A body executes SHM with an amplitude a. At what displacement from the mean positions, the potentail energy of the body is one-fourth of its total energy?

Options

A. `(a)/(4)`
B. `(a)/(2)`
C. `(3a)/(4)`
D. Some other fraction of a

Answer

To solve the problem, we need to determine the displacement \( x \) from the mean position where the potential energy \( U \) of a body executing Simple Harmonic Motion (SHM) is one-fourth of its total energy \( E \).

### Step-by-Step Solution:

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

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

3. **Setting Up the Equation**:
- According to the problem, we want to find \( x \) such that the potential energy is one-fourth of the total energy:
\[ U = \frac{1}{4} E \]

4. **Substituting the Expressions**:
- Substitute the expressions for \( U \) and \( E \) into the equation:
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{4} \left( \frac{1}{2} m \omega^2 a^2 \right) \]

5. **Simplifying the Equation**:
- Cancel \( \frac{1}{2} m \omega^2 \) from both sides:
\[ x^2 = \frac{1}{4} a^2 \]

6. **Finding the Displacement**:
- Taking the square root of both sides gives:
\[ x = \sqrt{\frac{1}{4} a^2} = \frac{a}{2} \]

7. **Conclusion**:
- The displacement \( x \) from the mean position where the potential energy is one-fourth of the total energy is:
\[ x = \frac{a}{2} \]

### Final Answer:
The displacement from the mean position is \( \frac{a}{2} \).

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