Question

If a stretched wire is vibrating in the second overtone, then the number of nodes and antinodes between the ends of the string are respectively

• A. 2 and 2
• B. 1 and 2
• C. 4 and 3
• D. 2 and 3
• E. 3 and 4

Hint:

For the \(n\)-th harmonic of a string fixed at both ends: - Total Nodes = \(n + 1\) - Nodes between ends = \(n - 1\) - Antinodes = \(n\)

Answer 1

The Correct Option is D

Concept: In a stretched string fixed at both ends, the vibration modes are defined as harmonics.
• Fundamental (1st Harmonic): The string vibrates in one loop. Nodes are at the ends.
• 1st Overtone (2nd Harmonic): The string vibrates in two loops.
• 2nd Overtone (3rd Harmonic): The string vibrates in three loops.

Step 1: Identify the mode of vibration for the second overtone.
For a string fixed at both ends, the \(n\)-th harmonic is the \((n-1)\)-th overtone. Therefore, the second overtone corresponds to the third harmonic. In the third harmonic, the string vibrates in exactly three loops.

Step 2: Count the nodes and antinodes between the ends.
For 3 loops in a string fixed at both ends:
• Total Nodes: There are 4 nodes in total (2 at the fixed ends and 2 in the middle).
• Nodes between the ends: Excluding the fixed boundary ends, there are 2 nodes.
• Total Antinodes: Each loop contains one antinode. For 3 loops, there are 3 antinodes.