To solve this question, you need a solid understanding of how sound intensity level (in decibels) relates to sound intensity (power per unit area), and how this intensity changes as we move away from a point sound source.
Sound intensity level, [tex]\( L \)[/tex], in decibels (dB) can be calculated from sound intensity [tex]\( I \)[/tex] using the following formula:
[tex]\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
where [tex]\( L \)[/tex] is the sound intensity level in decibels, [tex]\( I \)[/tex] is the sound intensity, and [tex]\( I_0 \)[/tex] is the reference sound intensity, typically taken as the threshold of hearing, [tex]\( 10^{-12} \)[/tex] W/m².
As per the inverse-square law for sound, the intensity of sound changes inversely with the square of the distance from the source. If you double the distance from a source, the intensity becomes one fourth of its original value.
Given in the question:
- Initial intensity level [tex]\( L_1 \)[/tex] is 110 dB at a distance [tex]\( d_1 \)[/tex] of 4.0 m.
- Final intensity level [tex]\( L_2 \)[/tex] is 100 dB, and we need to find the distance [tex]\( d_2 \)[/tex] where this occurs.
First step is to use the decibel formula to find the ratio of the sound intensities corresponding to [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex]:
[tex]\[ \Delta L = L_1 - L_2 \][/tex]
[tex]\[ \Delta L = 110 \, dB - 100 \, dB = 10 \, dB \][/tex]
The change in level of 10 dB corresponds to a tenfold change in intensity, as 10 dB = 10 log10(10I/I0)/log10(I/I0) = 10 * 1.
Now, using the inverse-square law:
[tex]\[ \frac{I_1}{I_2} = \left(\frac{d_2}{d_1}\right)^2 \][/tex]
Given that [tex]\( \frac{I_1}{I_2} = 10 \)[/tex] (since the sound intensity level difference is 10 dB), we can write:
[tex]\[ 10 = \left(\frac{d_2}{d_1}\right)^2 \][/tex]
Now solve for [tex]\( d_2 \)[/tex]:
[tex]\[ \sqrt{10} = \frac{d_2}{d_1} \][/tex]
[tex]\[ d_2 = d_1 \cdot \sqrt{10} \][/tex]
[tex]\[ d_2 = 4.0 \, m \cdot \sqrt{10} \][/tex]
[tex]\[ d_2 = 4.0 \, m \cdot 3.162 \, (approximately) \][/tex]
[tex]\[ d_2 ≈ 12.65 \, m \][/tex]
At a distance of approximately 12.65 meters from the point sound source, the sound intensity level will be 100 dB.
