Certainly! Let's solve the problem step-by-step:
1. Initial Conditions:
- Initial distance from the roaring truck engine: [tex]\( d_1 = 8 \)[/tex] meters
- Initial sound intensity at that distance: [tex]\( I_1 = 20 \frac{W}{m^2} \)[/tex]
2. Change in Distance:
- The student moves 4 meters closer to the engine.
- New distance: [tex]\( d_2 = 8 \, \text{meters} - 4 \, \text{meters} = 4 \, \text{meters} \)[/tex]
3. Relationship Between Sound Intensity and Distance:
- The sound intensity [tex]\( I \)[/tex] is inversely proportional to the square of the distance [tex]\( d \)[/tex]. Mathematically, this is represented as:
[tex]\[
I \propto \frac{1}{d^2}
\][/tex]
- Hence, we can write this relationship as:
[tex]\[
I_1 \cdot d_1^2 = I_2 \cdot d_2^2
\][/tex]
where [tex]\( I_1 \)[/tex] and [tex]\( d_1 \)[/tex] are the initial intensity and distance, and [tex]\( I_2 \)[/tex] and [tex]\( d_2 \)[/tex] are the new intensity and distance.
4. Solving for New Intensity [tex]\( I_2 \)[/tex]:
- Rearranging the proportion to solve for [tex]\( I_2 \)[/tex]:
[tex]\[
I_2 = I_1 \cdot \left( \frac{d_1}{d_2} \right)^2
\][/tex]
- Substituting the known values:
[tex]\[
I_2 = 20 \frac{W}{m^2} \cdot \left( \frac{8 \, \text{m}}{4 \, \text{m}} \right)^2
\][/tex]
- Simplify the fraction inside the parentheses:
[tex]\[
I_2 = 20 \frac{W}{m^2} \cdot (2)^2
\][/tex]
- Calculate the squared term:
[tex]\[
I_2 = 20 \frac{W}{m^2} \cdot 4
\][/tex]
- Finally, multiply:
[tex]\[
I_2 = 80 \frac{W}{m^2}
\][/tex]
So, the measured sound intensity at the new distance of 4 meters is [tex]\( 80 \frac{W}{m^2} \)[/tex].
