The intensity of sound is inversely proportional to the square of the distance from the source, according to the inverse square law. This means that if you move twice as far from the source of the sound, the intensity of the sound would be one-fourth. In mathematical terms, if [tex]\( I_1 \)[/tex] is the intensity at distance [tex]\( d_1 \)[/tex] and [tex]\( I_2 \)[/tex] is the intensity at distance [tex]\( d_2 \)[/tex], then
[tex]\[ \frac{I_1}{I_2} = \left(\frac{d_2}{d_1}\right)^2 \][/tex]
Given that the intensity of sound at a distance of 1 meter is 10 times as strong as at 8 meters, we're essentially saying:
[tex]\[ \frac{I_1}{I_8} = 10 \][/tex]
where [tex]\( I_1 \)[/tex] is the intensity at 1 meter and [tex]\( I_8 \)[/tex] is the intensity at 8 meters. According to the inverse square law:
[tex]\[ \frac{I_1}{I_8} = \left(\frac{8}{1}\right)^2 \][/tex]
[tex]\[ \frac{I_1}{I_8} = 64 \][/tex]
So, even if we are given that [tex]\( \frac{I_1}{I_8} = 10 \)[/tex], the intensity would actually increase by a factor of [tex]\( 64 \)[/tex] when moving from 8 meters to 1 meter if the intensity decreases solely by the inverse square law.
However, since the question already states that the intensity of sound is 10 times as strong at 1 meter as at 8 meters, and we are supposed to simplify this information, we accept it as a given condition without applying the inverse square law. Therefore, the simplified answer is:
The intensity of sound at a distance of 1 meter is 10 times greater than the intensity at a distance of 8 meters.
