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The first overtone of an open orgen pipe...

The first overtone of an open orgen pipe beats with the first ouertone of a closed orgen pipe with a beat frequency of `2.2 H_(Z)` . The fundamental frequency of the closed organ pipe is `110 H_(Z)` . Find the lengths of the pipes . Speed of sound in air `u = 330 m//s` .

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The correct Answer is:
A, B
Let `l_(2)` and `l_(2)` be the lengths of closed and open pipes respectively.
Fundamental frequency of closed organ pipe is give by
`f_(1) = (upsilon) / (4l)`
Where `upsilon` is speed in air and `330 m//s` equal to
But, `f_(1) = 110 H_(Z)` (given)
Therefore, `(upsilon)/(4l_(1) = 110 H_(Z)`
:`. l_(1) = (upsilon)/(4xx110) = (330)/(4xx 110) m = 0.75 m`
First overtone of closed organ pipe will be
`f_(3) = 3f_(1) = 3 (110) H_(Z) = 330 H_(Z)`
This produces a beat frequency of `2.2 H_(Z)` with first overtone of open pipe .
Therefore , first overtone frequency of open organ pipe is either
`(330 + 2.2) H_(Z) = 332. 2 H_(Z)`
or `(330 - 2.2 ) H_(Z) = 332. 2 H_(Z)`
If it is `332.2 H_(Z)` , then
`((upsilon)/(2l_(2))) = 332.2 H_(Z)`
or `l_(2) = (upsilon)/(332.2) = (330)/(332.2) m = 0.99 m`
and if it `327.8 H_(Z)` , then
`((upsilon)/(2l_(2))) = 332.2 H_(Z)`
or `l_(2) = (upsilon)/ (327.8) m = (330)/(327.8) m = 1.0067 m`
Therefore, length of the closed organ pipe is `l_(2) = 0.75 m` while length of open pipe is either
`l_(2) = 0.99 m` or `1.0067 m` .
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