Question

At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?

Answer 1

At a distance of 100 m from the source the intensity will be 40 dB.

Step-by-step explanation:

Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.

The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.

The conversion between intensity and decibels corresponds to:

`L=10*log(I)/(I0)`

where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.

In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:

`L1 - L2=10*log(I1)/(I0) - 10*log(I2)/(I0)`

`L1 - L2=10*(log(I1)/(I0) - *log(I2)/(I0))`

`L1 - L2=10*[log((I1)/(I0)(I0)/(I2))]`

`L1 - L2=10*[log((I1)/(I2))]`

Being L1= 70 dB and L2= 40 dB

`70 dB - 40 dB=10*[log((I1)/(I2))]`

`30=10*[log((I1)/(I2))]`

`(30)/(10) =log((I1)/(I2))`

`3=log((I1)/(I2))`

`10^(3) =(I1)/(I2)`

`1,000=(I1)/(I2)`

The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:

`(I1)/(I2) =(d2^(2) )/(d1^(2) )`

Then:

`1,000=(d2^(2) )/(d1^(2) )`

Being d1= 10 m

`1,000=(d2^(2) )/(10^(2) )`

`1,000=(d2^(2) )/(100)`

1,000*100= d2²

10,000= d2²

√10,000= d2

100 m= d2

At a distance of 100 m from the source the intensity will be 40 dB.