What Is the Ratio of the New Sound Level to the Old?

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The discussion focuses on calculating the ratio of sound intensity levels when the intensity decreases from 90 dB to 75 dB. The relevant formula used is Sound Intensity Level = 10 log (I/Io). A participant initially miscalculated the ratio but later clarified that a 15 dB difference corresponds to a 1.5-Bel difference, indicating a sound intensity ratio of 10^1.5. This means the new sound level is less intense, leading to a division by the calculated factor. The conversation emphasizes the importance of understanding the logarithmic nature of sound intensity levels.
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[SOLVED] Sound Level

Homework Statement



If sound intensity level is reduced from 90dB to 75 dB, what is the ratio of the new sound level to the old?

Homework Equations



Sound Intensity Level = 10 log (I/Io)


The Attempt at a Solution



10 log 90 = 1.954
10 log 75 = 1.875

1.875/1.954 = .956
This probably goes against every physical law that exists, but it's all I've come up with so far. Help, please!
 
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Its easy!

Its 1/10^2.5

Sorry that was for me!

Its 1/10^1.5 for you

90--->75

Mine was 90-65
 
Last edited:
Could you please explain why?
 
This 15 db difference corresponds to a 1.5-Bel difference. This difference is equivalent to a sound which is 10^1.5 more intense. Always raise 10 to a power which is equivalent to the difference in "Bels.

Or less intense in this case thus dividing by 1.
 
Thank you!
 
Now i sent you a pm! lol
 

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