Question:
A sonometer wire of length \( L \) between the two bridges vibrates in 3rd harmonic. Antinodes are formed at
A sonometer wire of length \( L \) between the two bridges vibrates in 3rd harmonic. Antinodes are formed at
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In harmonic vibration problems, use the formula for the position of the antinodes to determine the specific locations of maximum displacement.
Updated On: Feb 18, 2026
- \( \frac{L}{4}, L, \frac{7L}{4} \)
- \( \frac{L}{2}, L, \frac{3L}{2} \)
- \( \frac{L}{3}, \frac{2L}{3}, \frac{5L}{3} \)
- \( \frac{L}{4}, \frac{3L}{4}, \frac{5L}{4} \)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the harmonic vibration.
In the 3rd harmonic, the number of antinodes is 3. The positions of the antinodes for the 3rd harmonic are given by the formula: \[ x = \frac{L}{4}, \frac{3L}{4}, \frac{5L}{4} \] where \( L \) is the length of the sonometer wire. These are the positions where the maximum displacement occurs.
Step 2: Conclusion.
Thus, the antinodes are formed at \( \frac{L}{4}, \frac{3L}{4}, \frac{5L}{4} \), corresponding to option (D).
In the 3rd harmonic, the number of antinodes is 3. The positions of the antinodes for the 3rd harmonic are given by the formula: \[ x = \frac{L}{4}, \frac{3L}{4}, \frac{5L}{4} \] where \( L \) is the length of the sonometer wire. These are the positions where the maximum displacement occurs.
Step 2: Conclusion.
Thus, the antinodes are formed at \( \frac{L}{4}, \frac{3L}{4}, \frac{5L}{4} \), corresponding to option (D).
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