(a) For open pipe, fundamental frequency
`f = (v)/(2l) = (340)/(2 xx 0.5) = 340 Hz`
Possible frequencies of open pipe `= nf, n = 1,2,3,…`
`340,680,1020,1360,1700,2040 Hz`
Frequencies between `1000` and `2000 Hz`
`= 1020 Hz,1360 Hz,1700 Hz`
(b) `f = (v)/(2l)`
`l = (v)/(f)` , l will be maximum if `f` is minimum
`l_(max) = (340)/(2 xx 20) = 8.5 m`
(c) The difference in successive frequencies of a pipe
(open or closed) `= (v)/(2l)`
`2592 - 1944 = (324)/(2l)`
`l = (1)/(4) m = 0.25 m`