Question

By what factor does the sound intensity increase if the sound intensity level increases from 60 db to 61 db?

Answer 1

`(I_(2))/(I_(1)) = 10^(0.1)`

Step-by-step explanation:

The sound intensity level β can be mathematically defined as

`\beta(dB) = 10log_(10)((I)/(I_(o) ) )`

where
`I_(o)` is the reference intensity and is equal to
`10^(-12)`
`W/m^(2)`.

Now,

`\beta_(2) - \beta_(1) = 10 log(I_(2) )/(I_(o)) -  10 log(I_(1) )/(I_(o))\\62-60= 10log(((I_(2) )/(I_(o) ) )/((I_(1) )/(I_(o) ) ) )\\1=10log((I_(2) )/(I_(1) ) )\\\\(I_(2) )/(I_(1) )=10^(0.1)`