To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.
The expression for sound power is,
[tex]P = AI[/tex]
Here,
A = Area
I = Intensity
P = Power
At the same time the area can be written as,
[tex]A = \frac{\pi d^2}{4}[/tex]
Now the intensity is inversely proportional to the square of the distance from the source, then
[tex]I \propto \frac{1}{r^2}[/tex]
The expression for the intensity at different distance is
[tex]\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}[/tex]
Here,
[tex]I_1[/tex] = Intensity at distance 1
[tex]I_2[/tex] = Intensity at distance 2
[tex]r_1[/tex] = Distance 1 from light source
[tex]r_2[/tex] = Distance 2 from the light source
If we rearrange the expression to find the intensity at second position we have,
[tex]I_2 = I_1 (\frac{r_1^2}{r_2^2})[/tex]
If we replace with our values at this equation we have,
[tex]I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})[/tex]
[tex]I_2 = 1.11*10^{-4} W/m^2[/tex]
Now using the equation to find the area we have that
[tex]A = \frac{\pi (8.4*10^{-3}m)^2}{4}[/tex]
[tex]A = 5.5*10^{-5}m^2[/tex]
Finally with the intensity and the area we can find the sound power, which is
[tex]P = AI[/tex]
[tex]P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)[/tex]
[tex]P = 6.1*10^{-9}J/s[/tex]
Power is defined as the quantity of Energy per second, then
[tex]E = 6.1*10^{-9}J[/tex]
