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 :

Step-by-step solution

Fundamental frequency of string, n= 1 2 l T m where, m= mass per unit length of the string = r^2 n= 1 2 l T r^2 For string A first overtone, n_A= 1 l₁ T₁ m₁ = 1 l₁ T₁ r₁^2 For string B second overtone, n_B= 3 2 l₂ T₂ r₂^2 ₂ According to question, n_A=n_B 1 l₁ T₁ r₁^2 l₁ = 3 2 l₂ T₂ r₂^2 l₂ but T₁=T₂ 2 l₂ 3 l₁ = r₁^2 r₂^2 4 l₂^2 9 l₁^2 = r₁^2 r₂^2 ( l₁ l₂ )^2= 4 r₂^2 9 r₁^2 2 l₂ 3 l₁ = r₁ r₂ l₂ l₁ = 3 r₁ 2 r₂ l₁ l₂ = 2 r₂ 3 r₁ but r₁=2 r₂ l₁ l₂ = 2 r₂ 3 (2 r₂ ) l₁: l₂=1: 3