Two wires with different densities are joined at $x = 0$ . An incident wave $Y_{i} = A_{i} sin (ω t - k_{1} x)$ travelling from left to right is partially reflected and partially transmitted $(Y_{t} = A_{t} sin (ω t - k_{2} x))$ at $x = 0$ . If the amplitude of transmitted wave is $A_{t}$ then $^{A_{t}} /_{A_{i}}$ is

\frac{2 k_{1}}{k_{1} + k_{2}}

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\frac{2 k_{2}}{k_{1} + k_{2}}

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\frac{k_{1} - k_{2}}{k_{1} + k_{2}}

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\frac{\sqrt{k_{1}} - \sqrt{k_{2}}}{\sqrt{k_{1}} + \sqrt{k_{2}}}

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Solution

The correct option is A $\frac{2 k_{1}}{k_{1} + k_{2}}$
Let the equation of reflected wave be $A_{r} s i n (ω t + k_{1} x)$
The sign is reversed to include the reversed direction of the reflected wave.
From continuity of the wave at the interface $x = 0$ , we get,
$A_{i} + A_{r} = A_{t}$
Also the first derivative of the wave are continuous at the boundary, hence,
$k_{1} (A_{i} - A_{r}) = k_{2} A_{t}$
$⟹ A_{i} - A_{r} = \frac{k_{2}}{k_{1}} A_{t}$
Rearranging the two equations gives,
$\frac{A_{t}}{A_{i}} = \frac{2 k_{1}}{k_{1} + k_{2}}$

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Two wires with different densities are joined at x=0. An incident wave Yi=Aisin(ωt−k1x) travelling from left to right is partially reflected and partially transmitted (Yt=Atsin(ωt−k2x)) at x=0. If the amplitude of transmitted wave is At then At/Ai is