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Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is `324 m s^-1`.

Text Solution

Verified by Experts

The correct Answer is:
`0.25 m`
Let the pipe resonates in `n^(th) & (n+1)^(th)` harmonic
`rArr(n+1)1944=n(2592)`
`rArr n=4:.L=(324)/(1296)=0.25m`
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ALLEN -WAVES AND OSCILLATIONS-Part-1(Exercise-04)[A]
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