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A harmonic wave is travelling on string $1 .$ At a junction with string 2 it is partly reflected and partly transmitted. The linear mass density of the second string is four times that of the first string, and that the boundary between the two strings is at $x = 0$ . If the expression for the incident wave is
$y_{i} = A_{i} \cos (k_{1} x - ω_{1} t)$
If the average power carried by the incident wave , transmitted wave and reflected waves are $P_{i}$ , $P_{t} a n d P_{r} r e s p e c t i v e l y, t h e n$

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$P_{i} = P_{t} + P_{r}$

$P_{i} \neq P_{t} + P_{r}$

$P_{i} = P_{t} - P_{r}$

$P_{i} = P + P$

answer is A.

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Detailed Solution

The average power of a harmonic wave on a string is given by

$P = \frac{1}{2} ρ A^{2} ω^{2} S v = \frac{1}{2} A^{2} ω^{2} μ v (as ρ S = μ)$

Now,

$P_{i} = \frac{1}{2} ω_{}^{2} A_{}^{2} μ_{1} v_{1} \frac{}{}$ $\dots$ (i)

$A_{r} = \frac{v_{2} - v_{1}}{v_{1} + v_{2}} A_{i} = \frac{A_{i}}{3} a n d A_{t} = \frac{2 v_{2}}{v_{1} + v_{2}} = \frac{2 A_{i}}{3}$ ……..(ii)

$P_{r} = \frac{1}{2} ω_{2}^{2} {(- \frac{A_{i}}{3})}^{2} (μ_{1}) (v_{1}) = \frac{1}{18} ω_{1}^{2} A_{i}^{2} μ_{1} v_{1}$ … ..(iii)

and

$P_{t} = \frac{8}{18} ω^{2} {(A_{i})}^{2} μ_{1} V_{1} . . . . . . (i v)$

From Eqs. (i), (ii) (iii) and (iv) we can show that

$P_{i} = P_{t} + P_{r}$

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