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The length of a sonometer wire AB is 110...

The length of a sonometer wire AB is 110 cm. Where should the two bridges be placed from A to divide the wire in 3 segments whose fundamental frequencies are in the ratio of ` 1: 2 : 3` ?

Text Solution

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Let `l_(1) , l_(2) , l_(3)` be the lengths of three segments.
Given ` l_(1) + l_(2) + l_(3) = 1.10` (i)
From relation ` n = (1)/( 2l) sqrt ((T)/(mu))`
If tension `T` and mass per unit length ` mu` are fixed , then ` n prop 1//l`
So ` nl = constant`
` n_(1) l_(1) = n_(2) l_(2) = n_(3) l_(3)`
`:. l_(1) : l_(2) : l_(3) = (1)/( n_(1)) : (1)/( n_(2)) : (1)/( n_(3)) = (1)/(1) : (1)/(2) : (1)/(3) = ( 6 : 3 : 2)/( 6) = 6 : 3 :2`
`:. l_(1) = 6 k , l_(2) = 3k , l_(3) = 2k , k` being a constant .
From Eq. (i)
` 6 k + 3k + 2k = 1.10 or 11k = 1.10`
`:. k = 1.10 //11 = 0.1`
`:. l_(1) = 6 xx 0.1 = 0.6 m`
`l_(2) = 3 xx 0.1 = 0.3 m`
`l_(3) = 2 xx 0.1 = 0.2 m`
Therefore the bridges must be placed at distance `0.6 m and ( 0.6 + 0.3 ) = 0.9 m`
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