Question

How many times louder is the intensity of

sound at a rock concert in comparison with
that of a whisper, if the two intensity levels
are 130 dB and 2 dB, respectively?

Answer 1

Answer:
`6.309* 10^(12)`

Step-by-step explanation:

Given

Intensity level of rock concert
`=130\dB`

Intensity level of Whisper
`=2\dB`

Intensity of sound is given by

`I=10\log _(10)((I)/(I_o))`

where
`I_o=10^(-12)\ watts/m^2`

and difference in the intensity of the two sounds is
`db=130-2=128\ dB`

and

`db=10\log _(10) ((I_1)/(I_2))`

`12.8=\log _(10)((I_1)/(I_2))`

`(I_1)/(I_2)=10^(12.8)`

`(I_1)/(I_2)=6.309* 10^(12)`

Thus intensity of Rock concert is
`6.309* 10^(12)` times louder than that of a whisper