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Spring oscillations - F = -kx

  1. Apr 17, 2012 #1
    attachment.php?attachmentid=142547&d=1334684647.png
    Looking at the three diagrams we can see that there are three possible situations

    (a) mass is to the right of eqm
    (b) mass is at eqm
    (c) mass is to the left of eqm
    (eqm = equilibrium)

    Lets look at position (a)

    If we consider the tension in both springs:
    the tension in the spring on the left has increased by k1x
    the tension in the spring on the right has decreased by k2x

    So the overall restoring force SHOULD BE k1x - k2x
    (the spring on the left is trying to pull it back to the left and the spring on the right is trying to pull it the right hence the forces act in opposite directions)

    BUT according to my book the overall restoring force is k1x + k2x....how?

    Any help is greatly appreciated

    exact quote:
     
  2. jcsd
  3. Apr 17, 2012 #2

    Integral

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    The compressed spring is not pulling away from the eqm but pushing towards. IE it is under compression.
     
  4. Apr 17, 2012 #3
    oh right
    thanks
     
  5. Apr 17, 2012 #4
    why is tension not present here?
    (I don't understand the explanation given)?
     

    Attached Files:

  6. Apr 17, 2012 #5

    Doc Al

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    Staff: Mentor

    I don't understand the choices. There are two forces acting on the hanging mass: gravity and the force from the spring. Of course, a stretched spring exerts a tension. So I don't know the intention of having 'tension' vs 'spring' as separate choices: you could call that force the spring force or the tension force exerted by the spring.

    And the given explanation reads like gibberish. Where is this question from?
     
  7. Apr 18, 2012 #6
    Even if both springs are under tension all of the time, the restoring force would still be [itex] k_1x + k_2x [/itex]. There's an increased force to the left, because the left spring pulls harder and a decreased force to the right because the right sprint pulls less hard than in the equilibrium position. This will give the same effect as an increased force to the right.
     
  8. Apr 18, 2012 #7
    A linear spring pulled from equilibrium exerts a force k*x. A linear spring compressed from equilibrium exerts a force k*x. Thus, the force is k*x+k*x.

    I'm not sure where the issue lays
     
  9. Apr 18, 2012 #8
    By the way, that attached .png is garbage, no offence.
     
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