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 find the kinetic energy of a particle in Simple Harmonic Motion (SHM) when it is at half the amplitude from the mean position, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Total Energy in SHM**:
The total mechanical energy (E) 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. **Identify the Position**:
We need to find the kinetic energy when the particle is at half the amplitude from the mean position. This means:
\[ x = \frac{A}{2} \]

3. **Use the Kinetic Energy Formula**:
The kinetic energy (K.E) at a position \( x \) in SHM is given by:
\[ K.E = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

4. **Substitute the Position**:
Substitute \( x = \frac{A}{2} \) into the kinetic energy formula:
\[ K.E = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{A}{2}\right)^2\right) \]
Simplifying this gives:
\[ K.E = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right) \]
\[ K.E = \frac{1}{2} m \omega^2 \left(\frac{4A^2}{4} - \frac{A^2}{4}\right) \]
\[ K.E = \frac{1}{2} m \omega^2 \left(\frac{3A^2}{4}\right) \]
\[ K.E = \frac{3}{8} m \omega^2 A^2 \]

5. **Relate Kinetic Energy to Total Energy**:
We know from the total energy formula that:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Therefore, we can express \( K.E \) in terms of \( E \):
\[ K.E = \frac{3}{8} m \omega^2 A^2 = \frac{3}{4} \left(\frac{1}{2} m \omega^2 A^2\right) = \frac{3}{4} E \]

6. **Final Answer**:
Thus, the kinetic energy of the particle at half the amplitude from the mean position is:
\[ K.E = \frac{3}{4} E \]

### Conclusion:
The kinetic energy at half the amplitude from the mean position is \( \frac{3}{4} E \).