A sonometer wire has a total length of 1m between the fixed ends. Where should the two bridges be placed below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1 : 2 : 3 ?

Explanation

suppose the lengths of the three segments are L_1L_2` and L_3 respectively the fundmaental frequencies are.
`v_1 = (1)/(2L_1) sqrt(f//mu)`
`v_2 = (1)/(2L^2) sqrt(F//mu)`
`v_3 = (1)/(2L_3) sqrt(F//mu)`
so that `v_1L_1 = v_2 L_2 = v_3 L_3. (i)`
`As v_1 : V_2 : V_3 = 1 : 2 : 3, we have
`v_2 = 2v_1 and v_3 = v_1` so that by (i) `
`L_2 = (v_1)/(v_2) L_1 = (L_1)/(2)`
and `L_3 = (v_1)/(3) L_1 = (L_1)/(3)`
As `L_1 + L_2 +L_3 = 1m,`
`we get L_1(1 +(1)/(2)+(1)/(3)) = 1m`
or, L_1 =(6)/(11)m`
Thus, `L_2 = (L_1)/(2) = (3)/(11)m` and `L_3 = (L_1)/(3) = (2)/(11) m.`
One bridge should be placed at `(6)/(11)` m from one end and the other should be placed at `(2)/(11)m` from the other end.