• Support PF! Buy your school textbooks, materials and every day products Here!

Frequency of a bead on a parabolic wire

  • Thread starter azarue
  • Start date
  • #1
1
0

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..
 

Answers and Replies

  • #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:
  • #3
356
11
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.
 
  • #4
36
5
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.
 

Related Threads for: Frequency of a bead on a parabolic wire

  • Last Post
Replies
2
Views
433
  • Last Post
Replies
3
Views
13K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
15
Views
741
  • Last Post
Replies
1
Views
876
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
Top