Question

Two waves of amplitudes ${A_0}$ and $x{A_0}$ pass through a region. If $x > 1$, what is the value difference in the maximum and minimum resultant amplitude is:
A) $\left( {x + 1} \right){A_0}$
B) $\left( {x - 1} \right){A_0}$
C) $2x{A_0}$
D) $2{A_0}$

Answer 1

We know that if two similar waves are moving in the same direction, with the same frequency, wavelength and amplitude, the waves are added together in the same phase. Then we can apply the value of the waves and find the difference in amplitude.

Formula used:
Resultant amplitude,
${A_r} = \sqrt {A_1^2} + \sqrt {A_2^2} + 2{A_1}{A_2}\cos \phi$
Where,
If two waves of amplitude ${A_1}$ and ${A_2}$ phase difference between them is $\phi$
Resultant amplitude ${A_r}$

Complete step by step solution:
Given by,
Two waves of amplitudes ${A_0}$ and $x{A_0}$ pass through a region.
We find the value difference in the maximum and minimum resultant amplitude
We know that,
If two waves of amplitude ${A_1}$ and ${A_2}$ phase difference between them is $\phi$
Resultant amplitude ${A_r}$ is given by,
${A_r} = \sqrt {A_1^2} + \sqrt {A_2^2} + 2{A_1}{A_2}\cos \phi$
Case 1: at $\phi = {0^ \circ }$
We get,
Maximum resultant amplitude
Here, ${A_{\max }} = \left| {{A_1} + {A_2}} \right|$

Case 2:
We get, Minimum amplitude
Here, ${A_{\min }} = \left| {{A_1} - {A_2}} \right|$
Let as assume,
${A_1} = x{A_0}$ and ${A_2} = {A_0}$
Here, We substituting value in a above equation,
Maximum amplitude,
we get, ${A_{\max }} = \left| {x{A_0} + {A_0}} \right|$
On simplifying,
$\Rightarrow {A_{\max }} = \left| {\left( {x + 1} \right){A_0}} \right|$
Minimum amplitude, we get,
$\Rightarrow {A_{\min }} = \left| {x{A_0} - {A_0}} \right|$
On simplifying,
$\Rightarrow {A_{\min }} = \left| {\left( {x - 1} \right){A_0}} \right|$
The difference in resultant amplitude is ${A_{\max }} = \left| {\left( {x + 1} \right){A_0}} \right|$ and ${A_{\min }} = \left| {\left( {x - 1} \right){A_0}} \right|$
The difference is $2{A_0}$.

Hence, The option D is the correct answer.

Note: If two waves with the same frequency and a constant difference in phase. There is a reduction of energy gain, the energy is redistributed and the distribution shifts with the period the energy is redistributed and the distribution stays constant in time. We continue to follow the maximum and minimum value of the phase in this form of a problem and then find the difference.