Question #24914
Q3. Calculate the difference between sound levels measured at 10m and at 63 m from a perfect line source of sound.
Expert's answer
We know that for a line source of sound, the level decreasesby 3 dB per doubling of distance from it.
Thus, compared to the level at 10 m, we know the level will be 3 dB lower at 20
m, and 6 dB lower at 40 m, and 9 dB lower at 80 m. Expressing this as a
logarithmic dependence, the difference in levels is
difference = (3 dB) * log2(D2/D1),
where log2 is a logarithm with the base 2, D2 is the analyzed distance and D1
is the reference distance (in our case, 10 m).
To know the attenuation at 63 m,
difference = (3 dB) * log2(63/10) = 7.966 dB
Answer: the sound is 7.966 dB weaker at 63 m than at 10 m.
Thus, compared to the level at 10 m, we know the level will be 3 dB lower at 20
m, and 6 dB lower at 40 m, and 9 dB lower at 80 m. Expressing this as a
logarithmic dependence, the difference in levels is
difference = (3 dB) * log2(D2/D1),
where log2 is a logarithm with the base 2, D2 is the analyzed distance and D1
is the reference distance (in our case, 10 m).
To know the attenuation at 63 m,
difference = (3 dB) * log2(63/10) = 7.966 dB
Answer: the sound is 7.966 dB weaker at 63 m than at 10 m.
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