Length of a sonometer wire is `1.21 m`. Find the length of the three segments for fundamental frequencies to be in the ratio `1 : 2 : 3`.

Answer

`0.66 m, 0.33 m, 0.22 m`

Explanation

`n_(1) = (1)/(2L_(1)) sqrt((T)/(mu))`
`n_(2) = (1)/(2L_(2)) sqrt((T)/(mu))`
`n_(3) = (1)/(2L_(3)) sqrt((T)/(mu))`
`n_(1) : n_(2) : n_(3) : 1 : 2 : 3`
`n_(2) = 2n_(1)`
`n_(3) = 3n_(1)`
`L_(3) = (n_(1))/(n_(2)) L_(1) = ((L_(1))/(2))`
`L_(1) + L_(2) + L_(3) = 1.21`
`L_(1) ((6 + 3 + 2)/(6)) = 1.21`
`L_(1) ((11)/(6)) = 1.21`
`L_(1) = (6 xx 121)/(11 xx 100) = (6 xx 11)/(100) = 0.66m`
`L_(2) = (L_(1))/(L_(2)) = (L_(1))/(2) = (0.66)/(20) = 0.33 m`
`L_(3) = (L_(1))/(3) = (0.66)/(3) = 0.22m`