An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive nodes are formed at distances 16 cm and 46 cm from the open end. The speed of sound in air in the pipe is
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Successive nodes in a standing wave are separated by λ/2. Here 46−16=30 cm, so λ=60 cm=0.6 m. The wave speed is v=fλ=500×0.6=300 m/s.
Formula / Key Relation
Δx_nodes = λ/2
46 cm − 16 cm = 30 cm
λ = 60 cm = 0.60 m
v = fλ = 500 × 0.60
= 300 m/s
Detailed Explanation
Concept & Reasoning
Although an open pipe has displacement antinodes at an ideal open end, the spacing between successive nodes anywhere in the standing-wave pattern remains λ/2. Therefore the end correction is not needed when two node locations are directly given; only their separation matters.
How to Solve It in the Exam
Successive nodes/antinodes are λ/2 apart.
Important Exam Point
Use separation of the two nodes, not either absolute distance from the open end.
Related Revision Path
This question sits in the revision path Power System → Waves and Acoustics → Standing Waves .
Quick Trick
Successive nodes/antinodes are λ/2 apart.
Common Mistake
Taking 30 cm as the full wavelength.
Exam Tip
Use separation of the two nodes, not either absolute distance from the open end.
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