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 waves is set up in the composite wire, the joint is found to be a node. The ratio of the number of an tin odes formed in \(W_1\) to \(W_2\) is

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Answer & Solution

Correct Answer

(B) \(1 : 2\)

Detailed explanation

\(\begin{array}{l}
{n_1} = {n_2}\\
T \to Same\\
r \to Same\\
l \to Same
\end{array}\)

Frequency of vibration

\(\mathrm{n}=\frac{\mathrm{p}}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}^{2} \rho}}\)

As \(\mathrm{T}, \mathrm{r},\) and \(l\) are same for both the wires

\(n_{1}=n_{2}\)

\(\frac{p_{1}}{\sqrt{\rho_{1}}}=\frac{p_{2}}{\sqrt{p_{2}}}\)

\(\frac{p_{1}}{p_{2}}=\frac{1}{2} \quad \because \rho_{2}=4 \rho_{1}\)

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