At a distance of 2.00 m from a point source of sound, the intensity level is 80.0 dB. What will be the intensity level at a distance of 4.00 m from this source? The lowest detectable intensity is 1.0 × 10^-12 W/m². A) 77.0 dB B) 74.0 dB C) 60.0 dB D) 40.0 dB E) 20.0 dB

Question

Quizplus · Accepted Answer

We can use the inverse square law formula to solve this problem:
I2/I1 = (r1/r2)^2
where I1 is the intensity at distance r1, I2 is the intensity at distance r2, and r1 and r2 are the distances from the point source.

We can rearrange this formula to solve for I2:
I2 = I1 * (r1/r2)^2

In this problem, we know that I1 = 10^(8.0) W/m^2 (since intensity level in dB = 10 log(I/I0) where I0 is the threshold of hearing)
and r1 = 2.00 m.

We want to find I2 at r2 = 4.00 m. Plugging in the values:
I2 = 10^(8.0) * (2.00/4.00)^2 = 10^(8.0) * 0.25 = 10^(8.0-6.0) W/m^2 = 1.0 × 10^-6 W/m^2

Now we can find the intensity level at this new intensity:
IL2 = 10 log(I2/I0) = 10 log(1.0 × 10^-6/1.0 × 10^-12) = 10 log(10^6) = 60 dB

Therefore, the answer is B.

At a Distance of 2