A tube open at only one end is cut into two tubes of non equal lengths. The piece open at both ends has of fundamenal frequency of `450 Hz` and other has fundamental frequency of `675 Hz`. What is the `1^(st)` overtone frequency of the original tube.

Explanation

`450 = (v)/(2l_(1)) , 675 = (v)/(4l_(2))`
length of original tube `= (l_(1) + l_(2))`
its first obtained frequency,
`n_(1) = (v)/(4(l_(1) + l_(2))) = (v)/(4((v)/(900) + (v)/(675xx4))) = ((2700 xx 900))/(4(2700 + 900)) = 168.75`
`I^(st)` overtone `= 3n_(1) = 506.25`