A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio 2 : 3 : 4 . Their lengths must be

Step-by-step solution

The frequency produced by a stretched wire is given by f = p 2 l T m … (i) Where, p = number of loops formed in vibration, l = length of wire, T = tension in wire, and m = mass per unit length of wire So, form Eq (i), we get f ∝ 1 l Here, the ratio of frequencies of three segments is f 1 : f 2 : f 3 = 2 : 3 : 4 ⇒ l 1 : l 2 : l 3 = 1 2 : 1 3 : 1 4 = 6 : 4 : 3 The length of wire, L = 260 cm, so l 1 = 6 13 × 260 = 120 cm l 2 = 4 13 × 260 = 80 cm a n d l 3 = 3 13 × 260 = 60 cm