There are two types of energy in a particle executing simple harmonic motion.

• Potential energy which is due to the displacement of the particle from the mean position
• Kinetic energy which is due to the velocity of the particle

At any instance, the total energy of the particle is equal to the sum of the kinetic energy and the potential energy.

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

Accelaration of the particle at an instant t is v = d²y/dt² = -a ω² sin ωt = - ω²y

The negative sign shows that it is directed towards the mean position

Potential energy (PE):

Restoring force F = mass * acceleration = m * (- ω²y) = -ky where k = spring factor of SHM = m ω²

Work done for the additional displacement dy against the restoring force dW = -F dy = - (-ky) dy = k dy

Total work done in displacing the particle from the mean position to the position of displacement y = W = ∫ ky dy within the limits 0 to y = ½ ky²

Potential energy = PE = ½ ky² = ½ m ω²y² = ½ m ω²a²sin²ωt

Kinetic Energy (KE):

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 )

Total Energy:

Total energy of a particle at an instant t = PE + KE = ½ m ω²y² + ½ m ω² (a² – y² ) = ½ m ω² a²

Where m. a, ω are all constants. Hence, total energy of the system remains constant at all times

As ω = 2∏f, Total energy T = ½ m (2∏f) ² a² = 2m ∏²f² a² which shows the total energy is proportional to square of amplitude, a and frequency, f.