Finding angular frequency about the equilibrium position.

AI Thread Summary
The discussion revolves around finding the angular frequency of a particle oscillating in a potential energy well defined by the equation U(x) = E/β^4(x^4 + 4βx^3 - 8(β^2)x^2). The equilibrium positions are identified at x = β and x = -4β, determined through differentiation of U(x). Participants suggest using the second derivative test to confirm the nature of these equilibrium points and to derive the angular frequency. A Taylor expansion of U(x) around the equilibrium positions is recommended to simplify the calculations. The conversation highlights the importance of correctly applying calculus to analyze oscillatory motion in potential energy contexts.
Aesteus
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Homework Statement



Alright so I've got a potential energy equation U(x) = E/β^4(x^4+4βx^3-8(β^2)x^2) and U'(x) = E/β^4[4(x^3) + 12β(x^2) - 16(β^2)x] (where β and E are constants) that describes a particle of mass m which is oscillating in an energy well. I solved for where the system has equilibrium positions through simple differentiation of U(x). (an equilibrium is at x=β and x=-4β) Now I have to find the angular frequency about each equilibrium position and estimate how small the oscillations should be around the equilibrium position.

Homework Equations



N/A

The Attempt at a Solution



I figure that since dU/dt = ma, I can differentiate U(x), equate it to acceleration, and solve for ω. However, the equation for U(x) is rather messy, and I still don't know across which distance the particle is oscillating. Any ideas?
 
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pymn_nzr said:
equilibrium position\rightarrow \frac{\partial U}{\partial x}=0



find static position \left.\begin{matrix} \frac{\partial^2 U}{\partial x^2} \end{matrix}\right|_{X=x_{1},x_{2}}> 0
 
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But how does that help me solve for angular frequency? I have already found the equilibrium positions where U(x) is concave.
 
Aesteus said:
Well I'm trying to find the angular frequencies of the particle at its equilibrium positions.
I'm not quite sure where you go that quadratic equation. Could you explain in more detail?

Thanks
this is diagram U(x)
1-equilibrium position(STATIC)\left.\begin{matrix} \frac{\partial^2 U}{\partial x^2} \end{matrix}\right|_{X=x_{1},x_{2}}> 0

2-indfferent\left.\begin{matrix} \frac{\partial^2 U}{\partial x^2} \end{matrix}\right|_{X=x_{1},x_{2}}= 0

3-equilibrium position(not static)\left.\begin{matrix} \frac{\partial^2 U}{\partial x^2} \end{matrix}\right|_{X=x_{1},x_{2}}< 0
 

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Aesteus said:
U(x) = E/β^4[4(x^3) + 12β(x^2) - 16(β^2)x] (an equilibrium is at x=β and x=-4β)



an equilibrium is not at x=β and x=-4β
 
Arghgagh ... there should be a prime in the original potential energy equation above. I corrected it.
Thanks.

-But the values of x = β and -4β are still correct.
 
The equilibrium points are correct. Write up the Taylor expansion of U(x) around both of them and stop at the second order term.

ehild
 
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