Question:
Two wires \(W_1\) and \(W_2\) have the same radius \(r\) and respective densities \(\rho_1\) and \(\rho_2\) such that \(\rho_2=4\rho_1\). They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension \(T\). The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in \(W_1\) to \(W_2\) is : 
Two wires \(W_1\) and \(W_2\) have the same radius \(r\) and respective densities \(\rho_1\) and \(\rho_2\) such that \(\rho_2=4\rho_1\). They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension \(T\). The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in \(W_1\) to \(W_2\) is :
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Both parts vibrate at the same frequency; wave speed is \(\sqrt{T/\mu}\) and the segments have equal length.
Updated On: Nov 11, 2026
- \(1:1\)
- \(1:2\)
- \(1:3\)
- \(4:1\)
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The Correct Option is B
Solution and Explanation
Same radius and tension, so \(\mu\propto\rho\): \(\mu_2=4\mu_1\). Wave speed \(v=\sqrt{T/\mu}\Rightarrow v_1=2v_2\).
O is midway, so both segments have the same length \(L\); each segment has nodes at both ends, so \(L=n\frac{\lambda}{2}=\frac{nv}{2f}\), i.e. \(n=\frac{2fL}{v}\propto\frac1v\) (same \(f\), same \(L\)).
\(\frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac12\).
Answer: (2) \(1:2\)
O is midway, so both segments have the same length \(L\); each segment has nodes at both ends, so \(L=n\frac{\lambda}{2}=\frac{nv}{2f}\), i.e. \(n=\frac{2fL}{v}\propto\frac1v\) (same \(f\), same \(L\)).
\(\frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac12\).
Answer: (2) \(1:2\)
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