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Homework Help: Easy Problem w/ Spring

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    An object of mass m = 50 kg is placed on top of 10 vertical springs. After this, the springs come to a new equilibrium position, 0.05 m below the original one, so xeq = –0.05 m. The springs are stretched upward 0.15 m from the equilibrium position. The spring is released after being stretched, and simple harmonic motion takes place, meaning that, during each "bounce"—up and down—the mass returns to the same maximum height x = xeq + s = –0.05 m + 0.15 m = 0.1 m, and the same minimum height x´. Determine the force constant k of each spring, and then determine the minimum height x´ of the mass in order for simple harmonic motion to occur.

    2. Relevant equations

    Fx = kx
    Fg = mg
    Fx´ = kx´

    3. The attempt at a solution

    At the top of its motion,

    Fx = Fg
    10kx = mg
    k = (mg)/10x = [(50 kg)(9.8 N/kg)]/[(10)(0.1 m)] = 490 N/m

    [Did I disclude any negatives that should have been present?]

    At the bottom of its motion,

    Fx´ = –Fg
    10kx´ = –mg
    x` = (–mg)/(10k) = [–(50 kg)(9.8 N/kg)]/[10(490 N/m)] = –0.1 m

    So, x = –x´—did I do this correctly?
    Last edited: Nov 11, 2007
  2. jcsd
  3. Nov 11, 2007 #2


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    Are the springs connected length wise? Meaning one spring on top of the other? Or are they next to each other...
  4. Nov 11, 2007 #3
    The springs are all beside one another.
    Last edited: Nov 11, 2007
  5. Nov 11, 2007 #4


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    do they give any more information? did you state the problem exactly?

    it seems to me like the spring constant shouldn't make a difference... ie any spring constant would give the same result...

    I'm probably misunderstanding something... but any 10 springs with a mass on top of it... stretched upwards a distance 0.1m from the 0 stretch position... in the ideal frictionless case... will undergo harmonic motion returning to a distance of 0.1m above the 0 stretch position...

    I think what you did with:

    Fx = kx

    is wrong... because you're assumeing the net force is 0 at the top of the motion... but it isn't. actually both forces are acting downwards... the spring force is downwards and mg is acting downwards....

    let us know if there is more information...
  6. Nov 11, 2007 #5
    Look back over my problem because I put in more variables:

    Let's say that, after the mass is placed on top of the springs, the springs come to a new equilibrium position:

    xeq = –0.05 m

    This means that, if the springs are stretched 0.1 m above the reference level (the original equilibrium position), then the upward stretch was 0.15 m above the new equilibrium position; that is, s = 0.15 m, and x = xeq + s = (–0.05 + 0.15)m = 0.1 m

    Now I suppose I should determine the force constant k @ the new equilibrium position, when x = –0.05, and when the net force is zero:

    k = (mg)/(10x) = [(50 kg)(–9.8 N/kg)]/[10(–0.05)] = 980 N/m

    Now how do I get x´? I know that x´ = –0.15 m, but how do I show this mathematically?


    At x´,

    mg = –10kx´ (since gravity is acting against the force of the springs; I'm assuming that the net force acting on the object is zero here—is this assumption correct?)
    x´ = –(mg)/(10k) = –0.05 m—now this can't possibly be right, because x´ = –0.15 m


    By the way, thanks for responding to my problem.
    Last edited: Nov 11, 2007
  7. Nov 11, 2007 #6


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    Yes, k = 980 N/m as you calculated.

    And the minimum x' seems to me to be -0.05 - 0.15 = -0.20m

    The equilibrium position is -0.05m. So the max is -0.05+0.15m = 0.10m.

    And the minimum position is -0.05-0.15 = -0.20m. Is this the wrong answer?
    Last edited: Nov 11, 2007
  8. Nov 11, 2007 #7
    Oh yes, you're right—x´ = –0.20 m. But how do I prove this mathematically? I mean, I need to show that when the maximum is x = 0.10 m, the corresponding minimum must be x´ = –0.20 m in order for simple harmonic motion to occur. I could easily say that A = 1.5 m and that the two values must thus be x = 0.10 m and x´ = –0.20 m, but I need to include forces or something to be more convincing. I tried to prove this earlier but came out with the wrong answer (x´ = –0.05). :(

  9. Nov 11, 2007 #8


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    You can use energy conservation... During oscillation, the top and bottoms are when v = 0... ie at those points, kinetic energy is 0.

    For simple harmonic motion we need the mass to return to the same maximum height (that's what makes it simple harmonic motion)...

    at that maximum height its energy = mgx + 10*(1/2)kx^2 (here I'm defining the gravitational potential energy at xunstretched=0 to be 0. keeps the math simple).

    if energy is not conserved... then that means x, the max height changes (since m, g and k are all fixed, energy change implies x, ie max height changes)... ie: simple harmonic motion doesn't happen.

    So you've got to have energy conservation for simple harmonic motion.

    so you need mgx + 10*(1/2)kx^2 = mgx' + 10*(1/2)k(x')^2 (at the bottom also, kinetic energy = 0... so we just consider grav potential energy and elastic potnetial energy)

    plug in your x, m, g and k and solve for x'. you should get -0.20m (you'll get 2 solutions... one will just be x = 0.10 which is the max... the answer you need is the other solution with is x = -0.20m which is the min)
    Last edited: Nov 11, 2007
  10. Nov 11, 2007 #9
    Awesome, man. I plug x´ = –0.20 m into the right side, and both sides equal 98. Thanks a lot for the help, and good luck to you in your future endeavours (as lame as that sounds). lol
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