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

Step-by-step solution

The fundamental frequency relationship for a string fixed at both ends gives the n th overtone frequency as f_n = (n+1) v 2L , where v = T is the wave speed and = r^2 is the mass per unit length for a wire of radius r and density . Substituting yields f_n = (n+1) 2Lr T . For the first wire's first overtone ( n=1 ), f₁ = 1 L₁r₁ T , and for the second wire's second overtone ( n=2 ), f₂ = 3 2L₂r₂ T . Setting f₁ = f₂ and canceling the common factor T gives 1 L₁r₁ = 3 2L₂r₂ . Given r₁ = 2r₂ , substitution yields 1 2L₁r₂