Frequency of a bead on a parabolic wire

AI Thread Summary
The discussion revolves around calculating the frequency of a bead vibrating on a parabolic track defined by the equation y=Ax^2. The user initially applies Newton's second law to analyze the forces acting on the bead and considers a simpler circular path scenario using small angle approximations. They explore the possibility of using a line integral to calculate 'x' for the parabolic wire and are advised that employing the Lagrangian method could simplify the problem. The conversation includes deriving equations related to the forces acting on the bead and attempts to formulate a differential equation for further analysis. The user expresses uncertainty about solving the resulting differential equation and seeks assistance.
azarue
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Homework Statement


Calculate the frequency of a bead with a mass of m vibrating on a parabolic track equals to y=Ax2

Homework Equations


F=ma

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..
 
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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:
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.
 
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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