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An open organ pipe resonates with a tuning fork of frequency 500 Hz. Two successive nodes are observed at distances 16 cm and 46 cm from the open end. Find the speed of sound in the pipe.

1. 230 m/s
2. 300 m/s
3. 320 m/s
4. 360 m/s

Correct answer: 300 m/s

Solution

The distance between two successive nodes = lambda/2. lambda/2 = 46 - 16 = 30 cm, so lambda = 60 cm = 0.60 m. Speed of sound v = f * lambda = 500 * 0.60 = 300 m/s.

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