When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is:

Options

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

Explanation

To solve the problem, we need to determine the displacement \( x \) of a particle executing simple harmonic motion (SHM) when its potential energy \( U \) is one-fourth of its maximum potential energy \( U_{max} \).

### Step-by-step Solution:

1. **Understand the Potential Energy in SHM**:
The potential energy \( U \) of a particle in SHM is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

2. **Maximum Potential Energy**:
The maximum potential energy \( U_{max} \) occurs when the particle is at its maximum displacement, which is the amplitude \( A \). Therefore, we can express the maximum potential energy as:
\[ U_{max} = \frac{1}{2} k A^2 \]

3. **Set Up the Equation**:
According to the problem, the potential energy \( U \) is one-fourth of the maximum potential energy:
\[ U = \frac{1}{4} U_{max} \]
Substituting the expression for \( U_{max} \):
\[ U = \frac{1}{4} \left( \frac{1}{2} k A^2 \right) = \frac{1}{8} k A^2 \]

4. **Equate the Two Expressions for Potential Energy**:
Now, we can set the two expressions for potential energy equal to each other:
\[ \frac{1}{2} k x^2 = \frac{1}{8} k A^2 \]

5. **Cancel Out Common Terms**:
We can cancel \( \frac{1}{2} k \) from both sides:
\[ x^2 = \frac{1}{4} A^2 \]

6. **Solve for Displacement \( x \)**:
Taking the square root of both sides gives:
\[ x = \pm \frac{A}{2} \]
This means the displacement can be either \( \frac{A}{2} \) or \( -\frac{A}{2} \).

7. **Final Answer**:
Since the problem asks for the displacement in terms of amplitude \( A \) and typically considers the positive value, we conclude:
\[ x = \frac{A}{2} \]