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← IIT-JEE 2010 · Paper 1
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IIT-JEE 2010 · Paper 1, Physics question 70

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Paragraph for Questions 70 to 72

When a particle of mass m moves on the x-axis in a potential of the form V(x)=kx2V(x)=kx^2, it performs simple harmonic motion. The corresponding time period is proportional to mk\sqrt{\dfrac mk}, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x=0x=0 in a way different from kx2kx^2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x)=αx4V(x)=\alpha x^4 (α>0\alpha>0) for ∣x∣|x| near the origin and becomes a constant equal to V0V_0 for ∣x∣≥X0|x|\ge X_0 (see figure).

V(x) against x: zero at the origin, rising smoothly on both sides (like αx⁴) and levelling off at the constant V₀ for |x| ≥ X₀.

If the total energy of the particle is E, it will perform periodic motion only if

Choose one option
Answer and worked solution

Official answer: B

Bounded motion. The potential is 0 at the origin and rises to V0V_0 for ∣x∣≥X0|x|\ge X_0, where it is flat. The particle oscillates between turning points only if

  • E>0E>0 (otherwise it cannot move at all; E=0E=0 means rest at the origin), and
  • E<V0E<V_0 (otherwise there are no turning points and it moves off to infinity with constant speed).

So periodic motion needs V0>E>0V_0>E>0. The official key also accepts (B), since E>0E>0 is a necessary condition.

(C) (key: B or C)