A normal conversation involves sound intensities of about 3.0 𝘅 10⁻⁶ Wm⁻². What is the decibel level for this intensity? What is the intensity of the sound for 100 dB? Ans. (64.8 dB, 0.01 w m-2)


Part 1: To find the decibel level for the intensity of 3.0 𝘅 10⁻⁶ Wm⁻².?


Given Data:


Intensity of the sound ( I ) = 3.0 𝐱 10⁻⁶ w m⁻²

Intensity of faintest audible  sound ( I₀ ) = 10⁻¹² wm⁻²


Required:


Intensity Level = L - L₀ = ?



Solution:


We have the formula for intensity level as follows



L  − L₀ = 10 log `\frac {I}{I₀}` dB
 
by putting corresponding values 

L  − L₀ = 10 log `\frac {3.0 ｘ 10⁻⁶ w m⁻²}{10⁻¹² wm⁻²}` dB

w m⁻² will cancel with each other So,

L  − L₀ = 10 log ( 3 ｘ 10⁻⁶⁺¹²) dB

L  − L₀ = 10 log ( 3 ｘ 10⁶) dB

L  − L₀ = 10 ( log (3) + log(10⁶) dB

L  − L₀ = 10 ( log (3) + 6 log10) dB

L  − L₀ = 10 ( 0.47 + 6ｘ 1) dB

L  − L₀ = 10 ( 0.47 + 6) dB

L  − L₀ = 10 (6.47) dB

L  − L₀ = 64.7 dB---------------Ans.1


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Part 2: To find the intensity of the sound for a 100 dB intensity level


Given:


Intensity Level = L - L₀ = 100 dB

Intensity of faintest audible sound ( I₀ ) = 10⁻¹² w m⁻²

To Find:


The intensity of the sound ( I ) = ? w m⁻²



Solution: 


We have the formula for intensity level as follows

L  − L₀ = 10 log `\frac {I}{I₀}` dB

by putting the values of each given term

100 dB = 10 log `\frac {I}{10⁻¹² w m⁻²}` dB

by dividing both sides by 10 dB we get

10 = log `\frac {I}{10⁻¹² w m⁻²}`

Now taking antilog on both sides 

Antilog 10 = Antilog log `\frac {I}{10⁻¹² w m⁻²}`


10¹⁰ = `\frac {I}{10⁻¹² w m⁻²}` 

or


l  = 10¹⁰ ｘ 10⁻¹² w m⁻²

l  = 10¹⁰⁻¹² w m⁻²

l  = 10⁻² w m⁻²

or

l  = 0.01 w m⁻² --------------Ans.2

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