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Classical Mechanics

  • Thread starter danai_pa
  • Start date
A mass m is move in potential

U(x) = -a/x+b/x^2

I can solve this problem to find force F(x), stable point, turnning point
but i can't to find the equation of period of the mass for boundaring movement
please help
 
How about just using an analogy with the case for a spring:

F(x)=kx, etc. :smile:
 
Berislav said:
How about just using an analogy with the case for a spring:

F(x)=kx, etc. :smile:
w(omega) = sqrt(k/m) is correct?.
if correct how can to find values k?. please explain
 

robphy

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Following Berislav 's suggestion,
U(x)=(1/2)kx^2 for a spring.

Hopefully, you *know* what force law F(x) corresponds to this potential energy function. How do you get F(x) from U(x)?

What characterizes [in terms of U(x)] the position of the stable equilibrium point?

Given a suitable total energy E (a constant), what is the range of positions available to the particle? If you can setup a differential equation for the conservation of energy, you can obtain an expression for t as a function of E and U(x).

If you can successfully do this for this potential energy function, you should [in principle] be able to apply the same ideas to your potential energy function.
 
[itex]F= - \frac {dU}{dX}[/itex]

Put above equal to kx and get k , use some math.

BJ
 
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how would you get k from that? Seems like you need to know what x is... :smile: Is that what you mean by "use some math"?
 
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danai_pa:

Did you get an answer to this problem? I don't believe you will need to solve any cubic equations. Do what robphy said...set up the equation

E = (1/2)mv^2 + U(x), where v = dx/dt

and solve for dt. Integrate from one turning point to the other. That's half a period. This gives you the exact period even if the oscillation is not small. If you are allowed to assume the oscillation is small, you can expand the potential energy about a stable point and then apply the F = -kx technique.
 

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