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SHM problem involving an applied force

  1. Sep 3, 2005 #1
    Hi, I need advice as to whether or not I solved this question correctly. I would really appreciate the feedback, etc... I really need to know if I did this right

    QA: A 2kg object is attached to a spring and placed on a horizontal smooth surface. A horizontal force of 20N is required to hold the object at rest when it is pulled 0.2m from its equilibrium position (the origin of the x axis). The object is now released from rest with an initial position of xi=0.2m, and it subsequently undergoes simple harmonic oscillations. Find the force constant of the spring.<--- This is the one I'm having problems with

    you see, I use the 20N force, but made it NEGATIVE and made it equal to the force in the spring. Therefore Fs= -20N = -kx
    x=0.2m so I isolate and solve for k.
    Is this right? I get a value of 100N/m. I know how to solve all the other following qa but it's just the first one that's bothering me. Do I have the right idea or did I do something oh so horribly wrong. I would really appreciate it if someone could rid me of my anxieties. lol
    Thank you very much for your time.
     
  2. jcsd
  3. Sep 3, 2005 #2
    And I also made Amplitude= 0.2m
    Is that the right idea, I also made the general equation x(t)= Acos(wt)
     
  4. Sep 3, 2005 #3

    LeonhardEuler

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    Yes, that all looks right.
     
  5. Sep 3, 2005 #4

    Fermat

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    The spring constant is just the amount of force required for unit displacement, whether that displacement gives tensile or compressive force.
    So, it didn't matter about signs.
    You have 20N for a 0.2m displacement, so

    k = 20N per 0.2m
    k = 100N/m
    =========
     
  6. Sep 3, 2005 #5
    Ok, but in this situation, do I use x(t)= Acos(wt)
    or Asin(wt)?
    How do I know which one to use! I always get confused!!
     
  7. Sep 3, 2005 #6

    LeonhardEuler

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    If you get confused, use x(t)=Acos(wt) + Bsin(wt) and use the initial conditions to solve for the coefficients A and B. In this case the object is not moving when it is released, so x'(0)=0-->0=-Asin(0) + Bcos(0) = -A(0) + B(1) = B -->B=0 --> x(t) = Acos(wt)
     
  8. Sep 3, 2005 #7

    Fermat

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    It all depends what your initial condition is/was. Where did the particle start from?

    If you have x = Acos(wt), then at t=0, x = A, so the movement has started off with the particle at the far end of its movement.
    If x = Asin(wt) then at t=0, x = 0, so the particle is at its equlibrium position - right in the middle of its motion.

    Edit: which is pretty much what LeonhardEuler has said.
     
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