A particle executing SHM with an amplitude A. The displacement of the particle when its potential energy is half of its total energy is

Options

A. `(A)/(sqrt(2))`
B. `(A)/(2)`
C. `(A)/(4)`
D. `(A)/(3)`

Explanation

To find the displacement of a particle executing Simple Harmonic Motion (SHM) when its potential energy is half of its total energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy in SHM**:
- The total mechanical energy (E) in SHM is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
- Here, \( m \) is the mass of the particle, \( \omega \) is the angular frequency, and \( A \) is the amplitude.

2. **Potential Energy in SHM**:
- The potential energy (PE) at a displacement \( x \) from the mean position is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

3. **Condition Given in the Problem**:
- We need to find the displacement \( x \) when the potential energy is half of the total energy:
\[ PE = \frac{1}{2} E \]

4. **Substituting the Expressions**:
- From the condition \( PE = \frac{1}{2} E \):
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \left( \frac{1}{2} m \omega^2 A^2 \right) \]
- Simplifying this gives:
\[ m \omega^2 x^2 = \frac{1}{2} m \omega^2 A^2 \]

5. **Canceling Common Terms**:
- We can cancel \( m \) and \( \omega^2 \) from both sides (assuming \( m \neq 0 \) and \( \omega \neq 0 \)):
\[ x^2 = \frac{1}{2} A^2 \]

6. **Solving for Displacement \( x \)**:
- Taking the square root of both sides:
\[ x = \sqrt{\frac{1}{2}} A = \frac{A}{\sqrt{2}} \]

7. **Final Answer**:
- Therefore, the displacement of the particle when its potential energy is half of its total energy is:
\[ x = \frac{A}{\sqrt{2}} \]