The potential energy U of a body of unit mass moving in one dimensional conservative force field is given by U=x2−4x+3. All units are in SI. For this situation mark out the correct statement(s).

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The body will perform simple harmonic motion about x = 2 units.

The body will perform oscillatory motion but not simple harmonic motion.

The body will perform simple harmonic motion with time period 2πs.

If speed of the body at equilibrium position is 4 m/s, then the amplitude of oscillation would be 22m.

answer is A.

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Detailed Solution

U=x2−4x+3 and F=−dUdx=−(2x−4)At equilibrium position F = 0, so x = 2 m.Let the particle is displaced by Δx from equilibrium position, i.e. from x = 2, then restoring force on body is, F=−2(2+Δx)+4=−2Δx i.e., F ∝−Δx, so performs simple harmonic motion about x = 2 m.Time period, T=2πmk=2π12=2πsFrom energy conservation, mvmax22+Umin=Umax1×422+22−4×2+3=(A+2)2−4(A+2)+3where A is amplitude.Solving the above equation, we get A=22m

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