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`.