How many machines can a factory add before exceeding the 90-dB limit?

[j.c]

Homework Statement

a noisy machine in a factory produces a decibel rating of 80 dB. how manyidentical machines could you add to the factory without exceeding the 90-dB limit?

Homework Equations

I=P/A and B+10log(I/I(sub)0) ... (I think)

The Attempt at a Solution

I tried to find the intensity at 80 decibels and the intensity at 90 decibels by solving for I using B=10log(I/I0) and the inverse log equation and got 1 x 10^-4 and 1 x 10 ^-3 respectively. I had no idea where to go from here though.

 

[Curious3141]

Homework Statement

a noisy machine in a factory produces a decibel rating of 80 dB. how manyidentical machines could you add to the factory without exceeding the 90-dB limit?

Homework Equations

I=P/A and B+10log(I/I(sub)0) ... (I think)

The Attempt at a Solution

I tried to find the intensity at 80 decibels and the intensity at 90 decibels by solving for I using B=10log(I/I0) and the inverse log equation and got 1 x 10^-4 and 1 x 10 ^-3 respectively. I had no idea where to go from here though.

OK, you've calculated the intensities, so by what factor is the intensity corresponding to 90dB higher than that corresponding to 80dB?

This question can actually be solved more simply by observing that 90dB - 80dB = 10dB = a ___ fold increase. (I left that blank for you to think about).

 

[j.c]

so then... an increase of 10 dB means that the intensity of the sound is multiplied by a factor of ten right?

but I'm still confused about how to get the number of machines that can be added

 

[Curious3141]

so then... an increase of 10 dB means that the intensity of the sound is multiplied by a factor of ten right?

but I'm still confused about how to get the number of machines that can be added

Great! That's correct.

OK, so you can squeeze ten machines there (and still meet the limit), but you only have one. The number of machines you could *add* is ____.

 

[j.c]

oh ok. thank you! I came to the conclusion of nine after your first post but that seemed almost too simple and i didnt want to embarass myself with a wrong answer haha. :)

 

[Curious3141]

No worries. The only embarrassment is not to try.