Consider a particle of mass m, having linear simple harmonic motion with amplitude a and constant angular frequency ω. Suppose t seconds after starting from the mean position, the displacement of the particle is given by y = a sin ωt

Velocity of the particle at an instant t is v = dy/dt = a ω cos ωt

Time period – We know A = ω²y Hence, ω = (A/y)^1/2

T = 2 π / ω = 2 π (y/A)^1/2

⁼ 2 π (Displacement / Accelaration)^1/2

Kinetic Energy – KE = ½ mv² = ½ m (a ω cos ωt)²

                                                    = ½ m a² ω² cos² ωt

= ½ m a² ω² (1 - sin² ωt)

= ½ m a² ω² (1 – y² / a² )

= ½ m ω² (a² – y² )

Case 1 At an instant when t = 0, the particle is at the mean position, so y = 0, Hence, KE = ½ m ω² a²

Case 2 At an instant when t = T/4, the particle is at the extreme position, so y = a, Hence, KE = 0

Hence, when KE completes 1 vibration, the SHM completes half the vibration ; The count of vibrations is frequency and inverse of frequency is time period

This proves that the time period for Kinetic energy is time period of simple harmonic motion by 2 which is T/2