# Two wires W1 and W2 have the same radius r and respective densities ρ1 and ρ2 such that ρ2=4ρ1.  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 W1 to W2 is :   Option: 1  1 : 1 Option: 2  1 : 2 Option: 3  1 : 3 Option: 4 4 : 1

When the joint is found to be a node.

then the frequency is given as

$n=\frac{p}{2 l} \sqrt{\frac{T}{\pi r^{2} d}}$

Where p is also equal to the number of antinodes formed  in the wire

As

\begin{aligned} &n_{1}=n_{2}\\ &T \rightarrow \text { same }\\ &r \rightarrow \text { same }\\ &l \rightarrow \text { same } \end{aligned}

So

\begin{aligned} &\frac{p_{1}}{\sqrt{d_{1}}}=\frac{p_{2}}{\sqrt{d_{2}}}\\ &\frac{p_{1}}{p_{2}}=\frac{1}{2} \end{aligned}

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