The total energy of a particle, executing simple harmonic motion is.
where x is the displacement from the mean position, hence total energy is independent of x.

Options

A. proportional to X
B. proportional to `x^(1//2)`
C. independent of X
D. proportional to `x^(2)`

Explanation

To solve the problem regarding the total energy of a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation of Motion**:
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x(t) = A \cos(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Potential Energy in SHM**:
The potential energy \( U \) of a particle in SHM is given by:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant. Substituting for \( x \):
\[ U = \frac{1}{2} k (A \cos(\omega t + \phi))^2 = \frac{1}{2} k A^2 \cos^2(\omega t + \phi) \]

3. **Kinetic Energy in SHM**:
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} mv^2 \]
where \( v \) is the velocity. The velocity can be found by differentiating the displacement:
\[ v = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
Therefore, the kinetic energy becomes:
\[ K = \frac{1}{2} m (-A \omega \sin(\omega t + \phi))^2 = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) \]

4. **Total Energy in SHM**:
The total mechanical energy \( E \) in SHM is the sum of kinetic and potential energy:
\[ E = K + U \]
Substituting the expressions for \( K \) and \( U \):
\[ E = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) + \frac{1}{2} k A^2 \cos^2(\omega t + \phi) \]

5. **Using the Relation Between \( k \) and \( \omega \)**:
Recall that \( k = m \omega^2 \). Thus, we can rewrite the potential energy term:
\[ U = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \phi) \]
Now substituting back into the total energy equation:
\[ E = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) + \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi) \]

6. **Simplifying the Total Energy**:
Factor out \( \frac{1}{2} m A^2 \omega^2 \):
\[ E = \frac{1}{2} m A^2 \omega^2 (\sin^2(\omega t + \phi) + \cos^2(\omega t + \phi)) \]
Since \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ E = \frac{1}{2} m A^2 \omega^2 \]

7. **Conclusion**:
The total energy \( E \) of a particle executing simple harmonic motion is:
\[ E = \frac{1}{2} k A^2 \]
This shows that the total energy is independent of the displacement \( x \) and depends only on the amplitude \( A \) and the spring constant \( k \).