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

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



 
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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