What are the Decibel Levels of Two Combined 70 dB Sound Sources?

[MathematicalPhysicist]

im reading the text of dave benson about music and maths and i have 2 questions from the text which i need an answer to.
1. power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure).
2. (multiple choices) two independents 70 dB sound sources are heard together. how loud is the resultant sound to the nearest dB?

a. 140 dB
b. 76 dB
c. 73 dB
d. 70 dB
e. none of above

now I am stuck at it because i think it is either a or d.

thanks for any help.

 

[MathNerd]

Originally posted by loop quantum gravity
im reading the text of dave benson about music and maths and i have 2 questions from the text which i need an answer to.
1. power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure).
2. (multiple choices) two independents 70 dB sound sources are heard together. how loud is the resultant sound to the nearest dB?

a. 140 dB
b. 76 dB
c. 73 dB
d. 70 dB
e. none of above

now I am stuck at it because i think it is either a or d.

thanks for any help.

I'm not sure about the first question but I will happily answer the second. The formula for decibles is

B = 10*log(I/A)

where B is the number of decibles, A is the reference level which is 10^-12 watts per metre^2 and I is the intensity of the sound. Now a 70dB source has an intensity as follows

70 = 10*log(I/A)
10^7 = I/A
I = A 10^7

there are two 70dB sources in total so the above intensity will be doubled. Therefore the total intensity is

I = 2*10^7 A

putting this back into the decible formula

B = 10*log(I/A)
B = 10*log(2*10^7 A / A)
B = 10*[log(2)+log(10^7)]
B = 10*[log(2)+7]
B ≈ 73.0103dB

So (c) would be the correct answer!

 

[MathematicalPhysicist]

thanks, surprsingly i haven't met this formula in the text.

 

[MathNerd]

Originally posted by loop quantum gravity
thanks, surprsingly i haven't met this formula in the text.

It is quite stupid of the author then to set problems without teaching the machinary to solve them!

 

[MathematicalPhysicist]

indeed.
you can find the text I am talking about in here http://www.math.uga.edu/~djb/html/music-hq.pdf
the problems are at page number 11.

 

[MathNerd]

I haven't taken a look at the book but I will take a shot at the first question you asked. I'm not quite sure what "power intensity" is but I will make the assumption that power intensity = intensity = I. The question is restated below

"power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure)."

therefore I = k C^2, where C is the amplitude and k is the constant of proportionality.

it is easily shown that 4*I = k (C*2)^2, therefore doubling the amplitude quadruples the intensity. So using the formula I used in my last post

B = 10*log(I/A)

now doubling the amplitude quadruples the intensity so

B'= 10*log(4*I/A)
B'= 10*log(4)+10*log(I/A)
B'= 10*log(4)+B

therefore if my assumption is correct i.e. power intensity is just another word for intensity, then the number of decibles corrosponding to a doubling of the amplitude is 10*log(4)≈6.0206dB.

 

[MathematicalPhysicist]

Originally posted by MathNerd
I haven't taken a look at the book but I will take a shot at the first question you asked. I'm not quite sure how amplitude fits in with dB but I will make the assumption that power intensity = intensity = I. The question is restated below

"power intensity is proportional to the square of the amplitude. how many decibles represent a doubling of the amplitude of a signal?
(my answer is 4 but I am not sure)."

therefore I = k C^2, where C is the amplitude and k is the constant of proportionality.

it is easily shown that 4*I = k (C*2)^2, therefore doubling the amplitude quadruples the

tell me if I am wrong but when you multiply I=k*C^2 by 4 it should be 4I=4kC^2.

 

[MathNerd]

Originally posted by loop quantum gravity
tell me if I am wrong but when you multiply I=k*C^2 by 4 it should be 4I=4kC^2.

Notice that in my post I had the following
4*I = k (C*2)^2

notice that the C*2 is all squared

So, k (C*2)^2 = k C^2 2^2 = 4k C^2
So, 4*I = 4k C^2

therefore your expression is equivalent to mine.