Middle Ear Amplifies Sound Waves: 30dB Reduction

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Homework Help Overview

The discussion revolves around the amplification of sound waves by the middle ear, specifically addressing a scenario where the ossicles are dysfunctional, leading to a reduction in sound energy transferred to the cochlea. The problem presents multiple-choice answers regarding the extent of this reduction, measured in decibels (dB).

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of decibels and their relationship to sound intensity. There are attempts to clarify the implications of a 30 dB reduction and how it translates to sound energy transfer. Questions arise regarding the correct interpretation of the dB scale and its logarithmic nature.

Discussion Status

Participants are actively engaging with the problem, discussing definitions and attempting to clarify the relationship between dB and sound intensity ratios. Some guidance has been offered regarding the interpretation of decibels, but there is still uncertainty about the calculations and the implications of the amplification ratio.

Contextual Notes

There is mention of confusion regarding the definitions and calculations related to decibels, as well as the potential for misunderstanding the relationship between dB and intensity ratios. Participants are encouraged to refer to external resources for clarification.

ilona
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Homework Statement



Hei!
I have problem with this quest:
Middle ear amplifies the sound waves entering cochlea by 30dB. This means that in case of total dysfunction of ossicles the sound energy transferred into cochlea is reduced:
a)3 times b)30 times c)1000 times d)3000 times

I don know how should I start with it..which formula?
 
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welcome to pf!

hei ilona! welcome to pf! :smile:

total dysfunction means no amplification, so it's 30dB quieter

sooo … what is the definition of dB ? :wink:
 
logarithmic unit of ratio of intensity to standard threshold of hearing(10^-12)
dB=log^10*I/Io
 
ilona said:
logarithmic unit of ratio of intensity to standard threshold of hearing(10^-12)
dB=log^10*I/Io
That is almost the definition of loudness, but not quite.
Amplification is defined in terms of output intensity and input intensity. And you meant base 10, not log on the 10th power... and it is decibel, that "deci" means something...
See: http://en.wikipedia.org/wiki/Decibel#Definition

ehild
 
(try using the X2 and X2 buttons just above the Reply box :wink:)
ilona said:
logarithmic unit of ratio of intensity to standard threshold of hearing(10-12)
dB=log10*I/Io

isn't that B, not dB ? :confused:
 
tiny-tim said:
isn't that B, not dB ? :confused:

It is not anything. log without argument multiplied by I/Io...
 
L(B)=log10*I/Io

but i wonder if there is no amplification we have 1 dB ,haven't we?
so why is not it just reduced by 30 times?
 
ehild said:
No amplification is zero dB.

yes, ilona, if I/Io = 1, then log I/Io = … ? :smile:
 
  • #10
log1= 0
 
  • #11
right, if the amplification ratio is 1, then the dB difference is 0

and if the amplification ratio is 10, then the B difference is 1, and the dB difference (10 times as much) is 10

so if the amplification ratio is 100, then the B difference is … ? :smile:
 
  • #12
ok , i think answer will be 1000times
 
  • #13
but once again , cause I don't think I understand it..
I have this formula LB=log10p1/p0
so I have 30 dB so it is 3 B
and now why p1/po is 1?
 
  • #14
ilona said:
ok , i think answer will be 1000times

yes :smile:
ilona said:
but once again , cause I don't think I understand it..
I have this formula …

forget the formula :rolleyes:

(you'll never remember it in the exam, anyway)

just remember ('cos it's easy) …

10 dB = 10 times louder


then you know that 1 B = 10 times louder, so 2 B = 100 times etc, and finally convert to dB by always multiplying by 10 (same as converting grams to decigrams, or metres to decimetres) :wink:
 
  • #15
ok, thank you very much =)
 

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