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Frequency of a bead on a parabolic wire

  1. Dec 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the frequency of a bead with a mass of m vibrating on a parabolic track equals to y=Ax2

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    Looking at the bead at any point which isn't equilibrium, I have:

    1. may =N-mgcosθ
    2. max=mgsinθ

    I tried to look at a simpler scenario where the bead follows a circular path, in that case I can use small angle approximation to claim that ay=0 and also sinθ=θ. also, I defined x=lθ where l equals the radius of the circular path.using that info, I can get the d2θ/dt=(g/l)*θ and the frequency equals to ω=√(g/l).

    Now I'm trying to make some assumptions for the parabolic wire. so I'm pretty sure I can use small angle approximation same as above. as for x, I'm thinking about calculating 'x' using line integral, would that be the best way to go or am I should I look at this problem from a different angle?

    Thanks..
     
  2. jcsd
  3. Dec 26, 2015 #2
    You can write only forces that makes work: ## F_x = -mg\,\sin\phi\,\cos\phi ## and ## F_y = -mg\,\sin\phi\,\sin\phi ## where ## tan\phi = 2ax ##.

    But this calculation ignore direction change. As gamma said below, may use Lagrangian calculation.
     
    Last edited: Dec 26, 2015
  4. Dec 26, 2015 #3
    Hi,

    For these types of problems, using Lagrangian method works well. I don't know what level of education you are in, but if you have learned that, the solution might become simpler. You would need to write an expression for kinetic energy and potential energy.
     
  5. Dec 26, 2015 #4
    can you clarify how have you chosen the coordinate axes and what is θ in your diagram
    i took my axes such that tan(θ) is dy/dx for the parabola
    (sorry for the crudeness of my diagram )
    bead.png

    my equations were
    1. mg-Ncos(θ)=may (ay=second derivative of y wrt t)
    2. Nsin(θ)=max (ax= " '' " x wrt t)

    after finding tan(θ)=-(ay+g/ax) i found dy/dx=2ax and equated the two

    from them i got
    2ax{d2y/dt2+g}+ d2x/dt2=0 .........3.

    using the equation of parabola

    d2y/dt2=2a{ (dx/dt)2+d2x/dt2}

    substitute it back in eqn 3.

    we will get a differential equation in x and t .which i do not know how to solve. let me know if u solve it.
     
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