Let m be the mass of a particle executing simple harmonic motion. A is the amplitude and T is the time period of the motion. The particle possesses kinetic energy due to its velocity all along its path of motion except at the extremities. Again restoring force acts on the particle all along its path of motion except at the position of equilibrium. So, the work that is to be done to move the particle against the storing force remains stored in the particle as potential energy.

## Kinetic Energy of Simple Harmonic Motion:

At a displacement x from the position of equilibrium,

the velocity of the particle v = ω√(A^{2} – x^{2})

So, the kinetic energy of the particle of mass m at that instant,

K = 1/2 mv

^{2}= 1/2 mω^{2}(A^{2}– x^{2})

So, **the kinetic energy of a particle executing SHM is maximum at the position of equilibrium and zero at the two ends, at the two extremities of its path of motion**.

## Potential Energy of Simple Harmonic Motion:

When the particle is at a distance x from its position of equilibrium, the restoring force acting on it is, F = mω^{2}x.

Again, when the particle is just at the position of equilibrium, x = 0, the restoring force F = 0, and no force acts on the particle.

So, within the displacement from 0 to x, the average force acting on the particle = (0 + mω^{2}x)/2 = (1/2)mω^{2}x

Now, the potential energy of the particle, U = work done to move the particle through the distance x against this average force = average force x displacement acting on the particle.

potential energy (U) = (1/2)mω

^{2}x.x

= (1/2)mω^{2}x^{2}