When the string of a sonometer of length `L` between the bridges vibrates in the first overtone , the amplitude of vibration is maximum at

Options

A. `L//2`
B. `(L//4) and ( 3L//4)`
C. `(L//6),(3L//6) and ( 5L//6)`
D. `(L)/(8) , (3 L)/(8) , ( 5 L)/(8) ,(7 L)/(8)`

Explanation

To solve the problem, we need to understand the behavior of a vibrating string in the context of standing waves, particularly when it vibrates in the first overtone.

### Step-by-Step Solution:

1. **Understanding the Sonometer and Vibrations**:
- A sonometer consists of a wire stretched between two bridges. When the wire vibrates, it produces standing waves. The first overtone is the first mode of vibration after the fundamental frequency.

2. **Identifying the Wavelength**:
- In the first overtone, the string has one antinode at the center and two nodes at the ends. The length of the string (L) corresponds to one full wavelength (λ) of the first overtone, which is actually half of the wavelength. Therefore, the relationship is:
\[ L = \frac{3}{2} \lambda \]
- From this, we can deduce that:
\[ \lambda = \frac{2L}{3} \]

3. **Locating the Antinodes**:
- The first overtone has two points of maximum amplitude (antinodes) located at:
- \( \frac{\lambda}{4} \) from one end, and
- \( \frac{3\lambda}{4} \) from the same end.
- Substituting the value of λ:
- The first antinode is at:
\[ \frac{\lambda}{4} = \frac{1}{4} \times \frac{2L}{3} = \frac{L}{6} \]
- The second antinode is at:
\[ \frac{3\lambda}{4} = \frac{3}{4} \times \frac{2L}{3} = \frac{L}{2} \]

4. **Conclusion**:
- The points of maximum amplitude (antinodes) in the first overtone are located at \( \frac{L}{6} \) and \( \frac{L}{2} \) from one end of the string. Therefore, the amplitude of vibration is maximum at these points.

### Final Answer:
The amplitude of vibration is maximum at \( \frac{L}{6} \) and \( \frac{L}{2} \).