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Math Help - Decibel loss question

  1. #1
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    Decibel loss question

    If it was possible to have a sound source of 4225 db, i know that at 1m it looses 11dB then every time you double the distance it looses 6db.

    IE: 1m = 11, 2m = 17, 4m = 23, 8m = 29 etc etc

    Question:
    how far would it take in m for the dB level to go from 4225 to zero? working on the above formula?

    Edited for 1m and revised for above example.
    Last edited by 1LISTEN; December 28th 2009 at 04:21 AM.
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  2. #2
    Super Member Bacterius's Avatar
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    That means that the decibel loss for a distance in metres d is 6 \times \log_2(d) + 11.

    You want d such as 6 \times \log_2(d) + 11 > 4225 (when the decibel loss is greater than 4225). Thus :

    6 \times \log_2(d) > 4214

    \log_2(d) > \frac{4214}{6}

    Thus :

    d > 2^{702.3333...}

    Conclude : we would require many metres (around 2.65 \times 10^{211}) to dissipate the sound. That's a little more than 6.62 \times 10^{203} times the circumference of the earth.
    Last edited by Bacterius; December 28th 2009 at 04:26 AM. Reason: Edited because of the quick change of mind of the asker
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  3. #3
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    Bacterius,

    Sorry i needed to edit i made a mistake, can you look at my above edited post please?
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  4. #4
    Super Member Bacterius's Avatar
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    Quote Originally Posted by 1LISTEN View Post
    Bacterius,

    Sorry i needed to edit i made a mistake, can you look at my above edited post please?
    Done.
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  5. #5
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    Bacterius,

    Thank you very much!

    P.s: sorry i edited, my mistake.
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  6. #6
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    Bacterius,

    my calculator won't accept 2.65 x 10^211?? it comes up math error, can you tell me how many mile it is taking 1 mile = 1609.34 m please ?

    Thank you !
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  7. #7
    Super Member Bacterius's Avatar
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    Uhm ... how to tell you ... that's a 211-digit-long number. Here is the approached value :
    26500000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00
    I think it is better to keep it in scientific format.

    \mathfrak{I} \ \mathfrak{told} \ \mathfrak{you} \ \mathfrak{that} \ \mathfrak{was} \ \mathfrak{a} \ \mathfrak{lot} \ \mathfrak{of} \ \mathfrak{metres} ...
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  8. #8
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    Thanks for your kind help.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by 1LISTEN View Post
    If it was possible to have a sound source of 4225 db, i know that at 1m it looses 11dB then every time you double the distance it looses 6db.

    IE: 1m = 11, 2m = 17, 4m = 23, 8m = 29 etc etc

    Question:
    how far would it take in m for the dB level to go from 4225 to zero? working on the above formula?

    Edited for 1m and revised for above example.
    1. There is something funny about this, a source level is usually quoted at 1m.

    Lets ignore that and take it a face value as a souce level of 4214 dB re whatever.

    At range R the sound intensity level will be:

    IL(R)=4214-10 \log_{10}(R^2)

    (this is what it means by an additional loss of 6 dB per doubling in range)

    Now solving this gives R=10^{210.7}\text{ m} = 10^{207.7}\text{ km}\approx 5 \times 10^{207}\text{ km}.

    This is absurd since 1 light year is only \approx 9.5 \times 10^{12} \text{ km}.

    2. There is a maximum sound intensity that can be supported in an acoustic medium before the underlying assumptions become invalid and I am pretty sure that a source level of 4214 dB re the usual reference levels is way beyoun that.

    Are you sure that you are not missing a decimal point in the given level?

    CB
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  10. #10
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    CB,

    someone asked a question on a forum what would two levels of 65 dB be, i replied 68.01 dB .

    10 \log\left(10^{6.5} + 10^{6.5}\right) = 68.01\:dB

    someone else replied no! it's 65 x 65 dB = 4225

    so having the answer off here i replied how far they would hear the sound, lol!

    p.s, thanks for your help.
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