Sound intensity of a thunderbolt

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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


[/B]
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 r1 being 1 km and r2 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.
 

Answers and Replies

  • #2
TSny
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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.
Your calculations look correct. Yes, the answer obviously does not correspond to reality. An assumption was made in this problem that might be unrealistic.

See Table B.2 here: https://ccrma.stanford.edu/~jos/pasp/Air_Absorption.html

Or play with some values here: http://resource.npl.co.uk/acoustics/techguides/absorption/
 
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