# Sound diminishes from x to y over what distance?

## Homework Statement

At a rock concert, a dB meter registered 130 dB when placed 2.2 m in front of a loudspeaker on the stage. (a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air? (b) How far away would the sound level be a somewhat reasonable 85 dB?

## Homework Equations

I = 10^(ß/10)*I_0 where I_0 = 1*10^-12 W/m^2
I_1 = I_2/(4πr^2)

## The Attempt at a Solution

a)
I = 10^(ß/10)*I_0
I = 10^(130 dB/10)*10^-12
I_2 = 10 W/m^2

b)
Convert dB to W/m^2:
I = 10^(ß/10)*I_0
I = 10^(85 dB/10)*10^-12
I_1 = 3.16 * 10^-4 W/m^2

Solve for r in formula: I_1 = I_2/(4πr^2) >>> r = √(I_2/(4πI_1))
r = √(10 W/m^2/(4π(3.16 * 10^-4 W/m^2))
r = 501824 m

Part b) is obviously way too large to be right. I need a formula that incorporates the original distance. Any ideas?

berkeman
Mentor

## Homework Statement

At a rock concert, a dB meter registered 130 dB when placed 2.2 m in front of a loudspeaker on the stage. (a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air? (b) How far away would the sound level be a somewhat reasonable 85 dB?

## Homework Equations

I = 10^(ß/10)*I_0 where I_0 = 1*10^-12 W/m^2
I_1 = I_2/(4πr^2)

## The Attempt at a Solution

a)
I = 10^(ß/10)*I_0
I = 10^(130 dB/10)*10^-12
I_2 = 10 W/m^2

b)
Convert dB to W/m^2:
I = 10^(ß/10)*I_0
I = 10^(85 dB/10)*10^-12
I_1 = 3.16 * 10^-4 W/m^2

Solve for r in formula: I_1 = I_2/(4πr^2) >>> r = √(I_2/(4πI_1))
r = √(10 W/m^2/(4π(3.16 * 10^-4 W/m^2))
r = 501824 m

Part b) is obviously way too large to be right. I need a formula that incorporates the original distance. Any ideas?

An easier way to approach part (b) is to look at the difference in dB, and convert that into the ratio of the intensities. Since the intensity varys by 1/r^2, having the ratio of the intensities will give you the ratio of the r^2, and that lets you find the 2nd r value.

Give that a try, and show us your work...

130 dB - 85 dB = 45 dB
I = 10^(ß/10)*I_0 where I_0 = 1*10^-12 W/m^2
I = 10^(45/10)*10^-12 W/m^2
I = 3.16*10^-8 W/m^2

So (3.16*10^-8 W/m^2) / (10 W/m^2) = 3.16*10^-9
and (2.2 m) / (x m) = 3.16*10^-9 m
x = 6.96 * 10^8 m
which is even larger than the answer I got before >.<

Are you saying I'm actually going to use the formula I_1 = I_2/(4πr^2) here? Or 1/r^2 (which would still give an outrageous value if I plug the answer into that)?

After a quick search I found the following equation:

$I_{2} = I_{1}\frac{r_{1}^{2}}{r_{2}^{2}}$

See if that works?

berkeman
Mentor
130 dB - 85 dB = 45 dB
I = 10^(ß/10)*I_0 where I_0 = 1*10^-12 W/m^2
I = 10^(45/10)*10^-12 W/m^2
I = 3.16*10^-8 W/m^2

So (3.16*10^-8 W/m^2) / (10 W/m^2) = 3.16*10^-9
and (2.2 m) / (x m) = 3.16*10^-9 m
x = 6.96 * 10^8 m
which is even larger than the answer I got before >.<

Are you saying I'm actually going to use the formula I_1 = I_2/(4πr^2) here? Or 1/r^2 (which would still give an outrageous value if I plug the answer into that)?

Don't worry about plugging in the power numbers. They just ask you how much farther away you need to go to be 45dB quieter.

-45dB = 10 log (I2/I1)

What is the ratio of I2 to I1?

What does that imply for the ratio of the distances from the source speaker?

After a quick search I found the following equation:

$I_{2} = I_{1}\frac{r_{1}^{2}}{r_{2}^{2}}$

See if that works?

3.16*10^-8 W/m^2 = 10 W/m^2 ((2.2 m)^2 / r^2)
r = 39136.2 m

Hm, better, but still a huge distance =/

Don't worry about plugging in the power numbers. They just ask you how much farther away you need to go to be 45dB quieter.

-45dB = 10 log (I2/I1)

What is the ratio of I2 to I1?

What does that imply for the ratio of the distances from the source speaker?

-45dB = 10 log (I2/I1)

(3.16*10^-8 W/m^2) / (10 W/m^2) = 3.16*10^-9, or, ignoring the power numbers, 10 times the distance? So it would be 22 meters?

berkeman
Mentor
After a quick search I found the following equation:

$I_{2} = I_{1}\frac{r_{1}^{2}}{r_{2}^{2}}$

See if that works?

-45dB = 10 log (I2/I1)

Use the second equation to find the ratio of the Intensities (sound powers). Then use the first equation to find the ratio of the distances. That gives you the second distance...

Perhaps you mean I'm not supposed to plug in anything for I_1 and I_2? That's the only way it would make sense I suppose. Solving for that gets me a ratio of I_2 / I_1 of sqrt(10)/100000.
If I plug that into I_2 / I_1 = r_1^2 / r_2^2 with r_1 = 2.2 m, I get a distance of about 391.221 m, which looks reasonable.

So the answer is about 391.22 meters? The part I didn't understand was that I wasn't supposed to substitute anything in for I_2 / I_1 in your second equation, which wouldn't have given any variable to solve for. Thank you!

berkeman
Mentor
Perhaps you mean I'm not supposed to plug in anything for I_1 and I_2? That's the only way it would make sense I suppose. Solving for that gets me a ratio of I_2 / I_1 of sqrt(10)/100000.
If I plug that into I_2 / I_1 = r_1^2 / r_2^2 with r_1 = 2.2 m, I get a distance of about 391.221 m, which looks reasonable.

So the answer is about 391.22 meters? The part I didn't understand was that I wasn't supposed to substitute anything in for I_2 / I_1 in your second equation, which wouldn't have given any variable to solve for. Thank you!

You got it. The nice thing about dB is that you can represent very large ratios with them, and still have manageable numbers.

If they had asked for the distance where the intensity was 40dB down, that is 10^4 = 10,000 times less. Square root of that is 100, so you would have been 100 times farther away than 2.2m. Since this problem listed 45dB down, the distance was a bit further than 100 times farther away.