27. A particle moves under the potential 𝑈(𝑥) = 𝑒
+0.01𝑥 .
Which of the following describes the equilibrium points in the region −∞ < 𝑥 < ∞
a. One stable point, no unstable points
b. One unstable points, no stable points
c. Two stable points, one unstable point
d. Two unstable points, one stable point

Answer: de>Option (d) – Two unstable points, one stable point (though this option is incorrect for the given potential function).

Equilibrium Analysis

Equilibrium points occur where the force de>F = -dU/dx = 0, or equivalently, de>dU/dx = 0.

Given U(x) = e^0.01x + x, compute the derivative:

de>dU/dx = 0.01e^0.01x + 1

Since e^0.01x > 0 for all real x, the term de>0.01e^0.01x > 0. Therefore, dU/dx > 1 > 0.

No real solutions exist where de>dU/dx = 0, meaning no equilibrium points occur for this potential function.

Stability Consideration

To analyze stability, inspect the second derivative:

de>d²U/dx² = 0.0001e^0.01x > 0 everywhere, indicating a concave-up (minimum-like) curvature.

However, since no equilibrium points exist, classification into stable or unstable states does not apply.

Option Breakdown

• a) One stable, no unstable: Incorrect — no equilibria exist.
• b) One unstable, no stable: Incorrect — zero equilibrium points in total.
• c) Two stable, one unstable: Incorrect — equation de>dU/dx = 0 has no real solutions.
• d) Two unstable, one stable: Incorrect despite being structurally similar to expected outcomes; the actual count is zero. The question likely contains a misprint, possibly meant as de>U(x) = e^{-0.01x² + 0.01x}.

Concept Insight

For the potential de>U(x) = e^0.01x + x, the derivative never vanishes, indicating continuous positive slope and hence no equilibrium. This example is valuable for strengthening conceptual understanding of equilibrium and stability conditions in potential energy analysis.

Such problems are common in CSIR NET Physics equilibrium and mechanics topics, testing analytical reasoning about the relationship between potential energy, force, and curvature.

Equilibrium Analysis for U(x) = e^0.01x + x

The potential de>U(x) = e^0.01x + x describes a particle’s motion where equilibrium points occur when dU/dx = 0.

Compute the first derivative:

de>dU/dx = 0.01e^0.01x + 1 = 0

This equation has no real roots, because the exponential term de>e^0.01x remains positive for all de>x across the range de>−∞ < x < ∞. Therefore, de>dU/dx is always positive and never zero.

Stability Concept

For typical potentials such as double-well forms, stability is determined by examining the second derivative:

de>d²U/dx² > 0 → Stable (local minimum)
de>d²U/dx² < 0 → Unstable (local maximum)

In this case, the second derivative de>d²U/dx² = 0.0001e^0.01x is always positive, indicating upward concavity everywhere. However, since no equilibrium points (de>dU/dx = 0) exist, stability classification does not apply.

Practice Tip for CSIR NET

When analyzing equilibrium for given potentials:

• Check de>dU/dx for zeros to locate equilibrium points.
• Use de>d²U/dx² to determine stability (minima → stable, maxima → unstable).
• Graph or approximate numerically for complex forms.

Note that potentials combining linear and exponential terms rarely yield multiple equilibria unless negative coefficients or opposing slopes are introduced.