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, 201411 yr by Function

Isn't sound obeying inverse square law?

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

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, 201411 yr by Sensei

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)