A sonometer wire is vibrating in the second overtone. In the wire there are

Options

A. two nodes and two antinodes
B. one node and two antinodes
C. four nodes and three antinodes
D. three nodes and three antinodes

Answer

To solve the problem of determining the number of nodes and antinodes in a sonometer wire vibrating in the second overtone, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Fundamental Frequency:**
- The fundamental frequency (first harmonic) of a vibrating string or wire corresponds to the formation of one loop. In this case, the length of the wire (L) is equal to half the wavelength (λ) of the wave.
- Therefore, for the fundamental frequency, we have:
\[ L = \frac{\lambda}{2} \]

2. **Identifying the First Overtone:**
- The first overtone (second harmonic) corresponds to two loops being formed. Here, the length of the wire is equal to the wavelength (λ).
- Thus, for the first overtone, we have:
\[ L = \lambda \]

3. **Understanding the Second Overtone:**
- The second overtone (third harmonic) corresponds to three loops being formed. In this case, the length of the wire is equal to \( \frac{3}{2} \) times the wavelength.
- Therefore, for the second overtone, we have:
\[ L = \frac{3\lambda}{2} \]

4. **Counting Nodes and Antinodes:**
- In a vibrating wire, nodes are points where there is no displacement, and antinodes are points where the displacement is maximum.
- For the second overtone, we can visualize the formation:
- There are 3 loops formed, which means there are 4 nodes (one at each end and one between each loop).
- There are also 3 antinodes (one between each pair of nodes).

5. **Conclusion:**
- Therefore, in the sonometer wire vibrating in the second overtone, there are **4 nodes and 3 antinodes**.

### Final Answer:
In the wire, there are **4 nodes and 3 antinodes**.