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$\text{Hint: } U_{\text{max}} = K_0 \text{ at the mean position.}$ $\text{Step: Let's analyze each option one by one:}$ $\text{The kinetic energy } K \text{ is given as: } K = K_0 \cos^2 \omega t$ $\text{Maybe } K_0\text{:}$ $\text{The maximum value of kinetic energy is } K = K_0 \times 1 \text{ (when } \cos^2 \omega t = 1\text{)}$ $U_{\text{max}} = K_0$ $\text{Therefore, the potential energy may equal to } K_0.$ $\text{Must be } K_0\text{: This is not true because, in SHM, the maximum potential energy is not always equal to } K_0, \text{ because it depends on the } \cos^2 \omega t.$ $K = K_0 \cos^2 \omega t$ $\text{Here, the } \cos^2 \omega t \text{ may lie between the value } [0, 1]$ $\text{"Maybe more than } K_0\text{": When we change the reference point for potential energy (raising or lowering it) we are effectively adding or subtracting a constant from the potential energy. The equation for potential energy becomes:}$ $U' = U + U_{\text{shift}}$ $\text{the potential energy can exceed the initial } K_0. \text{ These are cases where the energy available to the system grows beyond its original configuration, allowing for } U_{\text{max}} \text{ to be greater than } K_0.$ $\text{Hence, option (4) is the correct answer.}$