The total length of a sonometer wire fixed between two bridges is 110 cm . Now, two more bridges are placed to divide the length of the wire in the ratio 6 : 3 : 2 . If the tension in the wire is 400 N and the mass per unit length of the wire is 0 . 01 kg m - 1 , then the minimum common frequency with which all the three parts can vibrate, is

Step-by-step solution

l 1 : l 2 : l 3       =   6 : 3 : 2 The frequancy in any n t h mode for a segment is f = n v 2 l = c o n s t a n t no of loops ∝ length of the segment so, no of loops are in the ratio 6:3:2 hence total loops =11 The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum λ 2 = 1 11 x 110 = 1 0 cm f = V λ = 1 λ · F μ = 1 0.2 4 0 0 0.01 = 1 0 0 0 Hz