Energy in Simple Harmonic Motion:

A particle executing SHM possesses two types of energy if a particle executes SHM, its kinetic energy changes into potential energy and vice-versa keeping total energy constant (if friction of air is neglected).

Kinetic energy:  

This is an account of the velocity of the particle.

K = ½ mv²

= ½ mA² ω²cos2 (ωt + φ) = ½ m ω² (A² –X²)

From this expression, we conclude that kinetic energy is maximum at the centre(x = ±A).

Potential energy:  

This is an account of the displacement of the particle from its mean position.

U = ½ m ω²x² = ½ mA² ω²sin² (ωt + φ)

Thus, potential energy has its minimum at the centre (x = 0) and increases as the   particle approaches either extreme of oscillation (x = ±A).

Thus, total energy = kinetic energy + potential energy

Or E = ½ m ω²A²

Obviously, the total energy is constant and is proportional to the square of amplitude (A) of motion. Figures show the variations of total energy (E), potential energy (U) and kinetic energy (K) with displacement(x).

Example: 

A body is executing simple harmonic motion at a displacement x its potential energy is E₁ and at a displacement y it potential energy is E₂. The potential energy (E) at displacement (x + y) is

Solution:

Potential energy of a body executing SHM

U = ½ m ω²y²

U₁ = ½ mω²y²

And U₂ = ½ mω²y²

At displacement (x + y)

U = ½ mω²(x + y) ²

√U₁ + √U₂ = √U

i.e., √E = √E₁ + √E₂