A particle performing `SHM` with amplitude `10cm`. At What distance from mean position the kinetic energy of the particle is thrice of its potential energy ?

Explanation

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

### 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) is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
where:
- \( m \) is the mass of the particle,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the distance from the mean position.

2. **Set up the equation**:
- We are given that the kinetic energy is three times the potential energy:
\[ KE = 3 \times PE \]
- Substituting the formulas for KE and PE, we have:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = 3 \times \frac{1}{2} m \omega^2 x^2 \]

3. **Simplify the equation**:
- Cancel out \(\frac{1}{2} m \omega^2\) from both sides (assuming \( m \) and \( \omega \) are not zero):
\[ A^2 - x^2 = 3x^2 \]
- Rearranging gives:
\[ A^2 = 4x^2 \]

4. **Substitute the value of amplitude**:
- Given that the amplitude \( A = 10 \, \text{cm} \):
\[ (10 \, \text{cm})^2 = 4x^2 \]
- This simplifies to:
\[ 100 \, \text{cm}^2 = 4x^2 \]

5. **Solve for \( x^2 \)**:
- Dividing both sides by 4:
\[ x^2 = \frac{100 \, \text{cm}^2}{4} = 25 \, \text{cm}^2 \]

6. **Find the value of \( x \)**:
- Taking the square root of both sides:
\[ x = \sqrt{25 \, \text{cm}^2} = 5 \, \text{cm} \]

### Final Answer:
The distance from the mean position where the kinetic energy is three times the potential energy is **5 cm**.