When the displacement of a simple harmonic oscillator is half of its amplitude, its potential energy is 3J. Its total energy is

Options

A. 6J
B. 12 J
C. 15 J
D. 20 J

Explanation

To find the total energy of a simple harmonic oscillator when its displacement is half of its amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Information**:
- Displacement \( x = \frac{A}{2} \) (where \( A \) is the amplitude)
- Potential Energy \( PE = 3 \, \text{J} \)

2. **Formula for Potential Energy in SHM**:
The potential energy \( PE \) of a simple harmonic oscillator is given by the formula:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
where \( m \) is the mass, \( \omega \) is the angular frequency, and \( x \) is the displacement.

3. **Substituting the Displacement**:
Substitute \( x = \frac{A}{2} \) into the potential energy formula:
\[ PE = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 \]
Simplifying this gives:
\[ PE = \frac{1}{2} m \omega^2 \cdot \frac{A^2}{4} = \frac{1}{8} m \omega^2 A^2 \]

4. **Relating Potential Energy to Total Energy**:
The total energy \( E \) of a simple harmonic oscillator is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
From the expression for potential energy, we can see that:
\[ PE = \frac{1}{4} E \]
This means that the total energy is four times the potential energy when the displacement is half the amplitude.

5. **Calculating Total Energy**:
Given that \( PE = 3 \, \text{J} \), we can find the total energy:
\[ E = 4 \times PE = 4 \times 3 \, \text{J} = 12 \, \text{J} \]

### Final Answer:
The total energy of the simple harmonic oscillator is \( 12 \, \text{J} \).

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