The total energy of a particle in SHM is E. Its kinetic energy at half the amplitude from mean position will be

Options

A. E/2
B. E/3
C. E/4
D. 3E/4

Explanation

To solve the problem, we need to determine the kinetic energy of a particle in Simple Harmonic Motion (SHM) when it is at a position that is half the amplitude from the mean position. Let's denote the amplitude as \( A \) and the total energy as \( E \).

### Step-by-Step Solution:

1. **Understanding Total Energy in SHM**:
The total energy \( E \) of a particle in SHM is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( m \) is the mass of the particle, \( \omega \) is the angular frequency, and \( A \) is the amplitude.

2. **Kinetic Energy Formula**:
The kinetic energy \( K \) of the particle at a position \( x \) in SHM is given by:
\[ K = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 \]
This equation states that the kinetic energy is equal to the total energy minus the potential energy at position \( x \).

3. **Substituting the Position**:
We need to find the kinetic energy when the particle is at \( x = \frac{A}{2} \):
\[ K = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 \]

4. **Calculating \( \left(\frac{A}{2}\right)^2 \)**:
Calculate \( \left(\frac{A}{2}\right)^2 \):
\[ \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]

5. **Substituting Back into the Kinetic Energy Equation**:
Substitute \( \frac{A^2}{4} \) into the kinetic energy equation:
\[ K = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 \frac{A^2}{4} \]

6. **Factoring Out Common Terms**:
Factor out \( \frac{1}{2} m \omega^2 \):
\[ K = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right) \]

7. **Simplifying the Expression**:
Simplify the expression inside the parentheses:
\[ A^2 - \frac{A^2}{4} = \frac{4A^2}{4} - \frac{A^2}{4} = \frac{3A^2}{4} \]
Thus, we have:
\[ K = \frac{1}{2} m \omega^2 \cdot \frac{3A^2}{4} \]

8. **Relating to Total Energy**:
Recall that the total energy \( E = \frac{1}{2} m \omega^2 A^2 \). We can express \( K \) in terms of \( E \):
\[ K = \frac{3}{4} \cdot \frac{1}{2} m \omega^2 A^2 = \frac{3}{4} E \]

### Final Answer:
The kinetic energy of the particle at half the amplitude from the mean position is:
\[ K = \frac{3E}{4} \]