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Two questions about sound

  1. Sep 13, 2003 #1
    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 im 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 im stuck at it because i think it is either a or d.

    thanks for any help.
     
  2. jcsd
  3. Sep 13, 2003 #2
    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!
     
    Last edited by a moderator: Sep 13, 2003
  4. Sep 13, 2003 #3
    thanks, surprsingly i havent met this formula in the text.
     
  5. Sep 13, 2003 #4
    It is quite stupid of the author then to set problems without teaching the machinary to solve them!
     
  6. Sep 13, 2003 #5
    indeed.
    you can find the text im talking about in here http://www.math.uga.edu/~djb/html/music-hq.pdf
    the problems are at page number 11.
     
  7. Sep 13, 2003 #6
    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 im 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.
     
    Last edited by a moderator: Sep 13, 2003
  8. Sep 13, 2003 #7
    tell me if im wrong but when you multiply I=k*C^2 by 4 it should be 4I=4kC^2.
     
  9. Sep 13, 2003 #8
    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.
     
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