Question

A particle is executing SHM linearly with an amplitude a the displacement of the particle from its mean position when it's potential energy is 1/3 of its maximum energy is what?

Answer 1

amplitude of SHM is A
      we know, general equation of SHM is y = Asin(ωt +_ Ф)
so, work done = F.dy
                    = -mω²y.dy cos180° [F = ma and a = -ω²y in SHM ]
                    = mω²y²/2 
we know, work done stored in system named as potential energy of system. 
hence, P.E = W = mω²y²/2 
            for maximum energy, y = A
hence, maximum energy = mω²A²/2    
                A/C to question,
 potential energy = maximum energy/3
  mω²y²/2 = mω²A²/2 × 3 
       y²  = A²/3 
      taking square root both sides,
  y = +_A/√3 
Hence, the displacement of particle from the mean position  is A/√3

Answer 2

Given :
let A be the amplitude.
Potential energy at displacement x from mean position is given by:

P.E=1/2 kx²

let total energy be E=1/2kA²

1/2kx²=E/3=1/6kA² [ According to given]

x²=1/3 A²x=A/√3

∴The displacement of particle from its mean position is A/√3