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Homework Help: Intensity of sound

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data
    At 1 m away from the source of a sound, the intensity of the sound is 90 dB. At 10 m away the intensity is…?

    The answer is 70 dB but I don't know how...

    2. Relevant equations

    I = Power/Area

    3. The attempt at a solution
    My thought was to set up the equations:

    90dB = P/4pi
    XdB = 4pir^2

    I'm not sure where to go from here.
  2. jcsd
  3. Dec 10, 2013 #2


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    Staff: Mentor

    You are on the right track.

    What is the ratio of the surface area of the two spheres?

    And what is the equation for power in dB based on the ratio of intensities?
  4. Dec 11, 2013 #3
    The ratio of the surface area of the two spheres would be

    4pi/4pi(10)^2 or

    I'm not sure how to set up the quation for power in dB based on the ratio of intensities

    I0/I = (p1/a1)/(p2/a2)?
  5. Dec 11, 2013 #4


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    Staff: Mentor

    You use dB to handle large ratios.

    For voltage, V(dB) = 20 * log(V/Vo) where Vo is some reference voltage

    For power P(dB) = 10 * log(P/Po) where Po is some reference power

    So for this question, you will be using the 2nd equation...
  6. Dec 11, 2013 #5


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    Homework Helper
    Gold Member

    First, allow me to offer some general guidance regarding decibels.

    Decibels always imply power ratios. Always. No matter what.

    The decibel is defined, relating the ratio of two powers P2 and P1 as:

    [tex] 10 \log_{10} \left( \frac{P_2}{P_1} \right) [/tex]
    So one might ask, "well then how can we use decibels with voltage? Voltage isn't power," or "where does the '20' come from when we use ratios of voltages?"

    The answer is that we make the pseudo-assumption that the input and output impedances of a given circuit are equal; i.e., Rin = Rout = R. (By the way, this is a realistic pseudo-assumption, since it is often the case for circuits that are optimized for maximum power transfer, signal to noise ratio, etc.)

    Also we know that the power through a resistor is V2/R.

    Plugging that into our decibel equation, and noting that log(x2) = 2logx,

    [tex] 10 \log_{10} \left( \frac{ \frac{V_2^2}{ R}}{ \frac{V_1^2}{ R}} \right) = 10 \log_{10} \left( \frac{V_2^2}{V_1^2} \right) = 10 \log_{10} \left( \left[ \frac{V_2}{V_1} \right]^2 \right) = 20 \log_{10} \left( \frac{V_2}{V_1} \right) [/tex]


    Now allow me to move on to something more specific: Sound/acoustics.

    Sound decibels are defined in terms of power flux* ratios.

    *(Power flux is power through a unit area -- you may think of the unit area as the cross sectional area of your ear canal if it helps visualize it -- or if you want to stick with SI units, the sound power propagating through a square meter.)

    The reference power flux is typically defined as the power flux that corresponds to the human threshold of hearing.

    If we define the reference power flux, that of the threshold of human hearing, as [itex] \Phi_{ref} [/itex], and the power of the source as P, then we know from the problems statement that

    [tex] 90 \ \mathrm{dB} = 10 \log_{10} \left( \frac{\frac{P}{A_1}}{\Phi_{ref}} \right) [/tex]
    where [itex] A_1 [/itex] is the surface area of a circle with a 1 meter radius. So now the question is, what is

    [tex] 10 \log_{10} \left( \frac{\frac{P}{A_{10}}}{\Phi_{ref}} \right) \ ,[/tex]
    where [itex] A_{10} [/itex] is the surface area of a circle with radius 10 meters?

    Hint: There is an easy way to do this such that you won't even need a calculator to solve this problem (you might even be able to solve it in your head). Do this by noting that log(x/y) = logx - logy, and asking yourself, Does the area of a circle increase proportionally with the radius, or does it increase proportionally with the square of the radius?" :wink:
    Last edited: Dec 11, 2013
  7. Dec 11, 2013 #6


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    Staff: Mentor

    Thanks for the clarification, collinsmark. Nice post :smile:
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