The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency `2.2 Hz` . The fundamental frequency of the closed organ pipe is `110 Hz`, find the lengths of the pipes . Take velocity of sound ` = 330 m//s`.

Explanation

Let `l_(1) and l_(2)` be lengths of open organ pipe and closed organ pipe respectively .
First overtone of open organ pipe ` = 2n_(1) = 2 xx (v)/( 2 l_(1)) = (v)/( l_(1))`
First overtone of closed organ pipe ` = 3 n_(2) = 3 xx (v)/( 4 l_(2))`
According to question , `(v)/(l_(1)) - ( 3v)/( 4 l_(2) = +- 2.2` (i)
As `n_(2)` is the fundamental frequency of closed organ pipe
`n_(2) = (v)/( 4 l_(2))`
`l_(2) = (v)/( 4 n_(2)) = (330)/( 4 xx 110) = 0.75 m`
From Eq.(i).
`(v)/(l_(1)) = ( 3v)/( 4 l_(2)) +- 2.2`
`(v)/(l_(1)) = 3 n_(2) +- 2.2`
Taking positive sign
`(v)/(l_(1)) = 3 xx 110 + 2.2 = 332.2`
`:. l_(1) = (v)/( 332.2) = (330)/(332.2) m = 0.993 m = 99.3 cm`
Taking negative sign
`(v)/( l_(1)) = 3 n_(2) - 2.2 = 3 xx 110 - 2.2 = 327.8`
`:. l_(1) = (v)/( 327.8) = (330)/(327.8) = 1.006 m = 100.6 cm`