question_answer

A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is : [JEE Main 8-4-2019 Morning]

A) 4 : 9                

B) 3 : 5

C) 1 : 4                

D) 1 : 2

Correct Answer: D

Solution :

Let mass per unit length of wires are\[{{\mu }_{1}}\]and \[{{\mu }_{2}}\]respectively. \[\because \]Materials are same, so density \[\rho \] is same. \[\therefore \]\[{{\mu }_{1}}=\frac{\rho \pi {{r}^{2}}L}{L}=\mu \] and \[{{\mu }_{2}}=\frac{\rho 4\pi {{r}^{2}}L}{L}=4\mu \] Tension in both are same = T, let speed of wave in wires are \[{{V}_{1}}\]and \[{{V}_{2}}\] \[{{V}_{1}}=\sqrt{\frac{T}{\mu }}=V;{{V}_{2}}=\sqrt{\frac{T}{4\mu }}=\frac{V}{2}\] So fundamental frequencies in both wires are \[{{\text{f}}_{01}}=\frac{{{V}_{1}}}{2L}=\frac{V}{2L}\And {{\text{f}}_{02}}=\frac{{{V}_{2}}}{2L}=\frac{V}{4L}\] Frequency at which both resonate is L.C.M of both frequencies i.e.\[\frac{V}{2L}.\] Hence no. of loops in wires are 1 and 2 respectively. So, ratio of no. of antinodes is 1 : 2.