Decibel Physics problem... I NEED HELP!!!!?

Question

At a rock concert, a sound meter registers 130dB when it is placed 2.80m directly in front of a loudspeaker. Recall that the surface area of a sphere is given by 4 pi r^2. The power is given as 985 Watts. Determine how far from the loudspeaker the meter must be in order that it would read only 90dB.

My professor started out the problem with this:

B (letter betta) = 10 log Intensity + 120.

Is the 120 supposed to be a 130? How would you solve this? Is my professor wrong using a 120. Maybe a typo from him?

Answer 1

Let us remember what is dB. It is the logarithm of relaton, at which 10dB = 1B = Lg(I/Io)=1 when I = 10 Io, That is when we enhance intensity 10 times, we get 10 dB more, when we enhance intensity 100 times, we get 20 dB more, and when we enhance intensity 10000 times, we get 40 dB more.

In acoustics, it is usually accepted to take the Io as the intensity at threshhold sound preddure of 20 microPascal.

Earnestly speaking, I do not remember what is 120, though I belive this is the deciBell Intensity of 1Watt/sq.m.

But you do not need this figure at all to solve the problem. You only need two figures for sound meter figures and one for the first distance to find the second distance.

At the initial distance from the center of loudspeakers Ri of 2.8 meters you have intensity Ii of 130 dB, and the total sound power would be W = 4PiRi^2Ii , the same power will be at the second distance (we assume that the sound power is not dissipated in air significantly). So, at distance Rx (this is the distance which we shall have to determine), the power W = 4PiRx^2Ix = 4PiRi^2Ii.
So, the relation Ii/Ix = Rx^2/Ri^2.

Let us now take into accont that
10 lg(Ii/Ix) = Bi - Bx = 130 - 90 = 40 dB
from this lg(Ii/Ix) = 4,
and from this Ii = 10000 Ix.

That is, the intensity at 90 dB is 10000 times less than at 130 dB.
And from here,
Rx^2 = 10000Ri^2,
hence Rx = 100 Ri = 280m
And this is the answer