EK Physics # 661

[betterfuture]

---

Members don't see this ad.

Sound emanating from source travels in all directions. This sound is measured at 20dB by an observer standing 40m away. In order to decrease the intensity level to 10dB, approximately how far from the source must the observer be?

Answer: 126m

Now I don't exactly understand how they worked it out. The book said to take the sq rt of 10 and multiply it by 40m. Can someone please work out and explain the algebra. I am not real good with my inverse sq rt proportions. Thank you very much.

 

[aldol16]

P = I*A, A = 4*pi*r^2 because sound spreads outward in a sphere. P = I*4*pi*r^2. Assuming constant power (a property of the sound source), I is inversely related to the square of r. For example, as r doubles, I is reduced by a factor of 4. Now, you need to figure out the factor changer in I. Change in dB = 10*log(I2/I0) - 10*log(I1/I0) = 10*log(I2/I1) = Thus, 10^change in dB/10 = I2/I1. In this case, 10^-10/10 = I2/I1, or I2/I1 = 1/10.

Now remember that I is inversely related to the square of r. Mathematically, this means that (I1)*(r1)^2 = (I2)*(r2)^2. Here, I2/I1 = 1/10, so (10)*(r1)^2 = (r2)^2. r2 thus = sqrt(10)*r1.

 

[betterfuture]

So that means if intensity increased by 4, the radius would have to decrease to 2 bc 2^2=4. Likewise if intensity decreased by 9 radius would increase to 3 bc 3^2=9. And here intensity decreased by 10 so radius would increase to 3.16 bc 3.16^2=10.

 

[aldol16]

So that means if intensity increased by 4, the radius would have to decrease to 2 bc 2^2=4. Likewise if intensity decreased by 9 radius would increase to 3 bc 3^2=9. And here intensity decreased by 10 so radius would increase to 3.16 bc 3.16^2=10.

If intensity increases by a factor of 4, I2/I1 = 4 and therefore (I1)*(r1)^2 = (I2)*(r2)^2 and (r2)^2 = (1/4)*(r1)^2, factoring to r2 = (1/2)*r1. So not 2 but 20.

 

[betterfuture]

If intensity increases by a factor of 4, I2/I1 = 4 and therefore (I1)*(r1)^2 = (I2)*(r2)^2 and (r2)^2 = (1/4)*(r1)^2, factoring to r2 = (1/2)*r1. So not 2 but 20.

But that would hold true for intensity only, right? Generally speaking, inverse sq rt proportions would be how like what I stated, correct. Sorry if I sound confused, I just really want to get a definite answer because this topic bothered me alot.

 

[aldol16]

But that would hold true for intensity only, right? Generally speaking, inverse sq rt proportions would be how like what I stated, correct. Sorry if I sound confused, I just really want to get a definite answer because this topic bothered me alot.

So that means if intensity increased by 4, the radius would have to decrease to 2 bc 2^2=4. Likewise if intensity decreased by 9 radius would increase to 3 bc 3^2=9. And here intensity decreased by 10 so radius would increase to 3.16 bc 3.16^2=10.

Ah, I see. You messed up by saying "decrease to 2." What you mean is decrease by a factor of two. In that case, you would be right, since it goes from 40 m to 20 m. So that applies with any inverse square law, from gravity to electrostatics.

 

[betterfuture]

Ah, I see. You messed up by saying "decrease to 2." What you mean is decrease by a factor of two. In that case, you would be right, since it goes from 40 m to 20 m. So that applies with any inverse square law, from gravity to electrostatics.

Thank you so much aldol16. I can kind of feel comfortable with inverse sq rts now!

 

[David513]

I think of it like...
20 to 10dB means decrease intensity by factor of 10.
You know that Intensity=Power/Area, and area is measured in m^2 .
Therefore, to decrease intensity by a factor of 10, you have to increase the distance by a factor of sqrt(10) so that when you square it to get area, you end up decreasing the intensity by a factor of 10.
Therefore, the distance is 40m, and you use your sqrt(10) as the factor by which you must increase the distance: 40*sqrt(10) = your answer.

 

[betterfuture]

I think of it like...
20 to 10dB means decrease intensity by factor of 10.
You know that Intensity=Power/Area, and area is measured in m^2 .
Therefore, to decrease intensity by a factor of 10, you have to increase the distance by a factor of sqrt(10) so that when you square it to get area, you end up decreasing the intensity by a factor of 10.
Therefore, the distance is 40m, and you use your sqrt(10) as the factor by which you must increase the distance: 40*sqrt(10) = your answer.

Thanks! That's exactly how I thought of it.