The length of a sonometer wire AB is 176 cm. Where should the two bridges be placed from end A to divide the wire into three segments whose fundamental frequencies are in the ratio 1:2:3 ?

Step-by-step solution

Let the lengths of the three segments be l₁ , l₂ , and l₃ . The fundamental frequency of a stretched string is given by f = v 2l . Since the tension and mass per unit length are constant for the wire, f 1 l . Given the ratio of fundamental frequencies is f₁ : f₂ : f₃ = 1 : 2 : 3 . Therefore, the ratio of their lengths is l₁ : l₂ : l₃ = 1 : 1 2 : 1 3 = 6 : 3 : 2 . Let l₁ = 6x , l₂ = 3x , and l₃ = 2x . The total length of the wire is l₁ + l₂ + l₃ = 176 cm . 6x + 3x + 2x = 176 11x = 176 x = 16 cm . The lengths of the