Let `l_(1)` be length of open organ pipe and `l_(2)` be the length of closed organ pipe.
Fundamental frequency of open pipe,
`(upsilon)/(2l_(1))=300`
`l_(1)=(upsilon)/(2xx300)=(330)/(600)=0.55m=55cm`
Now, frequency of first overtone of closed pipe `=` frequency of 1st overtone of open pipe
`(3upsilon)/(4l_(2))=(2upsilon)/(2l_(1))=2xx300`
`l_(2)=(3upsilon)/(4xx600)=(3xx330)/(2400)m=41.25cm`