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Finding spring constant of mass

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    The left side of the figure shows a light (`massless') spring of length 0.300 m in its relaxed position. It is compressed to 67.0 percent of its relaxed length, and a mass M= 0.190 kg is placed on top and released from rest (shown on the right).

    [URL]http://loncapa.gwu.edu/res/msu/physicslib/msuphysicslib/13_EnergyConservation/graphics/prob24_CompSpring.gif[/URL]

    The mass then travels vertically and it takes 1.30 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

    2. Relevant equations

    Fs=-kx
    Us=(1/2)kx^2

    3. The attempt at a solution

    first off this is the data that i use

    l(relax)=.3
    l(tensed)=.201
    so the X for spring = .3-.201 = .099
    x=.099
    m=.190
    V(final)=0
    t=1.3
    a=-9.81
    k=?

    i found V(initial)

    vFinal=vInitial+at
    vInitial = 12.753

    next i found the distance

    x = v(initial) + (1/2)at^2
    x= 8.2979

    then i put

    K+U(gravity)+U(spring)=Total energy
    15.45+15.466+.0049K=0
    K=6308.74

    and it is wrong... i'm really confused
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Oct 17, 2011 #2

    SammyS

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    Yes, it's wrong.

    x = v(initial)t + (1/2)at2 is for constant acceleration.

    Use y = -Acos(2πt/T) , where A is the amplitude of the motion and T is the period.

    You should be able to get A and T from the given information.
     
  4. Oct 18, 2011 #3
    I don't think i have learned how to get A and T yet man..

    i know T=2∏√(m/k)

    but how can I get the K if that's what the question is asking?

    beside T=2∏√(m/k) i believe i haven't learned any other equation yet that is relating to T, and I have no idea what A is..
     
  5. Oct 18, 2011 #4

    SammyS

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    T=2∏√(m/k) will work just fine. Solve it for k.
     
  6. Oct 18, 2011 #5
    I'm honestly quite confused on how to solve for k when I don't know the value of T
     
  7. Oct 18, 2011 #6

    SammyS

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    How would you describe on period of the motion of this spring & mass combination?

    (You are given information in this problem from which you can find T.)
     
  8. Oct 18, 2011 #7

    NascentOxygen

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    What is this x, and why do you need it? If you don't explain what you are doing, it's difficult to say whether you are on the right track or not.

    I used the same basic method as you, but I found k a bit under 1800 N/m

    Work done by spring = Integral from 0.099 to 0 of k.x

    Equate this to the (K.E. of the mass with your vInitial) + (mgh for 0.099m rise)
     
    Last edited: Oct 18, 2011
  9. Oct 18, 2011 #8

    NascentOxygen

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    It seems to me that you might be counting the same energy twice here.
    K.E. + U(gravity) are constant, once the mass has lost contact with the spring. You don't sum their maximum values. You either include K.E. when U is zero, or include U at the peak of the trajectory where speed is zero.
     
  10. Oct 19, 2011 #9

    NascentOxygen

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    After having pondered this puzzle some more, I have are a few comments.

    First, what I outlined above is not a complete answer. I should subtract from the 1.3 secs for flight that length of time during which the spring acts on the mass. It will be a small time, but that is no basis on which to overlook it. Unless we are being instructed to do so, viz., "Assume that the time required for the spring to reach its full extension is negligible."

    Other respondents seem to view the arrangement as a spring bob, with the mass attached to the spring and giving SHM. I read it as a spring flinging the mass into the air, and the two becoming separated early in the piece. It would have helped had Smartguy94 included the diagram he was given which undoubtedly would clear this up.
     
    Last edited: Oct 19, 2011
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