For a system executing simple harmonic motion,

total mechanical energy is conserved.

Let the displacement of the particle from the mean position be $x$ and the amplitude be $A$.

The kinetic energy of the system is given by:

$K = \dfrac{1}{2}m\omega^2(A^2 - x^2)$

Assuming the potential energy to be zero at the mean position, the potential energy is given by:

$U = \dfrac{1}{2}m\omega^2 x^2$

At the mean position ($x = 0$):

K_\max = \dfrac{1}{2}m\omega^2 A^2

U_\min = 0

At the extreme positions ($x = \pm A$):

K_\min = 0

U_\max = \dfrac{1}{2}m\omega^2 A^2

Comparing the maximum and minimum values, we get:

U_\max = K_\max = \dfrac{1}{2}m\omega^2 A^2

U_\min = K_\min = 0

Thus, the maximum potential energy is equal to the maximum kinetic energy, and the minimum potential energy is equal to the minimum kinetic energy.