Question:
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 :
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 :
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Same tension and same frequency; wave speed \(v = \sqrt{T/\mu}\) with \(\mu \propto r^2\), and each segment of length \(L\) holds a whole number of half-wavelengths.
Updated On: Nov 5, 2026
- \(3 : 5\)
- \(4 : 9\)
- \(1 : 2\)
- \(1 : 4\)
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The Correct Option is C
Solution and Explanation
Same material and same tension \(T\): \(\mu \propto r^2\), so \(\mu_B = 4\mu_A\).
\(v = \sqrt{T/\mu}\Rightarrow v_A = 2v_B\). The frequency \(f\) is the same, so \(\lambda_A = 2\lambda_B\).
Both ends and the joint are nodes, so each wire (length \(L\)) holds an integral number of loops: \(L = p\frac{\lambda_A}{2} = q\frac{\lambda_B}{2}\).
\(\frac{p}{q} = \frac{\lambda_B}{\lambda_A} = \frac{1}{2}\).
Answer: (3) \(1 : 2\)
\(v = \sqrt{T/\mu}\Rightarrow v_A = 2v_B\). The frequency \(f\) is the same, so \(\lambda_A = 2\lambda_B\).
Both ends and the joint are nodes, so each wire (length \(L\)) holds an integral number of loops: \(L = p\frac{\lambda_A}{2} = q\frac{\lambda_B}{2}\).
\(\frac{p}{q} = \frac{\lambda_B}{\lambda_A} = \frac{1}{2}\).
Answer: (3) \(1 : 2\)
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