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Question #42311

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass mm moves in one dimension x>0x > 0 under the influence of a conservative force field. The potential energy of the particle is given by U(x)=U0[ln⁡(xa)+cos⁡(xa)]U(x) = U_0 \left[ \ln\left(\frac{x}{a}\right) + \cos\left(\frac{x}{a}\right) \right] where U0U_0 and aa are positive constants. For large values of xx (x≫ax \gg a), the positions of the stable equilibrium points are approximately given by (where nn is a large positive integer):

Options

  • A.

    x≈2nπax \approx 2n\pi a

  • B.

    x≈(2n+12)πax \approx \left(2n + \frac{1}{2}\right)\pi a

  • C.

    x≈(2n+1)πax \approx (2n + 1)\pi a

  • D.

    x≈(2n+32)πax \approx \left(2n + \frac{3}{2}\right)\pi a

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Variable Force Integration|Equilibrium Stability|Graphical Concavity Significance

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze non-linear, transcendental potential energy surfaces.

A common trap is assuming that setting F=0F=0 and finding the asymptotic roots of sin⁡(x/a)=0\sin(x/a) = 0 is sufficient, which leaves students guessing between 2nπ2n\pi and (2n+1)π(2n+1)\pi.

Students often fail to apply the second derivative test (d2U/dx2>0d^2U/dx^2 > 0) to distinguish between stable and unstable equilibria.

Furthermore, the interaction between the decaying logarithmic potential and the oscillatory harmonic potential requires asymptotic reasoning (x≫ax \gg a) to isolate the dominant term in the concavity condition, effectively filtering out candidates who rely purely on algebraic manipulation without physical intuition.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptVariable Force Integration
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree