Sound Intensity: Reduce by 35 dB, Factor Calculation

AI Thread Summary
To determine the factor by which sound intensity is reduced when the level decreases by 35 dB, the equation Δβ=10Log(I/I°) is used. Substituting -35 dB into the equation leads to -3.5=log(I/I°), which simplifies to I/I°=10^-3.5. This calculation indicates that the sound intensity is reduced by a factor of approximately 3162. The correct interpretation is that the sound intensity is reduced by a factor of 3162, confirming the calculations are accurate.
Faraz
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Homework Statement


If you install special sound-reflecting windows that reduce the sound intensity level by 35.0 dB , by what factor have you reduced the sound intensity?

Homework Equations


Δβ=10Log(I/I°)

The Attempt at a Solution


I°=10^-12
I=I°-35
35=10^1.5440680
(I/I°)={(10^-12)-(10^1.5440680)}/{10^-12} =-3.5*10^13
But this is wrong
 
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Faraz said:

Homework Statement


If you install special sound-reflecting windows that reduce the sound intensity level by 35.0 dB , by what factor have you reduced the sound intensity?

Homework Equations


Δβ=10Log(I/I°)

The Attempt at a Solution


I°=10^-12
I=I°-35
35=10^1.5440680
(I/I°)={(10^-12)-(10^1.5440680)}/{10^-12} =-3.5*10^13
But this is wrong

Welcome to the PF.

I didn't follow some of what you did. Instead, start with substituting the -35dB into your first Relevant Equation:

-35dB=10Log(I/I°)

Now show the steps you use to solve for the fraction I/Io... :smile:
 
-3.5=log(I/I°)
10^-3.5=(I/I°)
but still wrong
 
Faraz said:
-3.5=log(I/I°)
10^-3.5=(I/I°)
but still wrong

It should be right, IMO. Do you get about 1/3162 as the answer?
 
Maybe the answer is supposed to be stated as "reducing the sound intensity by a factor of 3162..."?
 

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