- #1

Mulz

- 124

- 5

## Homework Statement

Find the sound intensity of a thunderbolt 1 km away. What sound intensity (dB) is it 1 km away? The thunderbolt can be heard up to 20 km away. The sound is spherical. The air absorbs nothing.

## Homework Equations

[tex] L = 10 \cdot lg(\frac{I_1}{I_0}) [/tex]

[tex] \Delta L = 10 \cdot lg(\frac{I_2}{I_1}) [/tex]

[tex] I = \frac{p_0 ^2}{2z} [/tex]

[tex] dim(I) = \frac{P}{4πr^2} [/tex]

## The Attempt at a Solution

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I went by the assumption that the sound dissipates at the distance of 20 km. I used equation 4 and found that

[tex] I_1 = \frac{p}{4πr_1^2} [/tex] and

[tex] I_2 = \frac{p}{4πr_2^2} [/tex] with r

_{1}being 1 km and r

_{2}20 km. I divided them and got [tex] \frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \frac{1}{400} [/tex] and implemented this in the second equation and got a difference of 26 dB.

This physically makes no sense. How can it be 26 dB difference at a 19 km difference? Whispering is louder. Assuming the sound is completely gone at 20 km, that would mean it's 26 dB at a distance of 1 km away from the lightning. This is obviously wrong and I cannot see the problem with my calculations.