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, \(\( L \)\), in decibels (dB) can be calculated from sound intensity \(\( I \)\) using the following formula:
\(\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \]\)
where \(\( L \)\) is the sound intensity level in decibels, \(\( I \)\) is the sound intensity, and \(\( I_0 \)\) is the reference sound intensity, typically taken as the threshold of hearing, \(\( 10^{-12} \)\) 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 \(\( L_1 \)\) is 110 dB at a distance \(\( d_1 \)\) of 4.0 m.
- Final intensity level \(\( L_2 \)\) is 100 dB, and we need to find the distance \(\( d_2 \)\) where this occurs.
First step is to use the decibel formula to find the ratio of the sound intensities corresponding to \(\( L_1 \)\) and \(\( L_2 \)\):
\(\[ \Delta L = L_1 - L_2 \]\)
\(\[ \Delta L = 110 \, dB - 100 \, dB = 10 \, dB \]\)
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:
\(\[ \frac{I_1}{I_2} = \left(\frac{d_2}{d_1}\right)^2 \]\)
Given that \(\( \frac{I_1}{I_2} = 10 \)\) (since the sound intensity level difference is 10 dB), we can write:
\(\[ 10 = \left(\frac{d_2}{d_1}\right)^2 \]\)
Now solve for \(\( d_2 \)\):
\(\[ \sqrt{10} = \frac{d_2}{d_1} \]\)
\(\[ d_2 = d_1 \cdot \sqrt{10} \]\)
\(\[ d_2 = 4.0 \, m \cdot \sqrt{10} \]\)
\(\[ d_2 = 4.0 \, m \cdot 3.162 \, (approximately) \]\)
\(\[ d_2 ≈ 12.65 \, m \]\)
At a distance of approximately 12.65 meters from the point sound source, the sound intensity level will be 100 dB.
