What is the optimal distance to increase sound intensity?

AI Thread Summary
To increase the sound intensity from 25.0 dB to 80.0 dB, a significant increase in intensity is required, specifically a 45 dB increase, which corresponds to over 10,000 times the original intensity. The relationship between distance and intensity follows the inverse square law, meaning that as distance decreases, intensity increases. Doubling the intensity results in a 3 dB gain, while a 20 dB increase requires 100 times the intensity. The calculations suggest that the optimal distance to achieve the desired sound level will be less than 2.5 meters from the source. Understanding the logarithmic nature of decibels is crucial for determining the necessary adjustments in distance.
MrButtPutts
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You are trying to overhear a juicy conversation, but from your distance of 25.0 m , it sounds like only an average whisper of 25.0 dB . So you decide to move closer to give the conversation a sound level of 80.0 dB instead.





how close should you come?

d= ?


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ok i just read what is expected of me when i post tha question, i am new at this. i have worked on this through my textbook and notes. all i got right now is d1/I1=d2/I2 1=d2/80. i don't know...
 
well two things you do need to know that decibels is a logarithmically based system, so that doubling the intensity results in a 3 dB gain. (log 2 times 10).
If you wanted a 20 db level increase, requires 100 times the intensity.

A 45 dB increase is over 10,000 times the intensity. For power which is proportional to amplitude squared, the figures scale logarithmically but such that a doubling in power is 6dB increase.

So the trick is to decide which is the quantity of interest and how it varies with distance. Then one can determine the change in distance. Off the top of my head guessing the answer will be less than 2.5m.
 
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