Answer :
Sure! Let's solve the problem step-by-step.
The intensity of sound follows the inverse square law, which states that the intensity of sound is inversely proportional to the square of the distance from the source. Mathematically, this can be expressed as:
[tex]\[ I \propto \frac{1}{d^2} \][/tex]
where:
- [tex]\( I \)[/tex] is the intensity of the sound,
- [tex]\( d \)[/tex] is the distance from the sound source.
Given:
- Distance for man 1 ([tex]\( d_1 \)[/tex]): 100 meters
- Distance for man 2 ([tex]\( d_2 \)[/tex]): 300 meters
To find the ratio of the sound intensities [tex]\( I_1 \)[/tex] (intensity at 100 meters) and [tex]\( I_2 \)[/tex] (intensity at 300 meters), we can use the inverse square law relationship.
Let's denote:
[tex]\[ I_1 = \frac{k}{d_1^2} \][/tex]
[tex]\[ I_2 = \frac{k}{d_2^2} \][/tex]
where [tex]\( k \)[/tex] is a constant.
To find the ratio [tex]\( \frac{I_1}{I_2} \)[/tex]:
[tex]\[ \frac{I_1}{I_2} = \frac{\frac{k}{d_1^2}}{\frac{k}{d_2^2}} \][/tex]
Since [tex]\( k \)[/tex] is constant, it cancels out:
[tex]\[ \frac{I_1}{I_2} = \frac{d_2^2}{d_1^2} \][/tex]
Now, substitute the given distances ([tex]\( d_1 = 100 \)[/tex] meters and [tex]\( d_2 = 300 \)[/tex] meters) into the equation:
[tex]\[ \frac{I_1}{I_2} = \frac{300^2}{100^2} \][/tex]
Calculate the squares:
[tex]\[ 300^2 = 90000 \][/tex]
[tex]\[ 100^2 = 10000 \][/tex]
Thus:
[tex]\[ \frac{I_1}{I_2} = \frac{90000}{10000} = 9 \][/tex]
So, the ratio of the sound intensity for a man standing 100 meters away to a man standing 300 meters away from the same bell is:
[tex]\[ \frac{I_1}{I_2} = 9 \][/tex]
Therefore, the sound intensity at 100 meters is 9 times greater than the sound intensity at 300 meters.
The intensity of sound follows the inverse square law, which states that the intensity of sound is inversely proportional to the square of the distance from the source. Mathematically, this can be expressed as:
[tex]\[ I \propto \frac{1}{d^2} \][/tex]
where:
- [tex]\( I \)[/tex] is the intensity of the sound,
- [tex]\( d \)[/tex] is the distance from the sound source.
Given:
- Distance for man 1 ([tex]\( d_1 \)[/tex]): 100 meters
- Distance for man 2 ([tex]\( d_2 \)[/tex]): 300 meters
To find the ratio of the sound intensities [tex]\( I_1 \)[/tex] (intensity at 100 meters) and [tex]\( I_2 \)[/tex] (intensity at 300 meters), we can use the inverse square law relationship.
Let's denote:
[tex]\[ I_1 = \frac{k}{d_1^2} \][/tex]
[tex]\[ I_2 = \frac{k}{d_2^2} \][/tex]
where [tex]\( k \)[/tex] is a constant.
To find the ratio [tex]\( \frac{I_1}{I_2} \)[/tex]:
[tex]\[ \frac{I_1}{I_2} = \frac{\frac{k}{d_1^2}}{\frac{k}{d_2^2}} \][/tex]
Since [tex]\( k \)[/tex] is constant, it cancels out:
[tex]\[ \frac{I_1}{I_2} = \frac{d_2^2}{d_1^2} \][/tex]
Now, substitute the given distances ([tex]\( d_1 = 100 \)[/tex] meters and [tex]\( d_2 = 300 \)[/tex] meters) into the equation:
[tex]\[ \frac{I_1}{I_2} = \frac{300^2}{100^2} \][/tex]
Calculate the squares:
[tex]\[ 300^2 = 90000 \][/tex]
[tex]\[ 100^2 = 10000 \][/tex]
Thus:
[tex]\[ \frac{I_1}{I_2} = \frac{90000}{10000} = 9 \][/tex]
So, the ratio of the sound intensity for a man standing 100 meters away to a man standing 300 meters away from the same bell is:
[tex]\[ \frac{I_1}{I_2} = 9 \][/tex]
Therefore, the sound intensity at 100 meters is 9 times greater than the sound intensity at 300 meters.
