Two successive resonance frequencies in an open organ pipe are 1800Hz and 2600Hz. Find the length of the tube if the speed of sound in air is 340 m/s.
A
20cm
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B
20.5cm
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C
21.25cm
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D
22.25cm
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Solution
The correct option is C21.25cm Modes of vibration in an open organ pipe are given by fn=nv2L where n=1,2,3...
So for nth harmonic fn=nv2L
and for (n+1)th harmonic fn+1=(n+1)v2L
From question data, fn+1−fn=2600−1800 ⇒(n+1)v2L−nv2L=800 ⇒v2L=800 ⇒L=v1600=3401600=21.25 cm