Let `l_(1)` and `l_(2)` are lengths of closed and open pipe
-respectively.
Fundamental frequency of closed pipe `= 110 Hz`
`(v)/(4 l_(1)) = 110`
`l_(1) = (v)/(4 xx 110) = (330)/(4 xx 110) = (3)/(4) m = 0.75 m`
`2 . (v)/(2 l_(2)) - 3 . (v)/(4 l_(2)) = +-5`
`(v)/(l_(2)) - 3 xx 110 = +-5`
`(v)/(l_(2)) = 330 +- 5 = 335 or 325`
`(v)/(l_(2)) = 335 rArr l_(2) = (v)/(335) = (330)/(335) = 0.98 m`
`(v)/(l_(2)) = 325 rArr l_(2) = (v)/(325) = (330)/(325) = 1.02 m`
Length of closed pipe `= 0.75 m`
Length of open pipe `= 0.98 m or 1.02 m`