Sound Intensity: Reduce by 35 dB, Factor Calculation

That would be the more common way to state it.Yes, I get about 1/3162 using logarithms.In summary, to find the factor by which the sound intensity is reduced when installing sound-reflecting windows that decrease the sound intensity level by 35.0 dB, one can use the equation Δβ=10Log(I/I°) and solve for the fraction I/I°, resulting in a factor of 1/3162.
  • #1
Faraz
16
0

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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  • #2
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
-3.5=log(I/I°)
10^-3.5=(I/I°)
but still wrong
 
  • #4
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?
 
  • #5
Maybe the answer is supposed to be stated as "reducing the sound intensity by a factor of 3162..."?
 

FAQ: Sound Intensity: Reduce by 35 dB, Factor Calculation

What is sound intensity and how is it measured?

Sound intensity is the amount of sound energy that passes through a unit area in a given amount of time. It is measured in decibels (dB) using a sound level meter.

Why would someone want to reduce sound intensity by 35 dB?

Reducing sound intensity by 35 dB is equivalent to reducing the volume of a sound by about 99.9%. This can be beneficial in situations where noise levels are too high and can potentially cause hearing damage or disturbance.

How is the factor for reducing sound intensity by 35 dB calculated?

The factor for reducing sound intensity by 35 dB can be calculated using the formula: Factor = 10^(dB/10), where dB is the desired reduction in decibels. In this case, the factor would be approximately 0.001.

Can sound intensity be reduced by more than 35 dB?

Yes, sound intensity can be reduced by more than 35 dB. The factor for reducing sound intensity can be calculated for any desired reduction in decibels, and the intensity can be reduced by that amount.

What are some common ways to reduce sound intensity by 35 dB?

There are various ways to reduce sound intensity by 35 dB, including using soundproofing materials, creating sound barriers, and using noise-cancelling technology. The most effective method will depend on the specific situation and source of the sound.

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