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Homework Help: Energy and Simple Harmonic Motion - Part II.

  1. Nov 13, 2006 #1
    A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall whle the other end is connected to a movable object. The spring and object are compressed by 0.064 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.5 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.037 m relative to its unstrained length?

    ... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

    ... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!
     
  2. jcsd
  3. Nov 13, 2006 #2

    Hootenanny

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    Use Hooke's law and conservation of energy.
     
  4. Nov 13, 2006 #3
    Do I set them equal to each other??

    ... I don't understand the .64 meter compression part ... does that equal "X" in Hooke's Law and in ????
     
  5. Nov 13, 2006 #4
    Hint -- Can you find the value of the spring constant?

    BTW -- you are right in that the compression WILL be X in hooke's Law.
     
  6. Nov 14, 2006 #5
    I figured it out ...

    A = 0.064m (given in the problem)
    omega = 11.5 (rad/s) - given in the problem
    x = 0.037m (given in the problem)


    V(max) = A(omega)
    V(max) = 0.064(11.5)
    V(max) = .736(m/s)

    V = V(max) * [the square root of the quantity of (1- {[x^2]/[A^2]})]
    V = .736 * [the square root of the quantity of (1- {[0.037^2]/[0.064^2]})]

    V = 0.601 (m/s)
     
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