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 .
A) 77.0 dB
B) 74.0 dB
C) 60.0 dB
D) 40.0 dB
E) 20.0 dB

Question

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  .
A) 77.0 dB 
B) 74.0 dB 
C) 60.0 dB 
D) 40.0 dB 
E) 20.0 dB

Quizplus · Accepted Answer

The intensity level (in dB) is given by the formula:

IL = 10 log(I/I0)

where I is the sound intensity in W/m^2 and I0 is the threshold of hearing, which is 1.00 x 10^-12 W/m^2.

Using this formula, we can find the sound intensity at a distance of 2.00 m:

80.0 dB = 10 log(I/1.00 x 10^-12)
8.00 = log(I/1.00 x 10^-12)
I/1.00 x 10^-12 = 10^8
I = 1.00 x 10^-4 W/m^2

Now, we can use the inverse square law to find the sound intensity at a distance of 4.00 m:

I2 = I1 (r1/r2)^2

where I1 is the sound intensity at distance r1, I2 is the sound intensity at distance r2, and r1 and r2 are the respective distances from the source.

Plugging in the values, we get:

I2 = (1.00 x 10^-4) (2.00/4.00)^2
I2 = 2.50 x 10^-5 W/m^2

Now, we can find the intensity level at a distance of 4.00 m:

IL = 10 log(I/1.00 x 10^-12)
IL = 10 log(2.50 x 10^-5/1.00 x 10^-12)
IL = 10 log(2.50 x 10^7)
IL = 74.0 dB

Therefore, the answer is B: 74.0 dB.

At a Distance of 2