Question

A particle of mass m is placed at lowest point of smooth

Answer 1

Answer:

Method I :-

⋆⋆ First let us prove that this a Simple Harmonic Motion for a small displacement :-

(Okay I am sorry for bad handwriting)

As seen in the above Free Body Diagram the restoring force is :-

⟹Frestoring=mgsinθ⟹Frestoring=mgsin⁡θ

⋆⋆ For small θθ, sinθ≈θ≈tanθsin⁡θ≈θ≈tan⁡θ :-

⟹tanθ=Slope=dydx=x2a⟹tan⁡θ=Slope=dydx=x2a

⋆⋆ Hence we get :-

⟹F=−mgx2a⟹F=−mgx2a

which is comparable to the equation of S.H.M. F=−kxF=−kx with k=mg2ak=mg2a

So this is an S.H.M and its Time Period(T)(T) will be given by :-

⟹T=2πmk−−−√⟹T=2πmk

⟹T=2π2ag−−−√⟹T=2π2ag

Method II :-

⋆⋆ If you already know that this is an S.H.M then find out the Potential Energy which will be :-

⟹P.E=mgy⟹P.E=mgy

⟹P.E=mgx24a⟹P.E=mgx24a

⋆⋆ Now comparing with the potential energy of a

standard S.H.M, P.E=12kx2P.E=12kx2 we get k=mg2ak=mg2a

⟹T=2π2ag−−−√⟹T=2π2ag