The intensity level of a sound is reported in unit decibel(dB). how does IL change if we increase a sound intensity by a factor of 10?
a)remains the same
b) increases by 1dB to 2dB
c)it increases by 2dB to 20dB
d)it increases by 20dB to 200dB
e) it decreases
equations...
IL=10(log I- logIo)
I am very confused as to how to go about solving this problem...any help would be greatly appreciated.
"None of the above" should be a choice
The sound intensity (power/area) increases by a factor of 10 whenever the I.L. increases by 10 dB
To solve this problem, you can use the equation for intensity level (IL) in decibels:
IL = 10(log(I) - log(Io))
where I is the new intensity and Io is the reference intensity.
Given that we are increasing the sound intensity by a factor of 10, we can write I as 10 times the reference intensity, Io. So, I = 10Io.
Substituting this value into the equation, we have:
IL = 10(log(10Io) - log(Io))
Now, let's simplify the equation:
IL = 10(log(10) + log(Io) - log(Io))
Since log(10) = 1, the equation becomes:
IL = 10(1) = 10
Therefore, the intensity level (IL) increases by 10 dB if we increase the sound intensity by a factor of 10.
So, the correct answer is:
b) increases by 1 dB to 2 dB.