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Q.

Two wires of different linear mass densities are soldered together end to end are then stretched under a tension . The wave speed in the first wire is twice in the second wire. Fraction of the incident power that is reflected at the junction is $\frac{1}{β}$ . Find the value of $β$ .

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answer is 9.

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Detailed Solution

We know fraction of energy transmitted $= 1 - {(\frac{A_{r}}{A_{i}})}^{2}$
$\begin{array}{l} ∴ f_{r} = Fraction of energy reflected = {(\frac{A_{r}}{A_{i}})}^{2} \\ A_{r} = {(\frac{v_{2} - v_{1}}{v_{1} + v_{2}})}^{2} \cdot A_{i} = {(\frac{v - 2 v}{2 v + v})}^{2} \cdot A_{i} = \frac{A_{i}}{3} ∴ f_{r} = {(\frac{- 1 / 3}{1})}^{2} = \frac{1}{9} \Rightarrow β = 9 \end{array}$

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