Derive the expression for total energy of a particle executing simple harmonic motion (SHM).

Explanation

Let the instantaneous displacement of the particle executing `S.H.mu" be ' x = A cos (phi t + phi)`
Potential energy `P.E = 1/2 kx^2`
`=1/2 k (A cos (omega t + phi))^2 = 1/2 KA^2 cos^2 (omega t + phi)`
Kinetic energy ` K.E = 1/2 mv^2`
` = 1/2 m ( - omega A sin (omega t + phi))^2`
` = 1/2 m omega^2 A^2 sin^2 (omega t + phi)`
` = 1/2 KA^2 sin^2 (omega t + phi) "where" k = m omega^2`
Total energy E = P.E + K.E
` = 1/2A^2 [cos^2 (omega t + phi) + sin^2 (omega t + phi) ]`
` = 1/2 KA^2 (1)`
` therefore E = 1/2 KA^2`