Fireworks Sound level at difference distances

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SUMMARY

The discussion centers on calculating the sound level of fireworks at various distances using the formula I=P^2/2dv. The user calculated an intensity of 92.77 dB, adjusting for environmental factors to arrive at a final estimate of 64.77 dB, which was noted to be lower than the expected 68.3 dB. The conversation highlights the importance of accounting for sound absorption over distance, specifically noting a 3.5 dB increase due to air absorption over 3.5 km. This adjustment aligns the calculated sound level with real-world observations.

PREREQUISITES
  • Understanding of sound intensity formulas, specifically I=P^2/2dv.
  • Familiarity with logarithmic calculations, particularly 10log(I/I0).
  • Knowledge of sound absorption principles in atmospheric conditions.
  • Basic physics concepts related to sound propagation and distance effects.
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  • Research sound intensity calculations using I=P/A and its applications.
  • Explore the effects of atmospheric conditions on sound propagation.
  • Learn about sound absorption coefficients and their impact on sound levels.
  • Investigate practical applications of sound level measurements in outdoor environments.
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Acoustics engineers, physicists, event planners, and anyone involved in sound measurement and analysis in outdoor settings.

norcal36
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Homework Statement
A firework goes off high above you, a distance d1=500m from your head, and you hear an acoustic pressure of 10Pa. Your sister is a distance d2=4.00x10^3 at the same time. If air absorbs 7.00 db/km in sound energy what sound level does your sister hear? Density of air=1.2kg/m^3....Speed of sound=343m/s.
Relevant Equations
I=pmax^2/2*density*velocity
I=power/Area
10log(I/I0)
This is the work I've done so far...

I=p^2/2dv...I=(10)^2/2(1.2)(343)=.12112

I(A)=P...(.12112)(4pi(500)^2)=380522.366

I=P/A...380522.366/4pi(4000)^2=.0018925

10log(I/I0)...10log(.0018925/10^-12)=92.77db

92.77-7(4)=64.77db

I am winging it and that's the closest I can get to the right answer which is 68.3db...

Any Guidance would be greatly appreciated!
 
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The 10 Pa at your place already had 3.5 dB absorption from the air in the first 0.5 km. You only need to consider 3.5 km more, that increases the sound level by 3.5 dB relative to your calculation and produces a perfect match.
 

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