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Platform on Springs, how to find amplitude?

  1. Aug 12, 2006 #1
    I have been working on this problem for an hour and can't seem to figure it out. I know there must be something fundamental that I must be doing wrong. Can someone give me a pointer?

    A platform of mass 0.8 kg is supported on four springs. A chunk of modeling clay of mass 0.6 kg is held above the table and dropped so that it hits the table with a speed of v = 0.9 m/s. The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 6 cm below its original position.

    With what amplitude does the platform oscillate immediately after the collison?

    In the first part of the problem I found that k=98.1 N/m.​

    So, initially, the E is entirely Kinetic (KE = .5mv^2) for the intial velocity of 0.9 m/s and the mass of the clay (0.6 kg). This means that intial E = initial KE = 0.243.

    Now, I don't know if the final E is kinetic and potential, or just potential? if I say it is kinetic too, then how do I find V right after the collision? I tried the equation v= (mvi+Mvi)/(m+M) and calculated (0.6*0.9+0)/(0.6+0.8)=0.3857 m/s. If I use this velocity for final KE, and solve for A in the final PE (PE=.5kA^2), I get the wrong answer (A=0.053m)

    If I don't incorporate final KE and simply say that final E = final PE, then I get A=0.0703m, which is wrong too.

    What am I doing wrong? Which one do I use? And why is it that both give me the wrong answer?
  2. jcsd
  3. Aug 12, 2006 #2


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    Homework Helper

    I'm not quite sure why you calculated the kinetic energy of the clay, as energy is not conserved during the collision.

    How did you get that equation?
    EDIT: Ah, you used vi for 2 different values.

    Do notice that the system is not at the "new" equilibrium point right after the collision.
    Last edited: Aug 12, 2006
  4. Aug 12, 2006 #3
    I don't understand this completely. Why isn't the energy of the clay conserved during the collision? Ei=KEi+PEi, right? And doesn't Ei=Ef?

    v= (mvi+Mvi)/(m+M)
    This equation gives the final velocity of 2 objects, of mass m and M with intial velocities vi1 and vi2, the collide and stick. It is more appropriately written vf= (mvi1+Mvi2)/(m+M) This is in my course text (Walker).
  5. Aug 12, 2006 #4


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    Homework Helper

    If the masses stick together, that's called an inelastic collision. Energy is not conserved, but momentum is. You can easily derive the formula you have in your book from the conservation of momentum.

    This I didn't quite understand. At the amplitude position, you're not supposed to have any kinetic energy.
  6. Aug 13, 2006 #5
    Okay, so, correct me if I'm wrong:

    The energy equation that should be used is KEi+PEi=PEf+KEf; However, there is no KE at the maximum amplitude (because it comes to rest), so the equation should thus be KEi+PEi=PEf.

    Well, KEi+PEi = .5*m*v^2 + .5*k*x^2 right?
    I assume I use the mass of both objects, m+M = 1.4 kg and the velocity calculated above through conservation of momentum (0.3857 m/s). The k=98.1 and x should be the distance the spring will compress, or, 0.06 m. That gives me 0.2807 on for KEi+PEi.

    On the right side of the equation, we have PEf, which is .5*k*A^2.
    Solving for A, I get 0.265 m, which is wrong.

    What am I doing wrong?
  7. Aug 13, 2006 #6

    Doc Al

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    Staff: Mentor

    Redo this last step. Everything else looks OK.
  8. Aug 13, 2006 #7
    Wow. After a dozen sheets of scrap paper and forgetting to incorporate ^2 a many times - the answer is 7.57 cm! Thank you!
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