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A composite wire of length 2L is made by joining two different wires A and B having the same length, made of the same material but of different radii r and 2r respectively. The composite wire is vibrating at such a frequency, that the junction of the two wires form a node. If the number of antinodes in the wire A is p and that in the wire B is q, then the ratio p:q is

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1:4

1:2

4:9

3:5

answer is A.

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Detailed Solution

1:2 $f_{1} = \frac{n}{2 l} \sqrt{\frac{T}{ρ A}}; f_{2} = \frac{m}{2 l} \sqrt{\frac{T}{ρ 4 A}}$ Given $f_{1} = f_{2}$ $\frac{f_{1}}{f_{2}} = \frac{n}{m} \times 2$ $\frac{n}{m} = \frac{1}{2}$

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