A particle undergoes simple harmonic motion having an amplitude of 10 cm . At what distance from the mean position will its kinetic and potential energies be equal?

Step-by-step solution

Let the distance from the mean position be x . The kinetic energy of a particle in SHM is given by K = 1 2 m ^2 (A^2 - x^2) . The potential energy is given by U = 1 2 m ^2 x^2 . Given that kinetic and potential energies are equal, K = U . 1 2 m ^2 (A^2 - x^2) = 1 2 m ^2 x^2 A^2 - x^2 = x^2 2x^2 = A^2 x = A 2 Substituting the amplitude A = 10 cm : x = 10 2 = 5 2 cm