Question

A particle free to move along the x – axis has potential energy given by
$U\left(x\right)=k[1-exp(-x^2\left)\right]for-\infty \le x\le +\infty$
Where k is a constant of appropriate dimensions. Then

• A. At point away from the origin, the particle is in unstable equilibrium
• B. For any finite nonzero value of x, there is a force directed away from the origin
• C. If its total mechanical energy is $\frac{k}{2}$ it has its minimum kinetic energy at the origin
• D. For small displacements from x = 0, the motion is simple harmonic

Answer 1

The correct option is D For small displacements from x = 0, the motion is simple harmonic

Figure 9.35 shows the plot of U(x) versus x.

At x = 0, potential energy$U\left(0\right)=k[1-exp(0\left)\right]=k(1-1)=0$ and it has a maximum value = k at $x=\pm \infty$ since
$U(\pm \infty )=k[1-exp(-\pm \infty {)}^2]=k(1-0)=k$
Since the total mechanical energy has a constant value = $\frac{k}{2}$ the kinetic energy will be maximum at x=0 and minimum at $x=\pm \infty$ At x = 0
${\left(\frac{dU}{dx}\right)}_{x=0}=\left[2kxexp\right(-x^2){]}_{atx=0}=0$
Hence the particle is in stable equilibrium at x = 0 (origin) and would oscillate about x = 0 (for small displacements) simple harmonically. Hence (d) is the only correct choice