How Does Sound Intensity Change with Fewer Firecrackers?

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SUMMARY

The sound level produced by two firecrackers firing simultaneously at a certain location is 95 dB. When one firecracker is used instead of two, the sound intensity decreases by 3 dB, resulting in a sound level of 92 dB. This conclusion is based on the principle that doubling the intensity of sound increases the decibel level by 3 dB. The formula used for calculations is dB = 10 LOG I + 120, where I represents intensity.

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  • Familiarity with logarithmic equations
  • Knowledge of the relationship between intensity and decibel changes
  • Basic algebra skills for solving equations
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Alicek
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Homework Statement


if two firecrackers produce a sound level of 95 dB when fired simultaneously at a certain place, what will the sound level be if only one is exploded?


Homework Equations



dB=10 LOG I + 120

The Attempt at a Solution


would I use 2I instead of I making an equation like 95 = 10 log 2I + 120 and then solve for one I?
Or set up the usual equation and then once I solve for I just divide it in half?
Also, I know that when intensity doubles, it goes up by 3 decibels, so do I just assume that two firecrackers is double the intensity of 1 and just subtract 3 decibels from the original 95?
 
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Alicek said:
Also, I know that when intensity doubles, it goes up by 3 decibels, so do I just assume that two firecrackers is double the intensity of 1 and just subtract 3 decibels from the original 95?

That's right.
 

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