How Does a 2.80 dB Difference Affect Sound Intensity?

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Homework Help Overview

The discussion centers around a problem involving sound intensity levels, specifically how a difference of 2.80 dB affects the intensity ratio between two sounds.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of the decibel equation and how to apply it to find the intensity ratio. There are attempts to express the relationship between the two sound intensities using logarithmic properties.

Discussion Status

Some participants have provided algebraic manipulations and suggested using logarithmic laws to simplify the expressions. There is an acknowledgment of confusion regarding the algebra, indicating that further clarification may be needed.

Contextual Notes

Participants are working within the constraints of the problem statement and are seeking to clarify their understanding of the logarithmic relationships involved in sound intensity levels.

Oijl
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Homework Statement


Two sounds differ in sound level by 2.80 dB. By what factor is the one intensity greater than the other?


Homework Equations





The Attempt at a Solution



Is the equation

[tex]B[/tex] = 10dB log(I/Io)

relevant?
 
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The equation is
Bdb = 10log(I/Io)
Now let B1 = 10log(I1/Io) and
B2 = 10log(I2/Io)
Take the difference and find the ratio of I1/I2.
 
Lets say sound 1 has an intensity of I1 and intensity level B, and sound 2 has an intensity of I2 with intensity level B+2.8

B = 10 log (I1 / I0)
B+2.8 = 10 log (I2 / I0)

See if you can use those

EDIT: rl.bhat beat me to it :P
 
I'm sorry, but the algebra is confusing me. Could you show the first few steps, please?
 
B1 = 10log(I1/Io)
B2 = 10log(I2/Io)
B1 -B2 = 10[log(I1/Io) - log(I2/Io)]
Use laws of logarithm to further simplification.
 

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