A particle executes SHM on a line 8 cm long . Its KE and PE will be equal when its distance from the mean position is

Options

A. 4 cm
B. 2 cm
C. `2sqrt(2)` cm
D. `sqrt(2)` cm

Explanation

To solve the problem of when the kinetic energy (KE) and potential energy (PE) of a particle executing simple harmonic motion (SHM) are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between KE and PE in SHM**:
- The total mechanical energy (E) in SHM is given by the sum of kinetic energy and potential energy:
\[ E = KE + PE \]
- The kinetic energy (KE) at a displacement \( x \) from the mean position is:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- The potential energy (PE) at a displacement \( x \) is:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

2. **Set KE equal to PE**:
- For the energies to be equal, we set:
\[ KE = PE \]
- This gives us:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

3. **Cancel common terms**:
- 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:
\[ A^2 = 2x^2 \]
- This can be rewritten as:
\[ x^2 = \frac{A^2}{2} \]

5. **Solve for \( x \)**:
- Taking the square root of both sides:
\[ x = \frac{A}{\sqrt{2}} \]

6. **Substitute the amplitude**:
- The problem states that the length of the line (which is the total distance of motion) is 8 cm. Therefore, the amplitude \( A \) is:
\[ A = \frac{8 \text{ cm}}{2} = 4 \text{ cm} \]

7. **Calculate the distance from the mean position**:
- Now substitute \( A \) into the equation for \( x \):
\[ x = \frac{4 \text{ cm}}{\sqrt{2}} = 4 \text{ cm} \cdot \frac{1}{\sqrt{2}} = 2\sqrt{2} \text{ cm} \]

### Final Answer:
The distance from the mean position when the kinetic energy and potential energy are equal is \( 2\sqrt{2} \text{ cm} \).

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