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Oscillating, hanging spring-block system

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A hanging spring stretches by 30.0 cm when an object of mass 500 g is hung on it at rest. We define its position as x = 0. The object is pulled down an additional 16.5 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later?


    2. Relevant equations
    Fs = -kx
    Fs = mg

    x(t) = Acos(ωt + ∅)

    w = (k/m)1/2
    w2 = k/m

    T = 2∏/w

    E = (1/2)kA2
    E = (1/2)mv2 + (1/2)kx2

    v = =- [ (k/m)(A2 - x2) ]1/2
    3. The attempt at a solution
    m = 0.5 kg

    All I can do right off the bat is solve for k

    Fs = (9.8 m/s2)(0.5kg) = 4.9N

    4.9N = -kx

    4.9N = (-k)(-0.3m)

    k = 16.33 N/m

    Now solve for ω ?
    w = [ (16.33 N/m) / 0.5 kg ]1/2

    ω = 5.7 s-1

    Not sure where to go from here regarding energy and I'm afraid that if I plug in to x(t) = Acos(ωt + ∅) it would be wrong?
     
  2. jcsd
  3. Jan 18, 2012 #2
    you know ω and t. Do you know how to find the amplitude-A?
     
  4. Jan 18, 2012 #3
    is A 16.5 cm ?

    so the problem is over if i use x(t) = Acos(ωt + ∅)
     
  5. Jan 18, 2012 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. (What will ∅ equal?)
     
  6. Jan 18, 2012 #5
    i would have to guess and say that ∅ = 0
     
  7. Jan 18, 2012 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Correct. (But you shouldn't have to guess!)
     
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