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Sound Intensity In Decibels | Sound Level In Decibels

Sound Intensity In Decibels | Sound Level In Decibels :- Human ear respond to sound intensities from 10^-12 Wm^-2to 1 Wm^-2. Here the minimum frequency 10^-12 Wm^-2is called threshold of hearing (I₀) or reference intensity of human ear and upper range 1 Wm^-2 is called threshold of pain.

We use logarithmic scale of intensity called the “sound level” which is defined as below :

$\displaystyle SL(in\text{dB})=10{{\log }_{10}}\left( \frac{I}{{{I}_{0}}} \right)$

Here I is the intensity of sound wave. Sound intensity level (SL) is expressed in decibels (dB).

For example a sound of intensity I₀has sound level :

$\displaystyle SL=10{{\log }_{10}}\left( \frac{{{I}_{0}}}{{{I}_{0}}} \right)=10{{\log }_{10}}\left( 1 \right)=10\times 0=0\text{ }dB$

Similarly, sound level of threshold of pain :

$\displaystyle SL=10{{\log }_{10}}\left( \frac{I}{{{I}_{0}}} \right)=10{{\log }_{10}}\left( \frac{1}{{{10}^{-12}}} \right)=10\times 12=120\text{ }dB$

Hence we can say :

Threshold of hearing = 0 dB and

Threshold of pain = 120 dB

Sr. No.	Source of sound	Intensity (in Wm^-2 )	Sound intensity level (SL) (in dB)
1.	Threshold of hearing	10^-12	0
2.	Rustle of leaves	10^-11	10
3.	Whisper	10^-10	20
4.	Quiet room	10^-8	40
5.	Normal conversation	3.2 × 10^-6	65
6.	Street traffic	10^-5	70
7.	Thunder	10^-1	110
8.	Threshold of pain	1	120
9.	Military jet aircraft (30 m away)	10²	140

Example :

Calculate the change in intensity level when the intensity of sound increases by 10⁶ times its original intensity.

Solution :

$\displaystyle \Delta SL=S{{L}_{f}}-S{{L}_{i}}=10{{\log }_{10}}\left( \frac{{{I}_{f}}}{{{I}_{0}}} \right)-10{{\log }_{10}}\left( \frac{{{I}_{i}}}{{{I}_{0}}} \right)$

$\displaystyle \Delta SL=10\left( {{\log }_{10}}\frac{{{I}_{f}}}{{{I}_{i}}} \right)=10\left( {{\log }_{10}}{{10}^{6}} \right)=10\times 6=60\text{ }dB$

Sound Intensity In Decibels | Sound Level In Decibels

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