Is this the correct approach? (finding frequency of oscillation)

  • Thread starter Sefrez
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Homework Statement


Find the frequency of small oscillations around the minimum of the potential
U(x)=1-e^(-x^2)


Homework Equations


Force is the negative of the gradient of the potential...


The Attempt at a Solution


Given the problem statement bit, "around the minimum," I take this as a hint to find the taylor expansion of U(x) to approximate the potential at the minimum.

In doing the taylor expansion at 0, I get that:
U(x) ≈ x^2

The force on a particle in this potential is given by:
F = -dU/dx = -2x.

And so we have that:
F + 2x = 0 => mx'' + 2x = 0. Solving this differential equation, we have something of the form:
x = A*cos(√(2/m)t - [itex]\phi[/itex])

So, we have, the angular frequency to be: ω = √(2/m).

Finally, the frequency is then: [itex]\nu[/itex] = ω/(2∏) = √(2/m)/(2∏) = (2m∏^2)^(-1/2)

Does this seem correct? I was a little confused that the frequency is dependent on the mass, but then I see that the potential given is independent of mass. But thats a bit odd. Thanks.
 

Answers and Replies

  • #2
rude man
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This all seems entirely correct to me.

Your teach can come up with any kind of potential he wants! :smile:
 

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