From the variation of potential energy in the direction of small oscillation of a simple pendulum, find the effective spring constant for the simple pendulum, where m is mass of the bob, l is length of the simple pendulum. 

detailed solution

Correct option is A

When a particle is executing SHM, at mean positionF=−dUdr=0. Also, its potential energy is minimum at equilibrium position. So, d2Udr2>0dUdr=−F⇒d2Udr2=−dFdr At equilibrium, dF=−d2Udr2r=0drComparing this standard SHM equation, i.e. At equilibrium, dF=−d2Udr2r=0drIn general equilibrium happens at r=r0( say )∴ keff=d2Udr2r=r0Also, at equilibrium F=−dUdr=0Now, coming to the situation given, when the bob is given a small horizontal displacement as shown.U=mgyy=l(1−cos⁡θ)=2lsin2⁡θ2∴ U=2mglsin2⁡θ2 As θ is small sin⁡θ2≃θ2So  U=2mglθ2⋅θ2 or U=12mglθ2θ≃xl⇒U=mg2lx2dUdx=mg2l⋅2x=mgl⋅x⇒d2Udx2=mgl ∴ keff=d2Udx2x=0=mgl