Following Berislav 's suggestion,
U(x)=(1/2)kx^2 for a spring.
Hopefully, you *know* what force law F(x) corresponds to this potential energy function. How do you get F(x) from U(x)?
What characterizes [in terms of U(x)] the position of the stable equilibrium point?
Given a suitable total energy E (a constant), what is the range of positions available to the particle? If you can setup a differential equation for the conservation of energy, you can obtain an expression for t as a function of E and U(x).
If you can successfully do this for this potential energy function, you should [in principle] be able to apply the same ideas to your potential energy function.
Did you get an answer to this problem? I don't believe you will need to solve any cubic equations. Do what robphy said...set up the equation
E = (1/2)mv^2 + U(x), where v = dx/dt
and solve for dt. Integrate from one turning point to the other. That's half a period. This gives you the exact period even if the oscillation is not small. If you are allowed to assume the oscillation is small, you can expand the potential energy about a stable point and then apply the F = -kx technique.
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