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Four springs Oscillation

  1. Mar 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Figure shows a particle of mass m attached with four identical springs each of length L0. Initial tension in each spring is F0.Neglect gravity calculate the period of small oscillations of the particle along a line perpendicular to the plane of figure. (Take F0 = (0.01 π2)N, m = 100 gm, L0 = 10cm)

    2. Relevant equations



    3. The attempt at a solution

    In the equilibrium state length of the spring L = L0 + z ,where L0 is the original length and z is the initial extension in the spring when in equilibrium.

    Initial tension in each spring when in equilibrium = kz = F0

    Suppose the mass is displaced by x amount upwards .In the attachment L' represents the new length.

    L'=√(L2+x2)

    Force by a spring = k(L'-L0) = = k(L'-L+z) = k(L'-L)+kz

    This force will have a a horizontal component and a a vertical component .The horizontal component will be cancelled by a similar component from the opposite string .

    The net force from all four springs will be F = 4k(L'-L0 )sinθ = 4k(L'-L)sinθ + 4kzsinθ

    sinθ = x/L'

    So net force =4kx(L'-L)/L' + 4kxz/L’

    =4kx(1-L/L') + 4kxz/L’

    I somehow feel I am not approaching the problem correctly .

    I would be grateful if somebody could help me with the problem.
     

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    Last edited: Mar 20, 2014
  2. jcsd
  3. Mar 20, 2014 #2

    ehild

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    It is correct so far. Now you consider the case when x is small with respect to L. Approximate the square root accordingly.


    ehild
     
  4. Mar 20, 2014 #3
    Hello ehild...

    L'=√(L2+x2)
    L'=L√(1+x2/L2)

    L' ≈ L(1+x2/2L2) when x<<L

    Is it correct ?

    I think it is not right .If we had x/L instead of x2/L2 ,then it would have been alright . x2/L2 term will be too small .But then if we don't keep it then L' ≈ L ,which again looks incorrect .Not sure when to keep which terms while making approximations.
     
    Last edited: Mar 20, 2014
  5. Mar 20, 2014 #4

    ehild

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    It is correct. You have to keep terms linear in x/L. Note that the square root is in the denominator. What is the force?

    ehild
     
  6. Mar 20, 2014 #5
    F = 4kx3/2L2(1+x2/2L2) + 4kxz/(1+x2/2L2)

    Right ?
     
  7. Mar 20, 2014 #6

    ehild

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    You made it very complicated...First: the force is opposite to x. Factor out 4kx. [tex]F=-4kx\left(1-\frac{L-z}{L \sqrt{1+(x/L)^2}}\right)[/tex]

    ehild
     
  8. Mar 20, 2014 #7
    Sorry for the mess up.

    Is [itex]L \sqrt{1+(x/L)^2}[/itex] ≈ [itex]L[/itex] ?

    If correct then [itex]F = \frac{-4kxz}{L}[/itex] or [itex]F = \frac{-4F_{0}x}{L}[/itex]

    Right?
     
    Last edited: Mar 20, 2014
  9. Mar 20, 2014 #8

    ehild

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    Yes, it is correct, but write L in terms of Fo and Lo, as they are given.

    ehild
     
  10. Mar 20, 2014 #9
    L = L0+z and kz=F0

    But ,neither z nor k is given in the problem ?

    I think there is something wrong in the question .The length given should be the stretched length of the spring in equilibrium not the natural length .

    What do you say ?
     
    Last edited: Mar 20, 2014
  11. Mar 20, 2014 #10

    ehild

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    Yes, Lo might be the initial length instead of the unstretched length.

    ehild
     
  12. Mar 20, 2014 #11
    Thanks...

    There is a similar problem of four springs just as in the OP.But in this case the springs are in their natural length i.e there is no initial tension.Here the oscillations are occurring in the plane of the figure , not perpendicular to it.We need to find the time period when the mass is displaced slightly towards the top vertical spring i.e the oscillations are occurring in the plane of the figure .Neglect gravity.

    My approach - For calculating forces by the left and the right spring ,we will approach exactly the way we have done in this question .Force due to left spring = Force due to right spring ≈ 0

    Force due to top spring + Force due to bottom spring = -2kx

    Net force F = -2kx

    Do I make sense ?
     
  13. Mar 20, 2014 #12

    ehild

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    The forces of the left and right springs are not zero if there is an initial tension in the springs. But the springs have their natural length this time, so the tension is zero. F=-2kx.

    ehild
     
  14. Mar 20, 2014 #13
    Thank you so much :)
     
    Last edited: Mar 20, 2014
  15. Mar 20, 2014 #14
    ehild...I found a few things not so obvious in the Original problem ?

    1) L' ≈ L , even when the springs are getting stretched .

    2) If the springs are not under initial tension in the OP ,then the mass would not be performing SHM , as net force will be zero .

    What do you think ?
     
  16. Mar 20, 2014 #15

    ehild

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    Not necessarily... Those springs can be stressed quite much. Think of rubber bands.

    The new force would not be zero, but it would not be linear in the displacement. The mass will move somehow, it would be interesting to see, how.

    ehild
     
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