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

Answer

To solve the problem, we need to understand the relationship between potential energy (PE), maximum potential energy, and displacement in simple harmonic motion (SHM).

### Step-by-Step Solution:

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

2. **Maximum Potential Energy**:
The maximum potential energy (PE_max) occurs when the particle is at its maximum displacement (amplitude \( a \)):
\[ PE_{\text{max}} = \frac{1}{2} k a^2 \]

3. **Given Condition**:
According to the problem, the potential energy is one-fourth of its maximum value:
\[ PE = \frac{1}{4} PE_{\text{max}} = \frac{1}{4} \left( \frac{1}{2} k a^2 \right) = \frac{1}{8} k a^2 \]

4. **Setting up the Equation**:
Now, we can equate the expression for potential energy at displacement \( x \) to the given condition:
\[ \frac{1}{2} k x^2 = \frac{1}{8} k a^2 \]

5. **Canceling \( k \)**:
Since \( k \) is a common factor, we can cancel it from both sides:
\[ \frac{1}{2} x^2 = \frac{1}{8} a^2 \]

6. **Solving for \( x^2 \)**:
Multiply both sides by 2 to isolate \( x^2 \):
\[ x^2 = \frac{1}{4} a^2 \]

7. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ x = \frac{a}{2} \]

### Final Answer:
The displacement of the particle from the equilibrium position, when its potential energy is one-fourth of its maximum value, is:
\[ x = \frac{a}{2} \]