The length of a sonometer wire between two fixed ends is 110cm. Where should the two bridges the placed so as to divide the wire into three segments, whose fundamental frequencies are in the ration` 1:2:3?`

Explanation

Figure-6.60 shows a sonometer wire between two fixed ends. Let the two bridgesbe placed at C and D such that the wire is divided into three segments (AC, CD and DB) of lengths `l_(1), l_(2)` and `l_(3)` respectively whose frequencies are in the ratio of `1 : 2 : 3`
For the length of string, we have
`AC + CD + DB = 100 cm`
Let `n_(1), n_(2)` and `n_(3)` be the fundamental frequencies of these segments respectively, then
`(n_(1))/(n_(2)) = (1)/(2)` and `(n_(2))/(n_(3)) = (2)/(3)`
We know that frequency is inversely proportional to the length of the segment because when tension remains constant
`n infty (I)/(l)`
As `sqrt((T)/(lambda))` is constant,
`n_(1)l_(1) = n_(2)l_(2) = n_(3)l_(3)`
`l_(1) = (n_(2))/(n_(1)) l_(2) = 2l_(2)`
`l_(3) = (n_(2))/(n_(3)) l_(2) = (2)/(3) l_(2)`
Substituting these values in equation - (6.199), we have
`2l_(2) + l_(2) + (2)/(3)l_(2) = 100`
`l_(2) = 27.27 cm`
From equation
`l_(1) = 2 xx 27.27` = 54.54 cm
`l_(3) = (2)/(3) xx 27. 27` = 18.18 cm