Question

A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F\left(x\right)=-kx+a{x}^3$. Here k and a are positive constants. For $x\ge 0$, the functional form of the potential energy U(x) of the particle is

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is D

The potential energy of the particle is given by
$U=-\int Fdx=-\int (-kx+a{x}^3)dx$
$\Rightarrow U=k\frac{x^2}{2}-a\frac{x^4}{4}=\frac{x^2}{4}(2k-a{x}^2).......\left(1\right)$
From Eq. (1) it follows that U = 0 at two values of x which are x = 0 and $x=\sqrt{\frac{2k}{a}}$. Hence, graphs (b) and (c) are not possible.
Also, U is maximum or minimum at a value of x given by $\frac{dU}{dx}=0$
$\Rightarrow \frac{d}{dx}(\frac{k{x}^2}{2}-\frac{a{x}^4}{4})=0\Rightarrow kx-a{x}^3=0$
$\Rightarrow x=\sqrt{\frac{k}{a}}$
At this value of x, U is maximum if $\frac{d^{2}U}{d{x}^2}<0$.
Now, $\frac{d^{2}U}{d{x}^2}=\frac{d}{dx}(kx-a{x}^3)=k-3a{x}^2$
At $x=\sqrt{\frac{k}{a}}$, $\frac{d^{2}U}{d{x}^2}=k-3a\frac{k}{a}$
$=k-3k=-2k$, which is negative.
Hence U is maximum at $x=\sqrt{\frac{k}{a}}$.
Hence, graph (a) is also not possible.
Moreover, U is negative for $x>\sqrt{\frac{2k}{a}}$.
Therefore, the correct graph is (d).