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## How can i determine the number of deibles for intensity?

The relationship betwee the number of deibels "B"and the intensity of sound I in watts per squre centemiter is given by
B=20log(base 10)(I/10^-16)

a)Determine the number of deibels for intensity of 10^-4 watts persqure centimeter.

B)Determine the number of deibels for intensity of 10^-6.

C)The intensity of sound in parts a) is 100 times as great as that in part b) .Is the number of deibels 100 times as great?

The term is decibel (dB), not deible.

a.) B(10^(-4)) = 20 log((10^(-4))/((10^(-16)))
= 20 log(10^12)
= 20*12*log(10)
= 240 dB

b.) B(10^(-6)) = 20 log((10^(-6))/((10^(-16)))
= 20 log(10^10)
= 20*10*log(10)
= 200 dB

c.) No, it's not. I.e., 240 ≠ 100*200.
B = 20 log (I/10^-16)

a) for I = 10^-4

B = 20 log(10^-4/10^-16) = 20 log (10^12) = 20 x 12 log 10 = 20 x 12 x 1 = 240 Db

b) For I = 10^-6

B = 20log(10^-6/10^-16) = 20 log 10^10 = 20 x 10 log 10 = 20 x 10 x 1 = 200 Db

c) No, sound intensity in part a is 240/200 = 1.2 times greater than sound intensity in part b