A particle executes SHM of amplitude A. At what distance from the mean position is its KE equal to its PE ?

Options

A. 0.71 A
B. 0.61 A
C. 0.65 A
D. 0.8 A

Explanation

To solve the problem of finding the distance from the mean position where the kinetic energy (KE) is equal to the potential energy (PE) for a particle executing simple harmonic motion (SHM) with amplitude A, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formulas for KE and PE in SHM**:
- The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- The potential energy (PE) of a particle in SHM is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

2. **Set KE Equal to PE**:
- We need to find the position \( x \) where \( KE = PE \):
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

3. **Eliminate Common Factors**:
- We can cancel \( \frac{1}{2} m \omega^2 \) from both sides (assuming \( m \) and \( \omega \) are not zero):
\[ A^2 - x^2 = x^2 \]

4. **Rearrange the Equation**:
- Rearranging gives us:
\[ A^2 = 2x^2 \]

5. **Solve for \( x^2 \)**:
- Dividing both sides by 2:
\[ x^2 = \frac{A^2}{2} \]

6. **Take the Square Root**:
- Taking the square root of both sides gives:
\[ x = \frac{A}{\sqrt{2}} \]

### Final Answer:
The distance from the mean position where the kinetic energy is equal to the potential energy is:
\[ x = \frac{A}{\sqrt{2}} \]