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Consider a particle of mass m moving in one dimensional potential V(x)=x4 – 2x2. The phase trajectory (x,p) of the particle is best represented by

This question was previously asked in
MH SET 2026 Paper 1 Question Paper (25-Oct-2026)
The correct answer is

Phase trajectories are the constant-energy curves \(\frac{p^2}{2m}+V(x)=E\), so their shape is fixed by the extrema of V(x).

\(V'(x)=4x^3-4x=4x(x^2-1)=0\) gives x = 0 and x = ±1. Since V(0) = 0 is a local maximum and V(±1) = –1 are minima, this is a symmetric double-well potential.

The minima (±1, 0) are stable equilibria (centres): for –1 < E < 0 the particle is trapped in one well, giving two separate families of closed orbits around x = +1 and x = –1.

The maximum (0, 0) is an unstable equilibrium (saddle point). The E = 0 curve, p = ±√(2m(2x² – x⁴)), is the separatrix – a figure-of-eight passing through the origin and enclosing both wells.

For E > 0 the trajectories are large curves enclosing both wells outside the separatrix. The portrait with two centres, a saddle at the origin and a separatrix is the one that matches; portraits made of concentric closed curves around a single centre describe a single-well (harmonic-type) potential, and straight lines crossing at the origin describe only an inverted-oscillator saddle.

Hence, the answer is Phase portrait (B) — refer to the image in the original question.

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    Codes:

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    (ii) Assembly language

    (iii) High level language

    (iv) Natural language

    Codes:

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