What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

Bridger Holden 2022-09-30 Answered
What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?
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Answers (2)

Tristin Durham
Answered 2022-10-01 Author has 6 answers
Consider sounds with intensities I 1 and I 2 with intensity levels β 1 and β 2 respectively. We have
β 1 = 10 log 10 ( I 1 I 0 )
and
β 2 = 10 log 10 ( I 2 I 0 )
Now subtracting the second equation from first,
β 1 β 2 = 10 [ log 10 ( I 1 I 0 ) log 10 ( I 2 I 0 ) ]
Fo
β 1 β 2 = 10 log e ( I 1 / I 0 I 2 / I 0 )
β 1 β 2 = 10 log 10 ( I 1 I 2 )
Let, for intensity I 1 , the sound intensity level β 1 = 90 d B
(a) For I 2 = 2 I 1 , the sound intensity level β 2 is given by
β 2 = β 1 10 log 10 ( I 1 I 2 ) = β 1 10 log 10 ( I 1 2 I 1 ) = 90 10 log 10 ( 1 2 )
β 2 = 90 10 × ( log 10 ( 1 ) log 10 ( 2 ) ) = 90 10 × ( 0.301 ) = 90 + 3.01 = 93.01 d B
(b) When I 2 = I 1 5 , the sound intensity level β 2 is given by
β 2 = β 1 10 log 10 ( I 1 I 1 / 5 ) = 90 10 log 10 ( 5 ) = 83.01 d B
Result:
(a)93.01dB (b) 83.01 dB
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aurelegena
Answered 2022-10-02 Author has 2 answers
Sound intensity level in decibels is:
β = 10 log ( I I 0 )
9 = log ( I I 0 )
Where I 0 = 10 12 W / m 2 , the limit of audiable sound. Ihe expression in the brackets can be written as:
10 9 = y
Where y is just a substitute symbol for the expression in the brackets. If we double the expression in the y and put it back in the equation, we get solution for the first part of the problem:
β a = 10 log 2 10 9
β a = 93 d B
Doing the same process for the second part of the problem, but this time instead of doubling y, we divide it with 5 and we get:
β b = 10 log 10 9 / 5
β b = 833 d B
Result:
β a = 93 d B
β b = 83 d B
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