Two uniform strings `A and B` made of steel are made to vibrate under the same tension. If the first overtone of `A` is equal to the second overtone of `B` and if the radius of `A` is twice that of `B`, the ratio of the lengths of the strings is

Options

A. `1 : 4`
B. ` 1 : 6`
C. ` 1 : 2`
D. ` 1 : 3`

Answer

D

Explanation

`n = 1/(2L) sqrt(T/m) = 1/(2L) sqrt(T/(pi r^(2) rho)) = 1/(2Lr) sqrt(T/(pi rho) ) `
Both wires have the same T and `rho`.
If `n_(A) " and " n_(B) ` are the fundamental frequencies of A and B , then first overtone of `A = 2n_(A)` and second overtone of
` B = 3n_(B)." " :' 2n_(A) = 3n_(B) `
`:. 2 [1/(2L_(A) r_(A))sqrt(T/(pirho))] = 3/(2L_(B) r_(B)) [T/(pi rho)]`
` 1/(L_(A) r_(A)) = 3/(2L_(B) r_(B)) " but " r_(A) = 2r_(B) `
` :. 1/(L_(A) xx 2r_(B)) = 3/(2L_(B) xx r_(B)) `
` :. 3L_(A) = L_(B) " or " L_(A)/L_(B) = 1/3`