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Simple harmonic motion

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Mass = 2.4 kg
    spring constant = 400 N/m
    equilbrium length = 1.5
    The two ends of the spring are fixed at point A, and at point B which is 1.9m away from A. The 2.4 kg mass is attached to the midpoint of the spring, the mass is slightly disturbed. What is the period of oscillation?


    3. The attempt at a solution
    So in this question the spring is stretching by .4m just to get from A to B, but then it is also being stretched and compressed when it oscillates, how would I incorporate that into my calculation? Also since the mass is in the middle would this change anything?
     
  2. jcsd
  3. Nov 3, 2013 #2
    A spring with a mass in the middle can be modeled as a mass between two springs.
     
  4. Nov 3, 2013 #3
    These springs have the constant 400, or 200 each?
     
  5. Nov 3, 2013 #4
    Recall the formulas for springs connected in series and in parallel. What is the case here?
     
  6. Nov 3, 2013 #5
    So the total spring constant is going to be the constant of those two springs added together, which is going to be 400(the original constant)? If this is the case than under the circumstance that the springs are not initially stretched past equilibrium ω = √((k1+k2)/m) and ωT = 2∏, but they are stretched, so this does make a difference but in what?
     
  7. Nov 3, 2013 #6
    It seems to me you are jumping to conclusions. Don't.

    What is the spring constant of each sub-spring?

    Then, let x = 0 be the position of the mass at the "loaded" equilibrium, i.e., the original spring is stretched between A and B, and the mass is stationary in the middle. What is the force due each sub-spring at the loaded equilibrium?

    Finally, let the mass be shifted a bit, so x is not zero. Again, what are the forces?
     
  8. Nov 3, 2013 #7
    Loaded Equilibrium F = 0 for each sub-spring
    Shifted F = 40 N for each sub-spring
     
  9. Nov 3, 2013 #8
    How is that possible? At the loaded equilibrium, the entire spring is stretched, and so are the sub-springs.

    The force is constant no matter what the shift is? How?

    You did not answer the question about the spring constant for sub-springs.
     
  10. Nov 3, 2013 #9
    Oh I didn't understand how you phrased it, you said x = 0, and I immediately thought F = -kx = 0, sorry. But the spring constant for the sub-springs will be the original/2. So k1 = 200 N/m and k2 = 200 N/m.
     
  11. Nov 3, 2013 #10
    No, that is not correct. How did you obtain that?
     
  12. Nov 3, 2013 #11
    I heard my professor say, when a spring is cut in half, its spring constant is cut in half too, so if that is wrong then maybe he made a mistake, but I probably misheard. Is it doubled?
     
  13. Nov 3, 2013 #12
    Yes, it is doubled and here is why. Assume you have two identical springs with constant k, connected to each other so they form a bigger compound spring. If one is stretched by d, it exerts F = kd on the other spring, and that force makes it stretch by F/k = d, too. So the compound spring is stretched by 2d, yet it exerts F, hence its constant is K = F/(2d) = (kd)/(2d) = k/2. In your case, the "compound" spring is the original spring, so the sub-springs constant is k = 2K, double the original.

    Now, back to the other questions in #6.
     
  14. Nov 3, 2013 #13
    The force made by each sub spring in the loaded equilibrium on the mass is 160N. But if you don't know how much it is shifted by then the force is going to be -800xN and 800xN.
     
  15. Nov 3, 2013 #14
    So the effective spring constant is 800 N/m, correct?
     
  16. Nov 3, 2013 #15
    Yes it is.
     
  17. Nov 3, 2013 #16
    So, do you have the answer?
     
  18. Nov 3, 2013 #17
    So is the angular frequency affected by the two springs being pre stretched? If not then the question is pretty easy.
     
  19. Nov 3, 2013 #18
    In #13, you found the total force on the mass. Because it is total, nothing else can affect the system.
     
  20. Nov 3, 2013 #19
    Isn't the total force 0 at equilibrium? I am thinking how I would incorporate the forces into the angular frequency, but I can't thinking of anything. Am I missing something?
     
    Last edited: Nov 3, 2013
  21. Nov 3, 2013 #20
    You are over-thinking the problem and are confusing yourself. You have already found the force. But let's do it one more time. Both sub-springs have stiffness k = 2K, where K is the stiffness of the original spring. Let's say the original string is stretched by 2d, then each sub-spring is stretched by d in the (loaded) equilibrium position.

    Let x be the shift from the loaded equilibrium. Then the extension of the left sub-spring is (d + x); of the right sub-spring, (d - x). What are the restoring forces due to the left sub-spring? due to the right sub-spring? the total force?

    Mind the signs.
     
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