Solving Motion along a Line: Finding Turning Points, Max U & Frequency

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The discussion focuses on analyzing the motion of a particle P of unit mass along the x-axis under the force F(x) = 2(x³ - x). The potential energy is defined as V(x) = x² - x⁴/2, with equilibrium points at x = 0 and x = ±1. The participant seeks to determine the turning points as a function of initial speed U and the maximum U for bounded motion, as well as to prove that small displacements from x = 0 lead to periodic oscillations with a frequency of √2.

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Homework Statement



A particle P of unit mass moves along an x-axis under the influence of the force

F(x)=2(x[itex]^{3}[/itex]-x)

Firstly, I find V(x)=x[itex]^{2}[/itex]-x[itex]^{4}[/itex]/2.
Where equilibrium points are F(x)=0 so x=0 with energy V(0)=0, x=1 with energy V(1)=V(-1)=1/2.
And I have also sketched the graph.

So here are the parts I'm stuck on:

1) Initially P is projected from the point x=1/2 with speed U. Using conservation of energy, find the turning points (where x'=0) as a function of U. Find the maximum value of U for which the resultant motion will be bounded.

2)Stating from Newton's second law, prove that a particle displaced by a small amount from x=0 will perform periodic oscillations with a frequency of [itex]\sqrt{2}[/itex]



Homework Equations



T=mx'[itex]^{2}[/itex]/2

T+V=E

The Attempt at a Solution



So for 1), I start with

U[itex]^{2}[/itex]/2+(1/2)[itex]^{2}[/itex]-(1/2)[itex]^{4}[/itex]/2=E

Giving E=U[itex]^{2}[/itex]+7/16

Then, because energy is conserved and x'=0 at turning points:

x[itex]^{2}[/itex]-x[itex]^{4}[/itex]/2=U[itex]^{2}[/itex]+7/16

And now I'm really stuck for how to find the turning points as functions of U?

I'm guessing I'm going about this the completely wrong way??

And I have no idea what I'm supposed to be doing for 2) :frown:

Sorry this is long, but any help will be greatly appreciated :smile:
 
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zebrastripes said:

Homework Statement



A particle P of unit mass moves along an x-axis under the influence of the force

F(x)=2(x[itex]^{3}[/itex]-x)

Firstly, I find V(x)=x[itex]^{2}[/itex]-x[itex]^{4}[/itex]/2.
Where equilibrium points are F(x)=0 so x=0 with energy V(0)=0, x=1 with energy V(1)=V(-1)=1/2.
And I have also sketched the graph.

So here are the parts I'm stuck on:

1) Initially P is projected from the point x=1/2 with speed U. Using conservation of energy, find the turning points (where x'=0) as a function of U. Find the maximum value of U for which the resultant motion will be bounded.

2)Stating from Newton's second law, prove that a particle displaced by a small amount from x=0 will perform periodic oscillations with a frequency of [itex]\sqrt{2}[/itex]



Homework Equations



T=mx'[itex]^{2}[/itex]/2

T+V=E

The Attempt at a Solution



So for 1), I start with

U[itex]^{2}[/itex]/2+(1/2)[itex]^{2}[/itex]-(1/2)[itex]^{4}[/itex]/2=E

Giving E=U[itex]^{2}[/itex]+7/16

Then, because energy is conserved and x'=0 at turning points:

x[itex]^{2}[/itex]-x[itex]^{4}[/itex]/2=U[itex]^{2}[/itex]+7/16

And now I'm really stuck for how to find the turning points as functions of U?

I'm guessing I'm going about this the completely wrong way??

And I have no idea what I'm supposed to be doing for 2) :frown:

Sorry this is long, but any help will be greatly appreciated :smile:

If v = velocity (= dx/dt), then by conservation of energy we have
[tex]\frac{1}{2} v^2 + V(x) = \frac{1}{2} U^2 + V(1/2) = <br /> \frac{1}{2} U^2 + \frac{7}{32},[/tex] since [itex]V(1/2) = 7/32.[/itex] At turning points, [itex]v = 0,[/itex] so you get an equation in [itex]x.[/itex]

For small |x| the potential V(x) is almost linear in x. If you neglect the x4 term in V(x) you get a simple harmonic oscillator.

RGV
 
Ray Vickson said:
For small |x| the potential V(x) is almost linear in x. If you neglect the x4 term in V(x) you get a simple harmonic oscillator.

RGV

Thanks RGV, but under what assumptions can I say for small |x|?

And would you mind giving me a hint about what to do for part 2)?

Thanks again!
 

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