A wave represented by $y = 100\sin \left( {ax + bt} \right)$ is reflected from a dense plane at the origin. If $36\% $ energy is lost and the rest of the energy is reflected then the equation of the reflected wave will be:
A. $y = - 80\sin \left( {ax + bt} \right)$
B. $y = - 8.1\sin \left( {ax + bt} \right)$
C. $y = - 10\sin \left( {ax + bt} \right)$
D. $y = - 8.2\sin \left( {ax + bt} \right)$

Hint: You can start by calculating what percentage energy of the original wave remains. Then calculate the intensity of the reflected wave by using the equation, i.e. ${I_2} = \dfrac{{64}}{{100}} \times {I_1}$ . Then use the equation $I \propto {A^2}$ to find the new amplitude of the wave. Remember that the direction of the reflected wave is opposite to that of the original wave.

Complete answer:
In the problem, we are given that when the wave (represented by $y = 100\sin \left( {ax + bt} \right)$ ) strikes a dense plane at the origin it is reflected. During the reflection $36\% $ of the energy of the wave is absorbed and the rest of the energy is reflected.

This means that the wave now has $\left( {100 - 36} \right)\% = 64\% $ its original energy.
The general equation for a wave is $y = A\sin \left( {ax + bt} \right)$ . So if we compare it with the equation given to us, we get
$A = 100$
Now, we know that the energy of a wave is represented by intensity and intensity is dependent on the amplitude
$I \propto {A^2}$
$I = k{A^2}$
Here, $I = $ The intensity of the wave
$A = $ The amplitude of the wave
$k = $ Proportionality constant

So, if the energy of a wave is absorbed then the amplitude of the wave would also reduce.
Let the amplitude of the original wave be ${A_1}$ and the amplitude of the reflected wave be ${A_2}$ . Also, let the intensity of the original wave be ${I_1}$ and the intensity of the new wave be ${I_2}$ .
So, for the original wave
${I_1} = kA_1^2$
And for the reflected wave
${I_2} = kA_2^2$
As we already know
$\Rightarrow {I_2} = \dfrac{{64}}{{100}} \times {I_1}$
$\Rightarrow kA_2^2 = \dfrac{{64}}{{100}} \times kA_1^2$
$\Rightarrow {A_2} = \dfrac{8}{{10}} \times {A_1}$
$\Rightarrow {A_2} = \dfrac{8}{{10}} \times 100$
$\Rightarrow {A_2} = 80$

As we discussed above the general form for the equation for the wave is $y = A\sin \left( {ax + bt} \right)$ .
Since the reflected wave is going in the opposite direction of the original wave, the equation of the reflected wave becomes

$y = 80\sin \left( { - ax + bt} \right)$
$y = - 80\sin \left( {ax - bt} \right)$

So, the correct answer is “Option A”.

Note:
This problem is based on a very important concept. While dealing with reflections or redfractions we generally assume that no energy of the wave is lost, but in reality, all reflections and refractions reduce the energy of waves as no material exists to date that can reflect or refract a wave with $100\% $ of its energy.