argh last physic question and i'm done :)

[Wheatmaster]

this one is stumping me:

The intensity level of 15 car engines in a garage is 100 dB.What is the intensit levelgnereated by ONE engine?

i thought to just take 100 and divide by 15 but there sounded too easy.

edit: here it is: A person stands in front of two stereo speakers driven by the same oscillator providing a 200 Hz tone. the speed of sound in the room is 330 m/s. The person is standing 1.65 meters from one speaker and 4.95 meters form the other. is the person standing on anode or an antinode?

 

[ndee]

that IS to easy 😉 Isn't sound logarithmic?

 

[TheBDB]

Haha no it isn't 100/15. dB is a log scale.

 

[Wheatmaster]

Originally posted by: TheBDB
Haha no it isn't 100/15. dB is a log scale.

so would it be the 15 car engines intensity and then plug it back into the formula to get the intensity level?

 

[TuxDave]

dB is log scale... so

dB = K*log(# of cars*sound per car)

 

[Wheatmaster]

Originally posted by: TuxDave
dB is log scale... so

dB = K*log(# of cars*sound per car)

what's k?

is they the way to do the problem using calculus? because i wont be doing that til next year 🙂

 

[dighn]

100 - db(15) i guess

 

[Wheatmaster]

ok i believe the answer is 88.23 db

how i did it:

100db=10log(I/10x^-12)
I=.01
take I, divided by 15= 6.67x10^-4

one engine db=10log(6.67x10^-4)/(10^-12)
one engine db=88.23

correct?

 

[dighn]

88.23.. looks right to me

 

[Wheatmaster]

bump