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November 23, 2014

November 23, 2014

Posted by **Joey** on Friday, November 28, 2008 at 12:23pm.

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 dont know what value to assign for I2 and I1, i tried both ways but its wrong.

Please Help!

Thanks In Advance

- Physics -
**drwls**, Friday, November 28, 2008 at 1:08pmThe 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 -
**Joey**, Friday, November 28, 2008 at 5:18pmHey sorry about the 3 posts, It was an accident.

Just a quick question, why do you need to find the sound level difference?

- Physics -
**Joey**, Friday, November 28, 2008 at 5:27pmOne 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 -
**drwls**, Saturday, November 29, 2008 at 2:15amI 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 -
**Sinisa**, Saturday, November 23, 2013 at 4:14pmhow do you get 6.50

and the final answers to be 16.9m & 93.1m?

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