How can i determine the number of deibles for intensity?

Question

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?

Answer 1

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.

Answer 2

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