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 mv ^{2}*

= *½ mA ^{2 }ω^{2}cos2* (

*ωt*+ φ) =

*½ m ω*(

^{2}*A*)

^{2 }–X^{2}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 ω ^{2}x^{2}* =

*½ mA*(

^{2}ω^{2}sin^{2}*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 ω ^{2}A^{2}*

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 *E _{1}* and at a displacement

*y*it potential energy is

*E*. The potential energy (

_{2}*E*) at displacement (

*x + y*) is

**Solution: **Potential energy of a body executing SHM

*U = ½ m ω ^{2}y^{2}*

*∴** U _{1} = ½ mω^{2}y^{2}*

And *U _{2} = ½ mω^{2}y^{2}*

At displacement (x + y)

U = ½ mω^{2}(x + y)^{ 2 }

∴ √U_{1} + √U_{2} = √U

i.e., √E = √E_{1} + √E_{2}