Question

3. A particle with unknown mass is placed into a region of space and subject to a potential energy given by U(x)=0.5*e*-9x® and found to oscillate about the global minimum potential with an angular frequency of 41 rad/sec. What is the frequency of its oscillations about its other local minimum/minima?

Answer 1

For any given potential the way to find the maxima/minima which are generally called the turning points is to solve for the derivate of the function and set it to zero, i.e, $U'(x) = 0 \implies \frac{dU(x)}{dx} = 0.$

For us,
U' (2) = exp (2) 2 3
Now solving for $\frac{\exp(x)}{2} - \frac{x^2}{3} = 0,$ gives us solutions.

Now the first two solutions are imaginary and we can discard them. So we see that there are indeed two turning points. Lets us plot the function and see,

10 5 -1.0 -0.5 1.0 -5

This is near the first turning point, which is clearly a maximum.

2000 1000 6 7 8 10 -1000 -2000 -3000
This is near the second turning point, which is clearly the minimum (global).

So how can we that the function has only this as the global minimum? Look at say for positive x. For large x, exp(x) dominates and you can ignore x^3. So exp(x) -> infinity and the function is bounded. For small x, exp(x) can be expanded and written as 1 + x + x^2 / 2 + x^3 / 6 + O(x^4). The fourth order term can be neglected since for small x (x < 1), x^4 -> 0. Hence we can say that the function has two turning points, one minimum one maximum.

Now for large x < 0, exp(x) -> and x^3 dominates and its grow and hence its bounded (as can be seen from the graph as well). For small x, again the same logic as above applies. Expansion and check. This is how one can find the turning points intutitively.

Now coming to the question, since the other turning point is a maximum, we can't possibly have any small oscillation there, as any tiny perturbation around that point and the mass will slide to the global minimum (and hence the name its an unstable fixed point).

So the time period is not defined for the mass at the other turning point since its a maximum.

tl;dr The system has two real turning points, One maximum, the other minimum for which the tiemperiod is well defined. The maximum has no well defined time period as any small perturbation will lead to the mass sliding off to the global minimum.

PS: If you say try to write the equation about the maximum you'll get, $\ddot{x} = +k \cdot x,$ instead of the usual $\ddot{x} = -k \cdot x,$ for small oscillation. So the system grows and falls for any small perturbation.