String 1 is connected with string 2. The mass per unit length in string 1 is ${\mu _1}$ and the mass per unit length in string 2 is $4{\mu _1}$ . The tension in the string is $T$ . A travelling wave is coming from the left. What fraction of the energy in the incident wave goes into string 2?

A) $1/8$
B) $4/9$
C) $2/3$
D) $8/9$

Hint: Some part of the wave will be reflected back and only some part of it will be transmitted to B. First calculate the amplitude of this transmitted wave, and then calculate the power in each of the wires and then their ratio. This ratio will be equal to their energy ratio as time will be the same for both and hence will get cancelled out. This will be the final answer.

Formula Used:
Amplitude of transmitted wave, ${A_t} = \dfrac{{2{v_B}}}{{{v_a} + {v_B}}}{A_i}$ where, ${v_B}$ is the velocity of wave in string B, ${v_A}$is the velocity of wave in string A, ${A_i}$ is the incident amplitude, or the amplitude of wave in string A
Velocity of a wave in a string, $v = \sqrt {\dfrac{T}{\mu }} $ where, $T$ is the tension in the string and $\mu $ is the mass per unit length of the string.
Power of a wave, $P = \dfrac{1}{2}\mu v{\omega ^2}{A^2}$ where, $\mu $ is the mass per unit length of the string, $v$ is the velocity of the wave in string, $\omega $ is the angular frequency of the wave, $A$ is the amplitude of the wave.

Complete step by step solution:
The wave coming from string A will have an amplitude ${A_i}$ , called the incident amplitude. This wave will travel through the string until it meets string B, where a part of it will be reflected back in string A in the opposite direction of the incident wave, and the other part will propagate through B. This wave is called the transmitted wave and its amplitude, ${A_t}$ is called transmitted amplitude.
From the formula, we have amplitude of transmitted wave, ${A_t} = \dfrac{{2{v_B}}}{{{v_a} + {v_B}}}{A_i}$
Also, we know that velocity of a wave in a string, $v = \sqrt {\dfrac{T}{\mu }} $
Using these two formulas, we get ${A_t} = \dfrac{{2\sqrt {\dfrac{{{T_B}}}{{{\mu _B}}}} }}{{\sqrt {\dfrac{{{T_A}}}{{{\mu _A}}}} + \sqrt {\dfrac{{{T_B}}}{{{\mu _B}}}} }}{A_i}$ where ${T_B}$ is tension in string B, ${T_A}$ is tension in string A, ${\mu _A}$ is mass per unit volume of string A, ${\mu _B}$ is mass per unit volume of string B
Since it is given that tension in the strings is equal, we can write above equation as
${A_t} = \dfrac{{2\sqrt {\dfrac{T}{{{\mu _B}}}} }}{{\sqrt {\dfrac{T}{{{\mu _A}}}} + \sqrt {\dfrac{T}{{{\mu _B}}}} }}{A_i}$
Now we are given in the question that mass per unit length of string A, ${\mu _A} = {\mu _1}$ and mass per unit length of string B, ${\mu _B} = 4{\mu _1}$
Substituting these in above formula, we get \[{A_t} = \dfrac{{2\sqrt {\dfrac{T}{{4{\mu _1}}}} }}{{\sqrt {\dfrac{T}{{{\mu _1}}}} + \sqrt {\dfrac{T}{{4{\mu _1}}}} }}{A_i} = \dfrac{{\sqrt {\dfrac{T}{{{\mu _1}}}} }}{{\sqrt {\dfrac{T}{{{\mu _1}}}} + \dfrac{1}{2}\sqrt {\dfrac{T}{{{\mu _1}}}} }}{A_i}\] (4 comes out of the square root and becomes 2)
\[{A_t} = \dfrac{{\sqrt {\dfrac{T}{{{\mu _1}}}} }}{{\sqrt {\dfrac{T}{{{\mu _1}}}} (1 + \dfrac{1}{2})}}{A_i} = \dfrac{2}{3}{A_i}\] (simplifying the equation)
Now, the power of a wave in string is given by $P = \dfrac{1}{2}\mu v{\omega ^2}{A^2}$
Power in string A, ${P_A} = \dfrac{1}{2}{\mu _A}{v_A}{\omega ^2}{A_i}^2 = \dfrac{1}{2}{\mu _i}\sqrt {\dfrac{T}{{{\mu _i}}}} {\omega ^2}{A_i}^2$ (substituting the respective values)
Power in string A, ${P_B} = \dfrac{1}{2}{\mu _B}{v_B}{\omega ^2}{A_t}^2 = \dfrac{1}{2}4{\mu _i}\sqrt {\dfrac{T}{{4{\mu _i}}}} {\omega ^2}{A_t}^2$ (substituting the respective values)
Now, \[\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{\dfrac{1}{2}{\mu _i}\sqrt {\dfrac{T}{{{\mu _i}}}} {\omega ^2}{A_i}^2}}{{\dfrac{1}{2}4{\mu _i}\sqrt {\dfrac{T}{{4{\mu _i}}}} {\omega ^2}{A_t}^2}} = \dfrac{{\dfrac{1}{2}{\mu _i}\sqrt {\dfrac{T}{{{\mu _i}}}} {\omega ^2}{A_i}^2}}{{\dfrac{1}{2}4{\mu _i}\sqrt {\dfrac{T}{{4{\mu _i}}}} {\omega ^2}{{(\dfrac{2}{3}{A_i})}^2}}}\] (substituting the value of ${A_t}$ as calculated before)
On cancelling like terms and solving roots, we get \[\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{\sqrt {\dfrac{T}{{{\mu _i}}}} }}{{4 \times \dfrac{1}{2} \times \sqrt {\dfrac{T}{{{\mu _i}}}} \times \dfrac{4}{9}}} = \dfrac{1}{{\dfrac{8}{9}}} = \dfrac{9}{8}\]
We know that Energy, $E = P \times t$ , where $P$ is power and $t$ is time.
Time taken for both wires will be equal and hence, power will be equal to energy.
$ \Rightarrow \dfrac{{{E_A}}}{{{E_B}}} = \dfrac{9}{8}$ or, $\dfrac{{{E_B}}}{{{E_A}}} = \dfrac{8}{9}$
Therefore, $\dfrac{8}{9}$ parts of energy in string 1 goes to string 2.

Hence option D is the final answer.

Note: In questions like these, it may not always occur in mind what to do next. Hence, you might need to memorize some important steps of these questions as their steps are not related to each other. It will help you solve this question and also, be faster. Memorizing the power conversion step, or the velocity conversion step will help.