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

Options

A. A
B. `A//4`
C. `A//2`
D. `3A//4`

Explanation

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 Relationship Between KE and PE**:
- We know that in SHM, the total mechanical energy \( E \) is the sum of kinetic energy and potential energy:
\[ E = KE + PE \]
- Given that the kinetic energy is three times the potential energy, we can express this as:
\[ KE = 3 \cdot PE \]

2. **Expressing Total Energy**:
- If we let \( PE = U \), then:
\[ KE = 3U \]
- Therefore, the total energy can be expressed as:
\[ E = KE + PE = 3U + U = 4U \]

3. **Using the Expression for Total Energy**:
- The total energy in SHM can also be expressed in terms of amplitude \( A \):
\[ E = \frac{1}{2} k A^2 \]
- Setting the two expressions for total energy equal gives us:
\[ 4U = \frac{1}{2} k A^2 \]

4. **Expressing Potential Energy**:
- The potential energy in SHM at displacement \( X \) is given by:
\[ PE = \frac{1}{2} k X^2 \]
- Substituting this into our equation gives:
\[ 4 \left(\frac{1}{2} k X^2\right) = \frac{1}{2} k A^2 \]
- Simplifying this, we have:
\[ 2 k X^2 = \frac{1}{2} k A^2 \]

5. **Canceling \( k \) and Solving for \( X^2 \)**:
- Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ 2 X^2 = \frac{1}{2} A^2 \]
- Multiplying both sides by 2 gives:
\[ 4 X^2 = A^2 \]
- Dividing both sides by 4 results in:
\[ X^2 = \frac{A^2}{4} \]

6. **Finding \( X \)**:
- Taking the square root of both sides, we find:
\[ X = \frac{A}{2} \]

### Conclusion:
Thus, the displacement from the mean position when the kinetic energy is three times the potential energy is:
\[ \boxed{\frac{A}{2}} \]