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Oscillation Displacement

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known
    On a very smooth horizontal table, a block of mass m=0.75 kg is attached to an ideal spring with a spring constant k= 242 N/m. The origin of the horizontal coordinate (x=0) is set at the equilibrium position of the block. The block is initially held at a negative position where it keeps the spring compressed. Then the block is released, and moves through position x1=0.105m with a speed v1=1.42 m/s.

    a) What was the initial position of the block, x0?

    b)What is the maximum speed of oscillation, vmax?


    2. Relevant equations
    omega = sqrt(k/m
    F=-kx
    Vmax=A*omega
    x=Acos(2pi*f*t)
    3. The attempt at a solution
    I thought about using conservation of energy by doing KE+PE=W
    W=Fs and KE=1/2mv^2 but I don't think this would give me displacement in terms of a negative/before equilibrium position
     
  2. jcsd
  3. Nov 5, 2016 #2

    gneill

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    Staff: Mentor

    Hi NY152,

    Welcome to Physics Forums.

    If you take a close look at the formulas for the components of the energy, both involve the squaring of a term. A negative value squared yields the same amount as a positive value of the same magnitude.
     
  4. Nov 5, 2016 #3
    Thanks for the input!! Do you think I'm on the right track then in terms of those formulas or am I missing something??
     
  5. Nov 5, 2016 #4

    gneill

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    Staff: Mentor

    You're on the right track. You're just missing stating the formula for the PE of a spring.
     
  6. Nov 5, 2016 #5
    I only have one attempt left, so I'm just going to post my work here in the hopes that someone might see if I'm doing it right or wrong.

    So I did: KE+PE=W W=Fs
    1/2mv^2+1/2kx^2=(-kx)(s)
    and found s to be =-0.0822
     
  7. Nov 5, 2016 #6

    gneill

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    Staff: Mentor

    No, the sum of the KE and PE is a constant, the total energy of the system. Better to write:

    E = KE + PE

    In an isolated system the KE and PE can trade energy back and forth, but their sum is always a constant.

    You are given a particular data point with both displacement and velocity. So you can find the value of E which will hold for every point throughout the cycles.
     
  8. Nov 5, 2016 #7
    Sorry for the late reply, so I solved for E and got 2.09, now do I just set that equal to PE + KE when the spring is retracted? In which case would velocity be zero? If so I get x=-0.131
     
  9. Nov 5, 2016 #8

    gneill

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    Staff: Mentor

    Yup. Good.
     
  10. Nov 5, 2016 #9
    Alright well it wasn't -0.131, but I got the max velocity right by setting x=0 so I'm not sure where I went wrong but thanks for the help!
     
  11. Nov 5, 2016 #10

    gneill

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    Staff: Mentor

    Your initial position should be correct. Did you remember to include the units?
     
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