Question-17 Two uniform wires of same material are vibrating under the same tension. If the 1^{st} overtone of 1^{st} wire is equal to the 2^{nd} overtone of 2^{nd} wire and radius of 1^{st} wire is twice the radius of 2^{nd} wire, the ratio of length of 1^{st} wire to that 2^{nd} wire is
(A) 1:3
(B) 3:1
(C) 2:3
(D) 3:5
Answer : (A)
Explanation:
Fundamental frequency of the first wire is
n=\frac{1}{2l_1}\sqrt{\frac{T}{m}}=\frac{1}{2l_1}\sqrt{\frac{T}{\pi r_1^2\rho}}=\frac{1}{2l_1 r_1}\sqrt{\frac{T}{\pi\rho}}The first overtone n_1=2n=\frac{1}{l_1 r_1}\sqrt{\frac{T}{\pi\rho}}
Similarly, the second overtone of the second wire will be,
n_2=\frac{3}{2l_2 r_2}\sqrt{\frac{T}{\pi\rho}}Given that n_1=n_2
∴ \frac{1}{l_1 r_1}\sqrt{\frac{T}{\pi\rho}}=\frac{3}{2l_2 r_2}\sqrt{\frac{T}{\pi\rho}}
∴ 3l_1 r_1=2l_2 r_2
∴ \frac{l_1}{l_2}=\frac{2r_2}{3r_1}
\frac{l_1}{l_2}=\frac{2r_2}{3(2r_2)} ….(∵ r_1=2r_2)
\frac{l_1}{l_2}=\frac{1}{3}