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A sonometer wire has a total length of 1...

A sonometer wire has a total length of 1 m 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 ?

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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.
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