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If the intensity of sound is doubled, by how many decibels does the sound level increase?
If the intensity of sound in watts per square metre is I, then the intensity level in decibel (dB) is given bydB= 10 log(I/I•),Where the base of logarithm is 10,I•=10^-12 watt per square metre( roughly the intensity at which audible sound can be heard)On the decibel scale the pain threshold is 1 watt per square metre and on decibel scale is:dB= 10 log(1 watt per square metre/10^-12 watt per square metre)dB= 10 log ( 1/10^-12)= 10* log (10^12)=10*12=120 dBIn the given problemdB= 10 log ( 2 I/I)= 10* log 2=10*0.3010=3.010dB
If the intensity of sound in watts per square metre is I, then the intensity level in decibel (dB) is given by
dB= 10 log(I/I•),
Where the base of logarithm is 10,
I•=10^-12 watt per square metre( roughly the intensity at which audible sound can be heard)
On the decibel scale the pain threshold is 1 watt per square metre and on decibel scale is:
dB= 10 log(1 watt per square metre/10^-12 watt per square metre)
dB= 10 log ( 1/10^-12)= 10* log (10^12)=10*12=120 dB
In the given problem
dB= 10 log ( 2 I/I)= 10* log 2=10*0.3010=3.010dB
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