A wire fixed at both ends in sonometer experiment is vibrating in the third overtone. There are

Options

A. two nodes, two antinodes
B. three nodes, three antinodes
C. four nodes, three antinodes
D. five nodes, four antinodes

Explanation

To solve the problem of a wire fixed at both ends vibrating in the third overtone, we need to understand the relationship between harmonics, overtones, nodes, and antinodes.

### Step-by-Step Solution:

1. **Understanding Overtone and Harmonic**:
- The fundamental frequency is the first harmonic (1st harmonic).
- The first overtone is the second harmonic (2nd harmonic).
- The second overtone is the third harmonic (3rd harmonic).
- The third overtone is the fourth harmonic (4th harmonic).

2. **Identifying the Number of Loops**:
- In the case of the third overtone (4th harmonic), the number of loops formed in the wire is equal to the harmonic number. Thus, for the third overtone, there are 4 loops.

3. **Counting Nodes and Antinodes**:
- In a standing wave, nodes are points where there is no movement (minimum amplitude), and antinodes are points where the amplitude is maximum.
- For a wire fixed at both ends:
- The number of nodes (N) is equal to the harmonic number + 1. Therefore, for the 4th harmonic (third overtone), the number of nodes is 4 + 1 = 5 nodes.
- The number of antinodes (A) is equal to the harmonic number. Therefore, for the 4th harmonic, the number of antinodes is 4.

4. **Conclusion**:
- Thus, in the third overtone, there are 5 nodes and 4 antinodes.

### Final Answer:
- **Number of Nodes**: 5
- **Number of Antinodes**: 4