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Question:

A wire of length 2L is made by joining two wires A and B of some 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?

1:4

4:9

3:5

1:2

Solution:

Correct option is D. 1:2
Let mass per unit length of wires are μ₁ and μ₂ respectively.
∵ Materials are same, so density ρ is same.
∴ μ₁ = ρπr²L/L = μ and μ₂ = ρ4πr²L/L = 4μ
Tension in both are same = T, let speed of wave in wires are V₁ and V₂
V₁ = √(T/μ) = V
V₂ = √(T/4μ) = V/2
So fundamental frequencies in both wires are
f₀₁ = V₁/2L = V/2L and f₀₂ = V₂/2L = V/4L
Frequency at which both resonate is L.C.M of both frequencies i.e. V/2L.
Hence no. of loops in wires are 1 and 2 respectively.
So, ratio of no. of antinodes is 1:2.