Can Heun's Method Accurately Predict Small Oscillations of a Nitrogen Atom?

AI Thread Summary
The discussion focuses on using Heun's method to compute the period of small oscillations for a nitrogen atom, with specific parameters provided. The user calculated the derivative of the potential energy function but received feedback questioning the accuracy of their derivative calculation. Suggestions were made to simplify the expression before differentiation and to verify the results using computational tools like Wolfram Alpha. The importance of confirming the derivative against manual calculations was emphasized to ensure correctness in the subsequent steps. Overall, the conversation highlights the need for careful verification in numerical methods.
akaPaul
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Homework Statement


Use the Heun method to compute the period of small oscillations about the equilibrium position of a nitrogen atom.
xi = 1.1
Um = 7.37
x0 = 1.2
alpha = 2.287
m = 2.325e-26

Homework Equations


[/B]
U(x) = Um((1-e^(-alpha(x-x0)))^2 - 1)

The Attempt at a Solution



I was told to take the derivative of dU/dx and got (-1/m)*(2*Um*alpha*(1-e**(-alpha*(x-x0)))*(e**(-alpha*(x-x0)))) and used that to calculate my heun step with a time step of 0.5e-15. With this I get a graph that looks like this:
1XxRKQK.jpg
which is a velocity vs. time graph. I'm wondering if this graph looks correct and if so why? I thought it would turn out to be something more like:
DampedSine.png



 
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akaPaul said:

Homework Statement


Use the Heun method to compute the period of small oscillations about the equilibrium position of a nitrogen atom.
xi = 1.1
Um = 7.37
x0 = 1.2
alpha = 2.287
m = 2.325e-26

Homework Equations


[/B]
U(x) = Um((1-e^(-alpha(x-x0)))^2 - 1)

The Attempt at a Solution



I was told to take the derivative of dU/dx and got (-1/m)*(2*Um*alpha*(1-e**(-alpha*(x-x0)))*(e**(-alpha*(x-x0)))) and used that to calculate my heun step with a time step of 0.5e-15. With this I get a graph that looks like this:
1XxRKQK.jpg
which is a velocity vs. time graph. I'm wondering if this graph looks correct and if so why? I thought it would turn out to be something more like:
DampedSine.png


If ##U(x) = U_m((1-e^{(-alpha(x-x0)))^2 - 1)}##, it's not clear how you calculated

$$U'(x) = (-1/m)*(2*Um*alpha*(1-e^{(-alpha*(x-x0)))}*(e^{(-alpha*(x-x0)))}))$$

I'm assuming that ##U_m## is a constant and not ##U ⋅ m##.

You might want to use some substitutions in the original expression for U(x) to simplify its form before taking the derivative. I would check the derivative using a tool like Wolfram Alpha before writing code with it.
 
SteamKing said:
If ##U(x) = U_m((1-e^{(-alpha(x-x0)))^2 - 1)}##, it's not clear how you calculated

$$U'(x) = (-1/m)*(2*Um*alpha*(1-e^{(-alpha*(x-x0)))}*(e^{(-alpha*(x-x0)))}))$$

I'm assuming that ##U_m## is a constant and not ##U ⋅ m##.

You might want to use some substitutions in the original expression for U(x) to simplify its form before taking the derivative. I would check the derivative using a tool like Wolfram Alpha before writing code with it.
Yea sorry about my bad formatting you're right it is a constant and I used python to compute the derivative.
 
akaPaul said:
Yea sorry about my bad formatting you're right it is a constant and I used python to compute the derivative.
I'm not sure how you do that, but I would check the derivative which Python gave against a pencil and paper derivative or Wolfram Alpha.
 
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