Question

In a standing wave formed as a result of reflection from a surface, the ratio of the amplitude at an antinode to that at node is x. The fraction of energy that is reflected is:
A. \[{\left[ {\dfrac{{x - 1}}{x}} \right]^2}\]
B. \[{\left[ {\dfrac{x}{{x + 1}}} \right]^2}\]
C. \[{\left[ {\dfrac{{x - 1}}{{x + 1}}} \right]^2}\]
D. \[{\left[ {\dfrac{1}{x}} \right]^2}\]

Answer 1

Hint: When two waves interfere then a stationary or standing wave is formed. In the question relation between the amplitude and x is given. As we know that the energy transported by any wave is directly proportional to the square of the amplitude. By using this concept, we can easily find the value for energy reflected.








Complete answer:
It is given that the ratio of the amplitude at an antinode to that at node is x.
 \[\dfrac{{{A_i} + {A_r}}}{{{A_i} - {A_r}}} = x\]
Where \[{A_i}\] is the amplitude of incident wave and \[{A_r}\] is the amplitude of reflected waves.
By applying componendo and dividendo on both the sides, we get
\[\dfrac{{{A_r}}}{{{A_i}}} = \dfrac{{x - 1}}{{x + 1}}\]
As we know that energy that is reflected is directly proportional to the square of the amplitude.
\[E \propto {A^2}\]
\[\dfrac{{{E_r}}}{{{E_i}}} = {\left( {\dfrac{{{A_r}}}{{{A_i}}}} \right)^2} = {\left( {\dfrac{{x - 1}}{{x + 1}}} \right)^2}\]
Or \[\dfrac{{{E_r}}}{{{E_i}}} = {\left( {\dfrac{{x - 1}}{{x + 1}}} \right)^2}\]
Therefore, the fraction of energy that is reflected is \[{\left( {\dfrac{{x - 1}}{{x + 1}}} \right)^2}\]
Hence option C is the correct answer






Note: The energy (E) transported by a wave is directly proportional to the square of the amplitude (A) that is \[E \propto {A^2}\] . So whenever change occurs in the amplitude the square of that effect impacts the energy. This means that a doubling of the amplitude results in a quadrupling of the energy. The amplitude of a wave is defined as the distance from the centre lines to the top of a crest to the bottom of a trough.