Question

What would be the intensity level of a 39-decibel sound after being amplified a million times?

Answer #1

**Given**

Intensity in Decibel 39

Increase in intensity million times

**Known values**

Threshold intensity of hearing I_{o} = 1 x
10^{-12} W/m^{2}

**Solution**

Intensity in Decibel I_{dB} = 10 log{(Intensity in
w/m^{2})/Threshold Intensity}

39 dB = 10 log(I/I_{o})

39 = 10log(I/I_{o})

3.9 = log(I/I_{o})

10^{3.9} = I/I_{o}

I = I_{o} x 10^{3.9}

After amplifying million times the new intensity

I’ = million times I

I’ = 1 x 10^{6} x I

I’ = 1 x 10^{6} x I_{o} x 10^{3.9}

I’ = I_{o} x 10^{9.9}

I’ in decibels

I’ dB = 10 log(I’/I_{o})

I’ dB = 10 log(I_{o} x
10^{9.9}/I_{o})

I’ dB = 10 log(10^{9.9})

I’dB = 10 x 9.9

I’ dB = 99 Decibels

After amplifying one million times the intensity level of 39 dB sound would be 99 dB

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