to left and right walls, respectively. The left spring is stretchedA and the right springs is compressed A
    Case C is two springs connected to one another with the mass on the far right and the wall on the left. It is stretched A to the right.
    Case D is two springs attached to the wall on the left and both attached to the mass on the right. It is stretchedA from equilibrium.

    Here is a quick crude paint drawing I made of it. Red represents equilibrium and the blue the distance A. A is the same in all 4 cases.

    2. Relevant equations

    F = -kx
    Us = .5kx^2

    3. The attempt at a solution

    a) I said that Cases B and D were equal and A and C are equal. B and D are the same because the masses each have two springs acting on them with the same force (-kA + -kA). However, Im not positive on Case C, I couldn't figure out if it was twice the force because there are two springs or if it was just the same as case A.

    b) for the time it takes, if it has more force acting on it, the mass should return faster. I ranked them the same B=D>A=C

    c) for energies, I said the same thing.

    Basically, can someone help me clarify case C? How do you add up springs in each of these cases?
  2. jcsd
  3. Dec 2, 2008 #2


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  4. Dec 2, 2008 #3
    Thanks for the link.

    So to find the forces, it would be something like this?:

    A) Total spring constant = k, F= -kx
    B) Total of springs: k + k = 2k, F= -2kx
    C) 1/k + 1/k = 2/k, (2/k)^-1 = k/2, F = -(k/2)x
    D) k + k = 2k, F = -2kx

    so order would be D=B>A>C?

    Same would apply to part B of the question, time, since its directly related to force: D=B>A>C?

    And just apply the above k's to part C, so it would be the same since the distance hasnt changed? D=B>A>C?
  5. Dec 2, 2008 #4


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    That looks better then.

    The Force comes from displacement of the springs, so for a given distance - Hooke's Law suggests that there will be a greater initial force and a quicker return to equilibrium. But be careful of signs, because Force is a vector, and sign matters.

    Remember that the potential energy is given by 1/2kx2. So when you square the displacement the Energy magnitude increases with displacement in either direction.

    And just to be careful on the time, the greater acceleration results in a quicker time. Your ordering is quickest to slowest hopefully and not in time needed to return to equilibrium.
  6. Dec 2, 2008 #5
    Thanks, I appreciate the help. The question was about time from quickest to slowest. A later question was about the period, which is opposite of the above order.
  7. Dec 3, 2008 #6
    So is quickest to slowest different than the time it takes for the block to return to equilibrium position?

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