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Conservation of Energy: Spring

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data

    A long, uniform spiral spring of total mass m and spring constant k is suspended vertically from a fixed point. A much larger mass M is attached at the other end and allowed to oscillate vertically. As the mass M moves up and down, the spring contracts and stretches approximately uniformly along its length. Show that the kinetic energy of motion of the spring is approximately mv2/6, where v is the speed of the mass M. Using conservation of total energy or otherwise, show that the period of the oscillating mass and spring is approximately:

    P=2[itex]\pi[/itex][itex]\sqrt{\frac{M+m/3}{k}}[/itex]

    2. Relevant equations

    PE=mgh
    EE=0.5kx2
    KE=0.5mv2

    3. The attempt at a solution

    I said the potential energy must equal the elastic energy when the spring is fully stretched because KE=0,so:

    (m+M)gh=0.5kx2

    But this is getting nowhere near getting: KE=mv2/6

    I've never done the conservation of energy in a spring with mass so any help here would be much appreciated.
     
    Last edited: May 20, 2014
  2. jcsd
  3. May 20, 2014 #2

    BiGyElLoWhAt

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    Is the spring mass m what you're supposed to show for KE=(mv^2) /6 ? I'd have to try to work it out, but I think it's fairly simple to do the period one. It can be shown that (omega) = sqrt(k/m) and you can use that along with some identities to show the period equals that.
     
  4. May 20, 2014 #3

    BiGyElLoWhAt

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    going afk for approx. 1 hour, will be back. good luck XD
     
  5. May 20, 2014 #4

    haruspex

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    Is the elastic energy zero when KE is at max?
     
  6. May 20, 2014 #5

    AlephZero

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    That is correct, but
    is wrong. The mass m of the spring does not all move a distance h. One end moves a distance h, but the other end is fixed.

    The velocity of the spring also varies along its length, from 0 at one end to v at the other.

    Because the mass M is much bigger than m, you can assume the tension is always constant along the length of the spring.
     
  7. May 20, 2014 #6
    Let μ the cumulative amount of spring mass between the fixed point and an arbitrary location along the spring. If the spring expands and contracts uniformly, the velocity at the arbitrary location is given by vμ/m. If the differential amount of mass between μ and μ +dμ is equal to dμ, what is the kinetic energy of the differential amount of mass dμ? What is the total amount of kinetic energy of the spring?

    Chet
     
  8. May 21, 2014 #7
    Hey haruspex. KE is at a max when spring is oscillating back up?

    AlephZero and Chestermiller, so is the total amount of kinetic energy:

    0.5[itex]\mu[/itex]v2+0.5d[itex]\mu[/itex]v2
     
  9. May 21, 2014 #8

    haruspex

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    Up or down, but what I asked was whether the PE was at zero (i.e. spring relaxed) when the KE is at its maximum. To ask it another way, if the system were not oscillating, just hanging at rest, would the spring be relaxed?
     
  10. May 21, 2014 #9
    Yes the spring would be relaxed so KE wuld be zero here too?
     
  11. May 21, 2014 #10

    haruspex

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    No.

    If the weight is hanging, but not oscillating, the spring will still be under tension. It has to support the weight. Call this the neutral position.

    Now consider what happens on an upward traverse when it oscillates. Below the neutral position the tension in the spring will be more than necessary to support the weight. So which way will it be accelerating?
    Above the neutral position the tension will be less than that needed to support the weight. Which way is it accelerating now? So at what position is the maximum speed reached?
     
  12. May 21, 2014 #11
    So below neutral position it is accelerating upwards and above neutral position its accelerating downwards? Max velocity as it passes through neutral position?
     
  13. May 21, 2014 #12
    No. It's the integral of [itex]\frac{1}{2}(\frac{μv}{m})^2dμ[/itex] from μ=0 to μ=m.
     
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