A particle performing S.H.M. starts horn mean position. The position of particle from its mean position where the difference of K E and P E is zero is (Take maximum displacement of particle is 8 m)

Options

A. 2 m
B. `2 sqrt 2 m`
C. `4 sqrt 2 m`
D. `sqrt 2 m`

Answer

To solve the problem, we need to find the position \( x \) of a particle performing Simple Harmonic Motion (SHM) at which the difference between its kinetic energy (K.E.) and potential energy (P.E.) is zero. Given that the maximum displacement (amplitude) \( A \) is 8 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energies in SHM**:
- The kinetic energy (K.E.) in SHM is given by:
\[ K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- The potential energy (P.E.) in SHM is given by:
\[ P.E. = \frac{1}{2} m \omega^2 x^2 \]

2. **Setting Up the Equation**:
- We need to find the position \( x \) where the difference between K.E. and P.E. is zero:
\[ K.E. - P.E. = 0 \]
- This implies:
\[ K.E. = P.E. \]

3. **Substituting the Energy Expressions**:
- From the above equations, we can write:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

4. **Simplifying the Equation**:
- Canceling \( \frac{1}{2} m \omega^2 \) from both sides (assuming \( m \) and \( \omega \) are not zero):
\[ A^2 - x^2 = x^2 \]
- Rearranging gives:
\[ A^2 = 2x^2 \]

5. **Substituting the Given Amplitude**:
- We know \( A = 8 \) m, so substituting this value:
\[ 8^2 = 2x^2 \]
\[ 64 = 2x^2 \]

6. **Solving for \( x^2 \)**:
- Dividing both sides by 2:
\[ x^2 = 32 \]

7. **Finding \( x \)**:
- Taking the square root of both sides:
\[ x = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \text{ m} \]

### Final Answer:
The position of the particle from its mean position where the difference of kinetic energy and potential energy is zero is:
\[ x = 4\sqrt{2} \text{ m} \]