For a particle performing S.H.M., the total energy is ' n ' times the kinetic energy, when the displacement of a particle from mean position is 3 2 A, where A is the amplitude of S.H.M. The value of ' n ' is

Step-by-step solution

A particle in simple harmonic motion with amplitude A and displacement x has total energy E = 1 2 m ^2 A^2 and kinetic energy KE = 1 2 m ^2 (A^2 - x^2) . Given the displacement x = 3 2 A , substitute into the kinetic energy equation: KE = 1 2 m ^2 (A^2 - ( 3 2 A )^2 ) = 1 2 m ^2 (A^2 - 3 4 A^2 ) = 1 2 m ^2 ( 1 4 A^2 ) = 1 8 m ^2 A^2 Since E = 1 2 m ^2 A^2 = 4 1 8 m ^2 A^2 , the total energy is exactly four times the kinetic energy at this displacement. The coefficient n such that E = n KE is therefore 4 .