Physics

posted by .

Hi I am absolutely stumped on this question.

A loudspeaker is placed between two observers who are 110 m apart, along the line connecting them. If one observer records a sound level of 60.1 dB and the other records a sound level of 74.9 dB, how far is the speaker from each observer?

What I did for this is

I found the the Intensities of both the sound levels.
Then using the formula r1= SQRT(I2/I1)*R2

I would solve for r1 and then do r1+r2=113 to find r2.

I also don't know what value to assign for I2 and I1, i tried both ways but its wrong.

• Physics -

The sound level difference is 14.8 dB. That means that the sound intensity ratio is 10^0.148 = 30.2

That means one speaker is sqrt 30.2 = 5.50 times farther way from the source than the other. If x is the shorter distance

x = 5.50 x = 6.50 x = 110 m.
x = 16.9 m The other distace is 93.1 m

• Physics -

Hey sorry about the 3 posts, It was an accident.
Just a quick question, why do you need to find the sound level difference?

• Physics -

One more thing,
how did you find 10^0.148 be 30.2? isn't it 1.406. I do see you used a ratio, but what was divided?

• Physics -

I meant 10^1.48 = 30.2. Sorry about the typo. The 30.2 is correct. 1.48 is 1/10 of the dB difference in received sopund power levels.

If it is a mystery where the 10^(0.1*dB difference)comes from, that is the definition of decibals in terms of intensity ratio

Log10(I2/I1) = 1/10 (dB2 - dB1)

10^[0.1(dB2-dB1)] = I2/I1

• Physics -

how do you get 6.50
and the final answers to be 16.9m & 93.1m?

• Physics -

I2/I1 = (r2)^2/(r1)^2
square root of (I2/I1) = r2/r1
This will be the difference between the two.
Therefore (distance between the two people/ square root of (I2/I1)) = smallest distance from loud speaker.
And distance between the two people minus the smallest distance is the largest distance.