If A is amplitude of a particle in SHM, its displacement from the mean position when its kinetic energy is thrice that to its potential energy

Answer

To solve the problem, we need to find the displacement \( x \) from the mean position when the kinetic energy \( KE \) of a particle in simple harmonic motion (SHM) is three times its potential energy \( PE \).

### Step-by-Step Solution:

1. **Understanding the Energy Relationships in SHM**:
- The total mechanical energy \( E \) in SHM is given by:
\[ E = KE + PE \]
- The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- The potential energy \( PE \) is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

2. **Setting Up the Equation**:
- According to the problem, we know that:
\[ KE = 3 \times PE \]
- Substituting the expressions for \( KE \) and \( PE \):
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = 3 \left( \frac{1}{2} m \omega^2 x^2 \right) \]

3. **Simplifying the Equation**:
- Cancel out the common terms \( \frac{1}{2} m \omega^2 \) from both sides:
\[ A^2 - x^2 = 3x^2 \]
- Rearranging gives:
\[ A^2 = 4x^2 \]

4. **Finding the Displacement**:
- Solving for \( x^2 \):
\[ x^2 = \frac{A^2}{4} \]
- Taking the square root of both sides:
\[ x = \frac{A}{2} \]

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