A particle executes simple harmonic motion with an amplitude A . If the total mechanical energy of the oscillator is 40 J, what is its kinetic energy when the displacement of the particle is half of its amplitude?

Step-by-step solution

Total mechanical energy of the particle in SHM is given by E = 1 2 k A^2 = 40 J. The potential energy U at a displacement x is U = 1 2 k x^2 . At x = A 2 , the potential energy is: U = 1 2 k ( A 2 )^2 = 1 4 ( 1 2 k A^2 ) U = E 4 = 40 4 = 10 J. According to the law of conservation of energy, the kinetic energy K at this position is: K = E - U K = 40 - 10 = 30 J. Answer: 30 J