1. Suppose that a jet engine at 50 meters has a decibel level of 130, and a normal conversation at 1 meter has a decibel level of 60. What is the ratio of the intensities of the two sounds?
we know that the decibel scale of loudness is $$L=10\log (\frac{I}{I_0})$$
$$I_0=10^{−12} \frac {watt}{m^2}$$ is the softest audible sound at 1000 hertz
2. Suppose that the intensity of sound is proportional to the inverse square of the distance from the sound. Based on this rule, calculate the decibel level of the sound from the jet at a distance of 100 meters
in the first question, do I need to have both decibels at the same amount of meters?
If I take as it is (130 and 60) the difference is 70, there fore $$70=10\log (\frac{I}{I_0})$$ $$7=\log (\frac{I}{I_0})$$ $$10^7=\frac{I}{I_0}$$ hence $10000000:1$
is it correct?
and in the second question. $$I=\frac{1}{d^2}$$, where d – distance
$$I=\frac{1}{10^4}=10^{-4}$$ $$L=10\log (\frac{10^{-4}}{10^{-12}})$$ $$L=10\log (10^{8})=10\cdot8=80$$
Best Answer
Let $I_i$ be the intensity at 50m and $I_f$ be the intensity at 100m.
The intensity is proportional to the inverse of the square, but not equal to it. At 50m, we have $L=10log(\frac{I_i}{I_0})=130$. From that, $I_i=10$.
Because it is proportional to the inverse of the square of the distance, $\frac{I_f}{I_i} = \frac{d_i^2}{d_f^2} = \frac{1}{4}$. Then, $I_f = \frac{10}{4}=2.5$.
Now, $L=10log(\frac{I_f}{I_0})=10log(\frac{2.5}{10^{-12}})$ which is roughly $124dB$.
Everything else looks fine to me.