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Spring on a xy plane

  1. Mar 28, 2008 #1
    A small metal block of mass m is placed on a smooth horizontal table and constrained to move along a frictionless, rectilinear groove, The block is attached to one end of a spring (of spring constant k) whose other end is fastened to a pin P.
    Let length L be the equilibrium length of the spring and the perpendicular distance to the groove, The spring is now pulled a distance Xo from the equilibrium position and released. Show that, if the displacement along the groove x « t. the restoring force on
    the block is proportional to x ^3 , so the motion is not simple harmonic although it will still be periodic). [Hint: For x « L, x^2 + L^2 "" L + (x^2 /2L).]




    General spring and harmonic , periodic motion formulas



    I thought the cosine of the F force on the spring is the responsible force i.e. restoring force and sine of F has no importance, therefore Fcosineø = -k x .. but from there I get nowhere :(
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 28, 2008 #2

    tiny-tim

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    Hi yasar! :smile:

    What diagram have you drawn?

    What is the total force in the spring at distance x along the groove?

    So what is the component along the groove?

    (Hint: cosine ~ x/(L + x^2/2L).)
     
  4. Mar 28, 2008 #3
    well, it's from "PHYSICS for Scientists..... by Fishbane Gasiorowicz Thornton" Tim, but I don't know how to upload a picture or a diagram :(

    If you could imagine it on a xy plane, x being the groove line and L is the distance from P point to the equilibrium at y axis, I fail to see the importance of y-component of F, if any.
     
  5. Mar 28, 2008 #4

    tiny-tim

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    ah - I hadn't realised the question provided you with a diagram!

    ok - then back to my two questions:

    What is the total force in the spring at distance x along the groove?

    So what is the component along the groove? :smile:
     
  6. Mar 28, 2008 #5
    >What is the total force in the spring at distance x along the groove?
    Fx^2+Fy^2=F^2


    >So what is the component along the groove?
    Fx= Fcosineø =F x/d = F x/seqroot(x^2+L^2)

    ..and the above should equal to (-kx)??
     
  7. Mar 28, 2008 #6

    tiny-tim

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    No no no … F isn't a constant, is it? … so that doesn't get you anywhere. :frown:

    The question wants you to assume that the spring has a coefficient, of µ, say, and to work out the force as a function of µ.

    (When you work out proportionality later on, the µ won't matter).

    So what is the total force in the spring at distance x along the groove, in terms of µ? :smile:
     
  8. Mar 28, 2008 #7
    it should be -kx cosineø (or -µx cosineø)
    then
    cosineø being: x/d=x/seqroot(x^2+L^2)
    so
    F=-(kx^2)/(seqroot(x^2+L^2))??
     
  9. Mar 28, 2008 #8

    tiny-tim

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    oops … I didn't see the letter k in the question. :redface:

    But shouldn't it be k times the increased length of the spring? You've used x, which is only the sideways displacement.

    (and don't forget to use the hint formula in the question.)
     
  10. Mar 29, 2008 #9
    first of all thank you for the help, I appreciate.
    ok, the force on the spring is F=-k(ɅL) (as ɅL is the increased length and if ɅL is zero so the force)
    and that force's cosine component makes the mass M moves along the x-axis on the groove:
    F=-k(ɅL)cosineø=-k(ɅL)x/seqroot(x^2+L^2)

    from here where do I go???
     
  11. Mar 29, 2008 #10

    tiny-tim

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    Hi yasar1967! :smile:

    (btw, what symbol did you intend to put before the L? It came out as a square on my iMac.)

    Well, you were asked to show that the restoring force on the block is proportional to x^3.

    So from now on, it's just algebra.

    You have force = -k(ɅL)x/√(x^2+L^2)

    So now you write ɅL in terms of L and x.

    Then you use the hint formula in the question, and you get … ? :smile:
     
  12. Mar 31, 2008 #11
    that was meant to be Delta: Ʌ
    :)

    thanks again for the help
     
  13. Mar 31, 2008 #12

    tiny-tim

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    You're very welcome! :smile:

    oh, and I get Delta on my iMac by pressing alt-j

    Can you read that? (copy it, if you like.)

    ∆∆∆∆∆∆ :wink:
     
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