Question

The potential energy of a particle along x-axis varies as, $U = - 20 + {\left( {x - 2} \right)^2}$ where $U$ is in joule and $x$ in meter. State the type of equilibrium and $x = 2m$.

Answer 1

In order to solve this question, we are first going to find the equilibrium position by differentiating the potential energy with respect to the position, then equating it to zero, we get the position equal to$x = 2m$. To find the type of equilibrium, we will double differentiate$U$and equate to zero.

Formula used:
For the particle to be in equilibrium,
$\dfrac{{dU}}{{dx}} = 0$
And for stable equilibrium of the particle,
$\dfrac{{{d^2}U}}{{d{x^2}}} > 0$
Where, $U$ is the potential and$x$is the position.

Complete answer:
It is given in the question that the potential energy of a particle along x-axis varies as,
$U = - 20 + {\left( {x - 2} \right)^2}$
Now simplifying this equation further, we get
U = {x^2} + 4 - 4x - 20 \\\ \Rightarrow U = {x^2} - 4x - 16 \\\

To find the equilibrium position, we need to differentiate the potential energy with respect to the position, i.e.$x$
$\dfrac{{dU}}{{dx}} = 2x - 4$
Putting this differentiation equal to zero, we get
$\dfrac{{dU}}{{dx}} = 2x - 4 = 0 \\\ \Rightarrow 2x = 4 \\\ \Rightarrow x = 2 \\\$
To find the type of the equilibrium, we again differentiate the first order differential with respect to position, i.e. $x$
$\dfrac{{{d^2}U}}{{d{x^2}}} = 2$
Here, we can see that
As $2 > 0$
Hence, we can deduce that
$\dfrac{{{d^2}U}}{{d{x^2}}} > 0$
Which shows that at $x = 2m$, the particle is in stable equilibrium.
The value of the minimum potential energy at the position $x = 2m$ is
$U = - 20$

Note: It is important to note that the less the potential energy of the particle more is the stable equilibrium, the minimum potential energy position is the equilibrium position and for it to be the stable equilibrium, the double differential of the potential energy has to be greater than zero.