Finding angular frequency about the equilibrium position.

In summary, the equation for U(x) is messy and I don't know across which distance the particle is oscillating. I need to find the angular frequencies of the particle at its equilibrium positions.
  • #1
Aesteus
18
0

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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  • #2
pymn_nzr said:
equilibrium position[tex]\rightarrow \frac{\partial U}{\partial x}=0[/tex]



find static position [tex]\left.\begin{matrix} \frac{\partial^2 U}{\partial x^2} \end{matrix}\right|_{X=x_{1},x_{2}}> 0[/tex]
 
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  • #3
But how does that help me solve for angular frequency? I have already found the equilibrium positions where U(x) is concave.
 
  • #4
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)[tex]\left.\begin{matrix} \frac{\partial^2 U}{\partial x^2} \end{matrix}\right|_{X=x_{1},x_{2}}> 0[/tex]

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

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

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  • #5
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
 
  • #6
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.
 
  • #7
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
 

Related to Finding angular frequency about the equilibrium position.

1. What is the definition of angular frequency about the equilibrium position?

The angular frequency about the equilibrium position is a measure of the rate at which an object oscillates or rotates around its equilibrium position. It is usually denoted by the symbol ω and is measured in radians per second.

2. How is angular frequency about the equilibrium position calculated?

Angular frequency about the equilibrium position can be calculated using the formula ω = 2π/T, where T is the period of the oscillation or rotation. The period is the time it takes for the object to complete one full cycle of oscillation or rotation.

3. What is the relationship between angular frequency and frequency?

Angular frequency and frequency are closely related. They are both measures of how fast an object oscillates or rotates. The main difference is that angular frequency is measured in radians per second while frequency is measured in Hertz (cycles per second). The two are related by the formula ω = 2πf, where f is the frequency.

4. How does changing the mass or stiffness affect the angular frequency about the equilibrium position?

Changing the mass or stiffness of an object can affect its angular frequency about the equilibrium position. Increasing the mass or stiffness will result in a decrease in the angular frequency, while decreasing the mass or stiffness will result in an increase in the angular frequency. This is because the period of oscillation or rotation is affected by the mass and stiffness, and the angular frequency is inversely proportional to the period.

5. Can the angular frequency about the equilibrium position be negative?

No, the angular frequency about the equilibrium position cannot be negative. It is a measure of the rate of oscillation or rotation, and therefore must always be positive. A negative value would indicate that the object is moving in the opposite direction, which is not possible in this context.

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