1. In simple harmonic motion (SHM), the displacement from mean position is denoted by $x$.

2. The potential energy of the oscillator is $U = \frac{1}{2}kx^2$ where $k$ is the force constant.

3. At the mean position, $x = 0$ so the potential energy becomes $U = \frac{1}{2}k(0)^2 = 0$ Hence, potential energy is minimum.

4. The total mechanical energy in SHM is constant: $E = \frac{1}{2}kA^2$ where $A$ is the amplitude.

5. Kinetic energy is given by $K = E - U$

   At the mean position, since $U=0$, $K = E$ which is the maximum possible value of kinetic energy.

6. Therefore, at the mean position:

   • kinetic energy is maximum
   • potential energy is minimum

7. Checking options:

   • A: kinetic minimum, potential maximum $\rightarrow$ incorrect
   • B: both maximum $\rightarrow$ incorrect
   • C: kinetic maximum, potential minimum $\rightarrow$ correct
   • D: both minimum $\rightarrow$ incorrect

Hence, the correct answer is C.