Answer:
See proof below.
Explanation:
For this case we need to show that [tex] A \propto \frac{1}{\sqrt{r}}[/tex]
Let's assume that we have two concentric circles. We need to assume that we have the same energy passing through the two concentric circles.
We know that the intensity is defined as:
[tex] I = \frac{P}{2 \pi r}[/tex]
We know that [tex] I =\frac{P}{A}[/tex] and [tex] P = \frac{Energy}{t}[/tex], so then:
[tex] I = \frac{E}{At}[/tex]
Where P represent the power and r the radius.
We know that the intensity is [tex] I \propto A^2[/tex] proportional to the amplitude square. So using transitivity we have that:
[tex] I \propto A^2 \propto \frac{P}{2\pi r}[/tex]
Since we are assuming that P is a constant and if we take the square root we got that:
[tex] A \propto \frac{1}{\sqrt{r}}[/tex]
