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Mass m suspended by two springs in series

  1. Oct 27, 2012 #1
    Hello guys i'm desperately trying to understand the solution of this assignement .
    What would be the differential equation og this Simple Harmonic Motion when you have a mass m suspended to 2 springs in series ?
     
  2. jcsd
  3. Oct 27, 2012 #2
    [kx - mg = m (dv/dt) = m(d sqaure x/dt)]
     
  4. Oct 27, 2012 #3

    haruspex

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    If the springs are considered massless, what do you think the relationship would be between the tensions in the two springs?
     
  5. Oct 27, 2012 #4
    yes the springs are massless, i got this answer from a textbook : the tension is uniform in the upper spring and has a magnitude of kx, hence the tension in the lower spring also has a magnitude of kx
    the displacement of M is 2x.
    the diff eq is :
    M2(dx²)/(dt) + kx = 0

    Why 2 in the first argument and not in the second since we have two springs.
     
  6. Oct 27, 2012 #5

    Doc Al

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    See if you can figure out what the effective spring constant is for the two springs in series.
     
  7. Oct 27, 2012 #6
    that's the problem my friend :( it's the reverse of series resistors on a circuit.
    K (equivalent) = (k1k2)/(k1 + k2)
    But i don't know how to get to it. maybe by moving backward
     
  8. Oct 27, 2012 #7

    Doc Al

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    Compare the force stretching the springs to the overall amount of stretch. How does the overall system stretch compare to that of each spring?
     
  9. Oct 27, 2012 #8
    How ?
     
  10. Oct 27, 2012 #9

    Doc Al

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    Imagine two springs in series, with spring constants k1 and k2. A force F stretches the two springs, thus F = k1x1 = k2x2.

    For the system as a whole, you have:
    F = k'(xtotal)

    See if you can solve for k'.
     
  11. Oct 27, 2012 #10
    how can i get rid of x1 and x2 ?
     
  12. Oct 27, 2012 #11

    Doc Al

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    Express them in terms of F and k.
     
  13. Oct 27, 2012 #12
    Thanks i did it with your big help, but i still don't understand this case when k1=k2 and x1=x2 because in my problem the two springs have the same k and since they are massless they'll be stretchend with the same x making a total of 2x in the displacement.
    Why in the first argument of the diff eq we have M2(dx²)/(dt) and in the 2nd we just have kx not 2kx ?
     
  14. Oct 27, 2012 #13
    It's okay now i totally got it thank you very much for your help :) please tell me which degree do you have ?
     
  15. Oct 27, 2012 #14

    Doc Al

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    When k1 = k2, k' = k/2.
    d2x/dt2 + k'x = 0

    d2x/dt2 + (k/2)x = 0

    Multiply both sides by 2!

    Edit: Looks like you figured it out while I was typing this.
     
  16. Oct 27, 2012 #15
    Thanks again
     
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