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

Answer

To solve the problem, we need to determine 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 of a Simple Harmonic Oscillator:**
The total energy (E) of a simple harmonic oscillator is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude.

2. **Identifying Halfway to the Endpoint:**
When the oscillator is halfway to its endpoint, the displacement \( x \) from the equilibrium position is:
\[ x = \frac{A}{2} \]

3. **Calculating the Potential Energy at Displacement \( x \):**
The potential energy (U) at a displacement \( x \) is given by:
\[ U = \frac{1}{2} k x^2 \]
Substituting \( x = \frac{A}{2} \) into the potential energy formula:
\[ U = \frac{1}{2} k \left(\frac{A}{2}\right)^2 \]
Simplifying this:
\[ U = \frac{1}{2} k \cdot \frac{A^2}{4} = \frac{1}{8} k A^2 \]

4. **Relating Potential Energy to Total Energy:**
We know from the total energy expression that:
\[ E = \frac{1}{2} k A^2 \]
Therefore, we can express \( k A^2 \) in terms of E:
\[ k A^2 = 2E \]

5. **Substituting into the Potential Energy Expression:**
Now substituting \( k A^2 = 2E \) into the potential energy expression:
\[ U = \frac{1}{8} (2E) = \frac{E}{4} \]

### Final Answer:
Thus, the potential energy when the oscillator is halfway to its endpoint is:
\[ U = \frac{E}{4} \]