Two wires A and B of lengths in the ratio 1: | vedclass

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(C) The fundamental frequency $f$ of a stretched wire is given by the formula:$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$where $T$ is the tension,$L$ is t

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$1: 1$

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(C) The fundamental frequency $f$ of a stretched wire is given by the formula:$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$where $T$ is the tension,$L$ is the length,and $\mu$ is the mass per unit length.Given that the tension $T$ is the same for both wires,the ratio of the fundamental frequencies of wires $A$ and $B$ is:$\frac{f_A}{f_B} = \frac{\frac{1}{2L_A} \sqrt{\frac{T}{\mu_A}}}{\frac{1}{2L_B} \sqrt{\frac{T}{\mu_B}}} = \frac{L_B}{L_A} \sqrt{\frac{\mu_B}{\mu_A}}$We are given the ratio of lengths $\frac{L_A}{L_B} = \frac{1}{2}$,so $\frac{L_B}{L_A} = \frac{2}{1}$.The mass per unit length $\mu$ is defined as $\mu = \frac{m}{L}$. Therefore,the ratio $\frac{\mu_A}{\mu_B}$ is:$\frac{\mu_A}{\mu_B} = \frac{m_A / L_A}{m_B / L_B} = \left( \frac{m_A}{m_B} ight) \left( \frac{L_B}{L_A} ight)$Given $\frac{m_A}{m_B} = \frac{2}{1}$ and $\frac{L_B}{L_A} = \frac{2}{1}$,we have:$\frac{\mu_A}{\mu_B} = \left( \frac{2}{1} ight) \left( \frac{2}{1} ight) = \frac{4}{1}$Thus,$\frac{\mu_B}{\mu_A} = \frac{1}{4}$.Substituting these values into the frequency ratio formula:$\frac{f_A}{f_B} = \left( \frac{2}{1} ight) \sqrt{\frac{1}{4}} = 2 	imes \frac{1}{2} = 1$Therefore,the ratio of the fundamental frequencies is $1: 1$.

Two wires $A$ and $B$ of lengths in the ratio $1: 2$ and masses in the ratio $2: 1$ are stretched by the same tension. The ratio of the fundamental frequencies of wires $A$ and $B$ is

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