If the total energy of a simple harmonic oscillator is E, then its potential energy, when it is halfway to its endpoint will be

Options

A. `(2)/(3)F`
B. `(1)/(8)E`
C. `(1)/(4)E`
D. `(1)/(2)E`

Answer

To solve the problem, we need to find the potential energy of a simple harmonic oscillator when it is halfway to its endpoint, given that the total energy is \( E \).

### Step-by-Step Solution:

1. **Understanding the Total Energy**:
The total energy \( E \) of a simple harmonic oscillator is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( A \) is the amplitude of the oscillation, \( m \) is the mass of the oscillator, and \( \omega \) is the angular frequency.

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

3. **Finding the Displacement at Halfway to the Endpoint**:
When the oscillator is halfway to its endpoint, the displacement \( y \) is:
\[ y = \frac{A}{2} \]

4. **Substituting \( y \) into the Potential Energy Formula**:
Substitute \( y = \frac{A}{2} \) into the potential energy formula:
\[ U = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 \]
Simplifying this gives:
\[ U = \frac{1}{2} m \omega^2 \cdot \frac{A^2}{4} = \frac{1}{8} m \omega^2 A^2 \]

5. **Relating Potential Energy to Total Energy**:
From the total energy equation, we know:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Therefore, we can express \( \frac{1}{8} m \omega^2 A^2 \) in terms of \( E \):
\[ U = \frac{1}{8} \cdot 2E = \frac{E}{4} \]

6. **Conclusion**:
Thus, the potential energy \( U \) when the oscillator is halfway to its endpoint is:
\[ U = \frac{E}{4} \]

### Final Answer:
The potential energy when the simple harmonic oscillator is halfway to its endpoint is \( \frac{E}{4} \).