In SHM , potential energy of a particle at mean position is `E_(1)` and kinetic enregy is `E_(2)` , then

Options

A. `E_(1)=E_(2)`
B. total potential energy at `x=(sqrt(3)A)/(2)` is `E_(1)+(3E_(2))/(4)`
C. total kinetic energy at `x=(sqrt(3)A)/(2)` is `(3E_(2))/(4)`
D. total kinetic energy at `x=(A)/(sqrt(2))` is `(E_(2))/(4)`

Explanation

To solve the problem, we need to analyze the potential energy (U) and kinetic energy (K) of a particle in Simple Harmonic Motion (SHM) at the mean position.

### Step-by-Step Solution:

1. **Understanding SHM Energy Components**:
- In SHM, the total mechanical energy (E) of the system is conserved and is the sum of kinetic energy (K) and potential energy (U).
- At the mean position (equilibrium position), the displacement (x) is zero.

2. **Potential Energy at Mean Position**:
- The potential energy (U) at the mean position is given as \( U = 0 \) because potential energy in SHM is calculated using the formula \( U = \frac{1}{2} k x^2 \) and at the mean position, \( x = 0 \).
- Therefore, \( U_{\text{mean}} = E_1 = 0 \).

3. **Kinetic Energy at Mean Position**:
- The total energy (E) in SHM is given by \( E = K + U \).
- Since \( U = 0 \) at the mean position, the total energy is entirely kinetic.
- Thus, \( K = E \) at the mean position.

4. **Maximum Kinetic Energy**:
- The maximum kinetic energy (K_max) occurs at the mean position and is equal to the total energy of the system.
- Therefore, \( K = E_2 \) at the mean position.

5. **Conclusion**:
- Given that \( E_1 = 0 \) (potential energy at mean position) and \( E_2 = K_{\text{max}} \), we can conclude that the kinetic energy at the mean position is equal to the total energy of the system.

### Final Answer:
- At the mean position, the potential energy \( E_1 = 0 \) and the kinetic energy \( E_2 \) is equal to the total energy of the system.

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