Question

A particle with total mechanical energy, which is small and negative, is under the influence of a one dimensional potential $U(x) = \dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}J$ where x is in meters. At time $t = 0\sec$ , it is at $x = - 0.5m$. Then at a later time it can be found.
(A) Anywhere on the x-axis
(B) Between $x = - 1.0m$ to $x = 1.0m$
(C) Between $x = - 1.0m$ to $x = 0.0m$
(D) Between $x = 0.0m$ to $x = 1.0m$

Answer 1

See the relationship between the total mechanical energy of a particle and its potential energy. Then differentiate the given equation of potential energy and find its roots. With the help of the roots find the minima and the maxima in which the particle will lie. Then, according to the given condition, find its position.

Complete Step By Step Answer:
We have been given that the total mechanical energy of a particle is small and negative,
We know total mechanical energy is the sum of potential energy and kinetic energy.
As kinetic energy is either zero or positive. So to satisfy the condition potential energy will always be negative so that mechanical energy comes negative.
Now taking the potential function
$U(x) = \dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}$
Find the maxima and minima of the above
Differentiate the above and its roots will be its minima and maxima
$\dfrac{{dU}}{{dx}} = {x^3} - x = 0$
$x = 0, + 1, - 1$ are the roots of the above.
Double differentiating the above equation we get,
$\dfrac{{{d^2}U}}{{d{x^2}}} = 3{x^2} - 1$
If $x = 0$ $\dfrac{{{d^2}U}}{{d{x^2}}} = - ve$ this is the point of maxima
If $x = 1$ $\dfrac{{{d^2}U}}{{d{x^2}}} = + 2( + ve)$ this is the point of minima
If $x = - 1$ $\dfrac{{{d^2}U}}{{d{x^2}}} = + 2( + ve)$ this is the point of minima
We have been given that the particle at time $t = 0\sec$ , it is at $x = - 0.5m$. Also, the range of potential energy will always be negative. So the particle will be in the range where its potential energy will be negative.
Therefore, it will lie between $( - 1,0)$
Hence, option C) between $x = - 1.0m$ to $x = 0.0m$ is the correct answer.

Note:
Since the term velocity, kinetic energy can never have a negative value (because mass is non-negative). When it comes to potential energy, it is determined by the reference line (that is, the line where zero potential energy is assumed). The difference in potential energy is independent of the reference line, whereas potential energy is reliant on it so you can consider zero potential energy, which results in negative Mechanical Energy. As a result, it is entirely dependent on your reference line.