Aparticle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :

Options

A. `sqrt2A`
B. `(1)/(sqrt2)A`
C. `2A`
D. `(1)/(2)A`

Answer

To solve the problem, we need to find the distance from the mean position when the kinetic energy (KE) of a particle executing simple harmonic motion (SHM) is equal to its potential energy (PE).

### Step-by-Step Solution:

1. **Understanding the Energy in SHM**:
In simple harmonic motion, the total mechanical energy (E) is constant and is the sum of kinetic energy (KE) and potential energy (PE). The expressions for KE and PE are:
- Kinetic Energy (KE) = \( \frac{1}{2} m v^2 \)
- Potential Energy (PE) = \( \frac{1}{2} k x^2 \)
where \( m \) is the mass, \( v \) is the velocity, \( k \) is the spring constant, and \( x \) is the displacement from the mean position.

2. **Relating KE and PE**:
We need to find the position \( x \) where KE = PE. Therefore, we set:
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]

3. **Using the Relation between \( v \) and \( x \)**:
The velocity \( v \) in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

4. **Substituting \( v \) into the KE Expression**:
Substituting \( v \) into the kinetic energy equation gives:
\[ KE = \frac{1}{2} m (\omega^2 (A^2 - x^2)) \]

5. **Setting KE Equal to PE**:
Now we can write the equation:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} k x^2 \]

6. **Using the Relation \( k = m \omega^2 \)**:
Since \( k = m \omega^2 \), we can substitute \( k \) into the equation:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

7. **Canceling Common Terms**:
We can cancel \( \frac{1}{2} m \omega^2 \) from both sides (assuming they are not zero):
\[ A^2 - x^2 = x^2 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ A^2 = 2x^2 \]

9. **Solving for \( x \)**:
Dividing both sides by 2:
\[ x^2 = \frac{A^2}{2} \]
Taking the square root:
\[ x = \frac{A}{\sqrt{2}} \]

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