Sound level at a distance

[Function]

Hello everyone!

 

In preperation for my approval exam to med school:

 

A sound source sends waves with a power of [math]100\text{ W}[/math]. The sound level, expressed in [math]\text{dB}[/math] at a distance of [math]10\text{ m}[/math] is:

A. 139

B. 129

C. 109

D. 10

 

I tried to solve this like this:

 

If [math]\left[i\right]=\text{Wm}^{-2}[/math], then [math]I=Px^{-2}[/math]:

[math]I=100\text{ W}\cdot 100^{-1}\text{ m}^{-2}=1\text{ Wm}^{-2}[/math]

 

[math]N=120+10\cdot\log{I}=120+10\cdot\log{1}=120\text{ dB}[/math]

 

Clearly, this isn't right. Can someone help me? The answer should be C.

Also tried to solve this by picking x² = 10²*pi, the result then was about 130 dB. But, it's answer C, so...?

 

Thanks!

 

F.

Edited June 22, 2014 by Function

[Sensei]

Isn't sound obeying inverse square law?

[Function]

What? I don't know, man.. Never saw this at school..

But, someone else taking the exam helped me: sound waves are spherical, the requested area is thus the surface area of a sphere with radius 10:

 

A = 4*pi*100 m²

 

I = P/A

...

N = 109 dB

[Sensei]

What? I don't know, man.. Never saw this at school..

But, someone else taking the exam helped me: sound waves are spherical, the requested area is thus the surface area of a sphere with radius 10:

 

 

That's exactly what I was talking about.

You couldn't not have it at primary school physics..

 

http://en.wikipedia.org/wiki/Inverse-square_law

http://en.wikipedia.org/wiki/Intensity_%28physics%29

Edited June 22, 2014 by Sensei

[SciLib]

Basically you need to calculate the ratio of the sound power at the source and the sound power at 10 m. (sound power at source/surface area of sphere 10m radius)