InterviewSolution
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InterviewSolution
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The potential energy U of a body of unit mass moving in one dimensional conservative force field is given by `U=x^2-4x+3`. All units are is SI. For this situation mark out the correct statement (s).A. The body will perform simple harmonic motion about `x=2` units.B. The body will perform oscillatory motion but not simple harmonic motion.C. The body will perform simple harmonic motion with time period `sqrt2pis`.D. If speed of the body at equilibrium position is `4(m)/(s)`, then the amplitude of oscillation would be `2sqrt2`m |
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Answer» Correct Answer - A::C::D `U=x^2-4x+3` and `F=-(dU)/(dx)=-(2x-4)` At equilibrium position `F=0`, so `x=2cm` Let the particle is displaced by `trianglex` from equilibrium position i.e, from `x=2`, then restoring force on body is `F=-2(2+trianglex)+4=-2trianglex` i.e., `Falpha-trianglex`, so performs simple harmonic motion about `x=2m` Time period `T=2pisqrt((m)/(k))=2pisqrt((1)/(2))=sqrt2pis` From energy conservation, `(mv_(max)^2)/(2)+U_(min)=U_(max)` `(1xx4^2)/(2)+(2^2-4xx2+3)=(A+2)^2-4(A+2)+3` Where A is amplitude solving the above equation we get `A=2sqrt2m`. |
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