JEE Mains · Physics · STD 11 - 14. waves and sound
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
- A \(1:1\)
- B \(1 : 2\)
- C \(1 : 3\)
- D \(4 : 1\)
Answer & Solution
Correct Answer
(B) \(1 : 2\)
Step-by-step Solution
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…
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