PHYSICS: Poschel-Teller Potential and Natural Frequency Derivation

[Void123]

Homework Statement

A particle of mass m moves in the one dimensional Poschel-Teller potential V(x). Find an expression for the natural frequency of small oscillations.

Homework Equations

V(x) = -V_{0}sech^{2}(x/\lambda)

The Attempt at a Solution

I am making the uncertain assumption that this would be the angular frequency. But, I do not know how to derive it based on just the potential alone. I have tried to determine the period as well, graphically, but this function doesn't seem to be the type associated with normal periodic motion.

 

[kuruman]

The potential that you have is not a harmonic potential, however any potential that is an even function can be expanded in Taylor series about its minimum and will have a leading term that is quadratic in x. The words "small oscillations" is a hint that you need to make this expansion. First plot your potential to make sure that it has a minimum, then expand about x = 0. Ignore the constant, that's the "zero of energy". The leading term will be

\frac{1}{2}\;\frac{d^{2}V}{dx^{2}}\big|_{x=0}\;x^{2}

That's a term that can be related to a harmonic potential of the form

\frac{1}{2}\;kx^{2}

whose frequency you can easily extract from the effective spring constant k.

 

[Void123]

The potential that you have is not a harmonic potential, however any potential that is an even function can be expanded in Taylor series about its minimum and will have a leading term that is quadratic in x. The words "small oscillations" is a hint that you need to make this expansion. First plot your potential to make sure that it has a minimum, then expand about x = 0. Ignore the constant, that's the "zero of energy". The leading term will be

\frac{1}{2}\;\frac{d^{2}V}{dx^{2}}\big|_{x=0}\;x^{2}

That's a term that can be related to a harmonic potential of the form

\frac{1}{2}\;kx^{2}

whose frequency you can easily extract from the effective spring constant k.

I figured it out. Thanks!

But just one tiny question, if the leading term is going to be a quadratic, then the original expansion has to be out to the fourth power?

 

[Void123]

I must not be doing something right, the whole thing went to zero.