The particle executing simple harmonic motion has a kinetic energy of \[{K_0}co{s^2}\omega t\]. The maximum values of the potential energy and the total energy are respectively
A) ${K_0}$ and ${K_0}$
B) 0 and ${K_0}$
C) ${K_0}/2$ and ${K_0}$
D) ${K_0}$ and $2{K_0}$

Hint: The law of conservation of energy states that the total energy of the system will remain constant. For simple harmonic motion, the kinetic energy and the potential energy exchange into each other depending on the displacement of the particle from the mean position.

Complete step by step answer:
We’ve been given that the particle executing an SHM has a kinetic energy of \[{K_0}co{s^2}\omega t\]. The maximum value of the kinetic energy of the SHM will hence correspond to $\theta = 0^\circ $ which will be
$E = {K_0}$. The particle will have maximum kinetic energy when it is passing through the mean position of the oscillation.
At this position, the particle will only have kinetic energy and no potential energy since the particle is at the mean position. Hence the total energy of the particle will be due to kinetic energy only and it will be ${K_0}$.
The total energy of the particle will remain constant due to the law of conservation of energy. Hence the maximum value of the total energy will be ${K_0}$ and will stay constant in the entire motion of the SHM.
The potential energy of the particle will be maximum when the particle is at a distance equal to the amplitude of the SHM from the mean position. At this point, the total energy of the SHM will only be due to potential energy. So, the maximum potential energy at this point will also be ${K_0}$ as the total energy will all be due to potential energy.

Hence option (A) is the correct choice.

Note: Here we must remember that the total energy of the system should remain constant. Then the fact that the energy of the Simple harmonic oscillator only interchanges its energy completely into one form to another will help us in answering the question.