The particle executing simple harmonic motion has a kinetic energy `K_(0) cos^(2) omega t`. The maximum values of the potential energy and the energy are respectively

Options

A. 0 and `2K_(0)`
B. `(K_(0))/(2)` and `K_(0)`
C. `K_(0)` and `2K_(0)`
D. `K_(0)` and `K_(0)`

Answer

To solve the problem, we need to find the maximum values of potential energy and total energy for a particle executing simple harmonic motion (SHM) with the given kinetic energy expression \( K(t) = K_0 \cos^2(\omega t) \).

### Step-by-step Solution:

1. **Understanding Kinetic Energy in SHM**:
The kinetic energy \( K(t) \) of the particle is given by:
\[ K(t) = K_0 \cos^2(\omega t) \]
The maximum value of \( \cos^2(\omega t) \) is 1, which occurs when \( \cos(\omega t) = 1 \).

2. **Finding Maximum Kinetic Energy**:
The maximum kinetic energy \( K_{\text{max}} \) can be calculated as:
\[ K_{\text{max}} = K_0 \cdot 1^2 = K_0 \]

3. **Total Energy in SHM**:
In simple harmonic motion, the total mechanical energy \( E \) is constant and is the sum of kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
At the mean position (where the kinetic energy is maximum), the potential energy is zero. Therefore, the total energy at this point is simply:
\[ E = K_{\text{max}} + U_{\text{min}} = K_0 + 0 = K_0 \]

4. **Finding Maximum Potential Energy**:
At the extreme positions of SHM, the kinetic energy is zero, and all the energy is potential. Thus, the maximum potential energy \( U_{\text{max}} \) is equal to the total energy:
\[ U_{\text{max}} = E = K_0 \]

5. **Conclusion**:
The maximum values of potential energy and total energy are both equal to \( K_0 \).

### Final Answer:
- Maximum Potential Energy \( U_{\text{max}} = K_0 \)
- Total Energy \( E = K_0 \)