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Sound Intensity pig calling contest

  1. Apr 9, 2008 #1
    Sound Intensity "pig calling contest"

    In a pig-calling contest, a caller produces a sound with an intensity level of 110db. How many such callers would be required to reach the pain level of 120db?

    The equation I think I have to use is (db_1 - db_2)=10*log(power_1 / power_2)

    The part that I'm confused on is how to get the two powers onto one side. Can the log be rewritten as (power_1 / power_2)log_10 ? then divide by both sides by log_10? Which would come out to 10=(power_1 / power_2) and I don't know how I would go from there. Provided that everything else is right.

    Please help!!
  2. jcsd
  3. Apr 9, 2008 #2
    You cannot 'divide' anything by a log to cancel it out since it is a function. What you need is,

    I = 10log_{10}(P)

    for the first case, you are given the intensity i.e. 110db. using the formula above, find the power. Once you've found the power for a single caller, assume that 'x' callers are needed and the power now will be xP.

    Substitute it in the formula again, this time taking the Intensity as 120 db. Find 'x'.
  4. Apr 9, 2008 #3
    I know asking this is pretty much just asking for you to tell me the answer but how do I solve the formula? 110=10log_10(P) or how do I move the log to isolate (P), if I do at all?
  5. Apr 9, 2008 #4
    ok... so that is what you wanted. sorry for overlooking that part.. my bad.. totally.

    To solve a logarithmic function:

    b = log_a(x)


    x = a^b

    in our case, a = 10.
  6. Apr 9, 2008 #5
    Thanks, so P=10^110 ? That seems like a very large number that my calculator won't calculate? If that is right does it come down to 10log_10(10^110 / 10^120) = x ?
  7. Apr 9, 2008 #6
    Hold on mate, decibels is not intensity. Decibels is the sound level. You are trying to find the intensity.

    [tex]\beta _1 - \beta _2 = 10log(\frac{I_1}{I_2})[/tex]

    You can find the ratio of I_1 to I_2, and then you can figure out how many callers you need for 120 dB.
  8. Apr 9, 2008 #7
    I am sorry for using the word 'Intensity'.. but decibel is basically a way of expressing ratios. And it may not necessarily include the ratio of Intensities, but Power or any other variable.

    And I used power.. but as we both agree on, it's just the ratio of these variables that matter. Power and Intensity are both dependent on the square of the amplitude of the wave. Hence, if the amplitude becomes twice, power and intensity becomes four times.

    Hence, if the Power initially is 'P', and n sources contribute [in phase, same frequency and amplitude], the amplitude becomes 'n' times and hence the power now is [itex]n^2P[/itex]. Hence now you need to solve for:

    \beta _1 - \beta _2 = 10log\left(\frac{P_1}{P_2}\right)

    where, [itex]P_2 = n^2P_1[/itex]. Now solve for 'n'.

    And again.. i am sorry for the factual inaccuracy of calling it Intensity. I just got confused with the title of the post.

    @Snazzy: Thanks for the correction :D
    Last edited: Apr 9, 2008
  9. Apr 9, 2008 #8
    I think I'm as lost as ever now. So I have B_1 - B_2=10log(P_1 / n^2P_1) and do the P's cancel out of the equation then? So then your left with n=(10^10)^1/2 ? Which equals 100,000, that I'm guessing isn't right.
  10. Apr 9, 2008 #9
    Well.. you sure should be. This has been a confusing thread.. so i'm gonna sum it up for you.

    When only one person shouts, let the power dissipated be 'P'. The amplitude of the sound wave be [itex]A = s_m[/itex]. Now, when 'n' no. of people shout, the amplitude of sound will become 'n' times the previous amplitude [since no other additional data is given, i'm assuming all the sound waves are in phase and of similar frequency]. Hence, now: [itex]A = ns_m[/itex].

    Power is proportional to the square of the amplitude. Hence:

    P_1 = ks_{m}^2

    and also,

    P_2 = k(ns_m)^2 = kn^2s_{m}^2

    Hence the ratio:

    \frac{P_1}{P_2} = \frac{1}{n^2}

    in the formula,

    \beta_1 - \beta_2 = 10log_{10}\left(\frac{P_1}{P_2}\right)

    While solving this, do remember to take the '10' on the other side as you normally would. I think you are confusing it with the base 10 and hence are getting the huge numbers.

    you should get something like:

    log_{10}\left(\frac{P_1}{P_2}\right) = -\frac{120 - 110}{10}

    hope this helps you.
    Last edited: Apr 9, 2008
  11. Apr 9, 2008 #10
    So does it just work out to be 10^1=1/n^2 and then I just take square root of each side? But how do I get the "n" to be in the numerator?
  12. Apr 9, 2008 #11
    Well doing how I think it should be done I get 3.16 people or 4 people to reach that level. Is that correct?
  13. Apr 9, 2008 #12
    well.. seems right to me. what does ur assignment/text-book/whatever say?
  14. Apr 9, 2008 #13
    I don't know the answer for it yet, I will in about an hour.

    Thanks again for your help and helping me through my stupidity, it means a lot to me.
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