If 'm' be the mass of the particle executing SHM and v its instantaneous velocity.
Then, kinetic energy `K = 1/2 mv^2`

`= 1/2 m [a omega cos ( omega t + phi )]^2`

`=1/2 ma^2 omega^2 cos^2 (omegat +phi)...................(1)`

The above equation expresses the kinetic energy in terms of time t.

The K.E. of the particle can also be expressed in terms of displacement `x`.

`K = 1/2m [ omega^2 (a^2 - x^2)] .... (2)`

The above equation reveal that the velocity (and hence the kinetic energy) keep on changing with time/displacement and is maximum at the mean position and minimum at extreme position.

If `alpha` be the acceleration in terms of displacement `x`, then we have

`alpha= -omega^2 x`

Multiplying both sides by mass 'm' of the particle executing SHM, we get

Force `vecF= -(omega^2m)vecx`

Since the force acting is a variable one, let us find the work done by the force, between the displacement `y` to `y + dy` (where `dy` is the elemental change in displacement)

`dW = vecF*vec(dy) =- momega^2ydy`

Total work done by the force between displacement `y = 0` to `y = x` , can obtained by integrating

`W = int_0^x -m omega^2 y dy = - m omega^2 x^2/2`

We know that, the increase in potential energy from `y = 0` to `y = x` , is the negative of the work done by the forces

`U(x) - U(0) = -W = 1/2 m omega^2 x^2`

Where `U(x)` denotes the P.E. of the system at a displacement x for the particle.

Taking `U(0)` to be zero, we get

`U (x) = 1/2 m omega^2 x^2 .... (2)`
For equation (4), the displacement 'x' can
substituted in terms of `t`

`U = 1/2 ma^2 omega^2 sin^2 ( omegat + phi) .... (3)`

The above equation reveal that the potential energy of the system, varies with time/displacement with minimum (zero, assume value) at the mean position and maximum `(1/2 ma^2 omega^2)` at the extreme position.


Total mechanical energy (E) can be obtain by adding `K` and `U` both in term of time (t) or both in term of displacement (x).

`E = K + U = 1/2 ma^2 omega^2 ..... (4)`

The above equation reveals that the total energy `(E)` is independent of time or position of the particle and is always constant.

In deriving the above expression for P.E or T.E at mean position is assumed to be zero. However if the system has
some non zero P.E.
At mean position, It should be taken into account to determine P.E. and T.E.

The expression for K.E. is independent of the choice of P.E. at mean position. Figure below shows the graphical variation of

K.E., P.E. and T.E. with displacement.

The curves of `K` and `U` intersect at `x` given by

`1/2 m omega^2(a^2-x^2) = 1/2 m omega^2 x^2`

`2x^2 = a^2=>x = 0.707 a`

and the corresponding `K` or `U` will be

`K = U = 1/2 [E] = 1/2 m omega^2 a^2`