Middle Ear Amplifies Sound Waves: 30dB Reduction

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The discussion revolves around understanding how the middle ear amplifies sound waves by 30 dB and the implications of total ossicle dysfunction. Participants clarify that a 30 dB reduction means sound energy entering the cochlea is significantly diminished. They explain the logarithmic nature of decibels, emphasizing that a 30 dB reduction corresponds to a sound intensity ratio of 1000 times less. The conversation highlights the importance of grasping the definition of decibels and their relationship to sound intensity. Ultimately, the consensus is that without amplification, the sound is effectively reduced by a factor of 1000.
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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 treshold of hearing(10^-12)
dB=log^10*I/Io
 
ilona said:
logarithmic unit of ratio of intensity to standard treshold 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 treshold 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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