05holtel said:
You are 10m away from a sound source. If you want the intensity in DB to double, you must move to a distance...?
Hello 05holtel,
Could you just double check your problem statement, and make sure you copied it over correctly?
I ask because it's very strange for anyone to want something to double
in dB. It's just something that doesn't happen often, if at all. As a matter of fact, depending on how you look at it doesn't even make sense.
It's
very common to want the power, energy, or power flux, etc., to double (in non-dB terms such as Watts, Joules, Watts/(square meter), etc.), But in terms of dB, you do that by
adding 3 dB.
For examples of dB respresentation, you could represent a power in dB in reference to 1 Watt.
P_{dBW} = 10 \ \mbox{log} \left( \frac{P}{1 \ \mbox{W}} \right)
You could also represent a power in terms of milliwatts.
P_{dBm} = 10 \ \mbox{log} \left( \frac{P}{1 \ \mbox{mW}} \right)
(Both of these concepts are used
widely in engineering professions, by the way.)
So let's make a very specific example, and calculate 1 Watt, in terms of P_{dBW} and P_{dBm}.
P_{dBW} = 10 \ \mbox{log} \left( \frac{1 \ \mbox{W}}{1 \ \mbox{W}} \right) = 0 \ \mbox{dBW}
P_{dBm} = 10 \ \mbox{log} \left( \frac{1000 \ \mbox{mW}}{1 \ \mbox{mW}} \right) = 30 \ \mbox{dBm}
Keep in mind that 1 Watt, 0 dBW, and 30 dBm are all
equal.
They are all the same thing!
But if we were to double our 1 Watt power example, in
terms of dBW, it wouldn't change the actual intensity at all. How can you double 0? It's still 0.
If we were to double the value
in terms of dBm, we get 60 dBm. Doing a reverse calculation, that corresponds to 1000 Watts!
My point of all of this is that it just doesn't make practical sense for wanting a value
in dB to double. In terms of dB, adding 3 dB doubles the power intensity. Subtracting 3 dB halves the power intensity. But
multiplying a dB value by something isn't so common.
Being specific to this problem, in terms of sound pressure level, suppose the intensity of the sound source was at the human auditory threshold (just barely audible). In that case, the sound intensity is measured to be 0 dB. There's no way to double that! Or suppose the sound intensity was just below the human threshold of hearing. In that case, the person would have to move backwards (to a distance greater than 10 m) to double it!
What I fear is that your instructor is trying to get you to recognize that
2 \times 10 \ \mbox{log} \left( \frac{I}{I_0} \right) = 10 \ \mbox{log} \left( \left[\frac{I}{I_0} \right]^2 \right)
and somehow relate that to the sound intensity being proportional to 1/r
2. This can be done (although I have no clue as to why anybody would ever want to), but there needs to be more problems details to do it, such as the specific sound intensity at 10 m for example. (And even though such things are possible to do mathematically, in practice, situations don't come up where somebody would ever really want to do that.)
So I just want to be sure that the problem statement is phrased correctly.
