Is this correct. Intensity and sound level.

[afcwestwarrior]

Homework Statement

A certain sound wave has a measured intensity of 40 decibels at a distance of 125 m from the point source where the waves originated. At what distance from the source would the intensity be increased to 70 decibels.

Homework Equations

Sound Level (B) = 10 decibels log ( I/ Io)

I = Ps / 4 pi (r^2)

The Attempt at a Solution

40/ 10 = log (I/Io)

10^4 = I / Io

10 ^4 (Io) = I
10 ^4 (Io) = Ps / 4 pi (r^2)

Ps = 10 ^4 (Io) 4 pi (r^2)

so we then we find the new radius

70/10 = log (I/Io)

10^ 7 ( Io) = I

10^ 7 ( Io) = Ps / 4 pi (r^2)

solve for r 2

r2 = √ (Ps / 4 pi (10^7) I o)

r2 = √ (10 ^4 (Io) 4 pi (r^2) ) / (4 pi (10^7) I o)

= 3.9 m


Shouldn't the radius increase by much more. Because 70 Decibels is much louder than 40 Decibels.
I'd figure that it would be by 50 meters or something.

 

[Jhero]

I understand your confusion and concern about the seemingly small increase in distance. However, it is important to note that decibels are measured on a logarithmic scale, meaning that a small change in decibel level actually represents a much larger change in intensity. In fact, an increase of 10 decibels corresponds to a 10-fold increase in intensity. So, while the increase in decibels from 40 to 70 may seem small, the corresponding increase in intensity is actually quite significant.

To put it in perspective, at 70 decibels, the sound wave would have an intensity that is 10,000 times greater than at 40 decibels. This means that the sound would be 10,000 times louder and more intense at the new distance. So, even though the increase in distance may seem small, the corresponding increase in intensity is actually quite significant.

I hope this explanation helps to clarify any confusion. Keep up the good work as a scientist!