A source of sound (1000 Hz) emits uniformly in all directions. An observer 3.19 m from the source measures a sound level of 39.2 dB. Calculate the average power output of the source.

I = 39.2/1.0 x 10^-12

= 3.92 x 10^-11 W/m^2 x 4 pi (3.19^2m^2)
= 5.01e-9 W

i got the wrong answer

To calculate the average power output of the source, we will need to use the formula that relates sound intensity (I) to power (P):

I = P / 4πr²

where I is the intensity, P is the power output, and r is the distance from the source.

To find the power output, we need to rearrange the formula and solve for P:

P = I * 4πr²

First, let's convert the sound level from dB to intensity:

I = 10^(L/10)

where L is the sound level in dB.

In this case, the sound level is 39.2 dB, so:

I = 10^(39.2/10) = 10^(3.92)

Now, we can substitute the values into the formula:

P = (10^(3.92)) * 4π(3.19)²

Simplifying this expression, we get:

P = 10^(3.92) * 4 * 3.14159 * 3.19^2

P = 15849.54 * 40.004

P ≈ 633,996.16

Therefore, the average power output of the source is approximately 633,996.16 Watts.

Use the decibel level definition to compute the intensity in units of Watts per area. Multiply that number by 4 pi R^2 (the area of a sphere of radius R) to get the total power output.

You did not use the correct logarithmic formula that relates decibels to intensity.

To review it, see:

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/sound/u11l2b.html

At 39.2 dB, the sound intensity level is 10^3.92 = 8318 times above the reference (hearing threshold) value of 10^-12 W/m^2. So the value of intensity you must use is 8.32*10^-9 W/m^2