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Homework Help: DB ear plug reduction

  1. Mar 10, 2010 #1
    1. When you buy earplugs at the store, they have a "dB rating". What it tells you is the reduction of SIL that the plugs will provide, if properly worn. Suppose you're helping some friends, working the stage at a rock concert wearing earplugs that provide a reduction of 40 dB. What is the ratio of the sound intensity just outside the plugs to the intensity reaching your eardrums? (40 dB may not seem huge, the music IS still pretty loud, but look what an impact this has on the energy being deposited in your ears!)



    2. Relevant equations
    -an increase in sound intensity by a factor of 10 increases the dB rating by 10.
    -dB measure a ratio.



    3. The attempt at a solution
    - I tried 40db, thinking that since dB is a ratio, this was the answer. I was wrong.
    - I tried 1E-8 as an answer because that is the intensity of 40 dB, but that was incorrect as well.

    how do i slove this problem?
     
  2. jcsd
  3. Mar 11, 2010 #2
    I would first write out the equation that relates intensity to decibels before you start guessing and checking.
     
  4. Mar 11, 2010 #3

    mgb_phys

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    "-an increase in sound intensity by a factor of 10 increases the dB rating by 10."
    Correct - but whats the relationship between power and dB?

    "-dB measure a ratio."
    Almost - but it's not the ratio you are thinking of, do you know about logs and powers of 10 ?
     
  5. Mar 11, 2010 #4
    Which by the way is
    dB = 10 log (I/I')
    where I' is constant reference intensity equal to 10^(-12) W/m^2.
     
  6. Mar 11, 2010 #5
    we arn't learning about logs in this class. It's a physics of sound and music course and isn't math based so we only learn basics... So, no. Unfortunately I don't know how to use logs in this case. Is there another way?
     
  7. Mar 11, 2010 #6
    Ok, then you are correct about "an increase in sound intensity by a factor of 10 increases the dB rating by 10."

    But you don't increase the dB by a factor of 10, you only increase it by adding 10 more dB.

    So for every 10 dB you add or subtract, you increase or decrease the intensity a factor of 10.
     
  8. Mar 11, 2010 #7
    Right, the intensity of the sound is 1E-8 W/m^2... so is that also the ratio of sound intensity just outside the plugs?

    The question asks for the ratio of the sound intensity just outside the plugs to the intensity reaching your eardrums.
     
  9. Mar 11, 2010 #8

    mgb_phys

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    dB is pretty fundementally about logs, but it's very simple.

    Each 'B' is a change in the power level by a factor of 10.
    And a dB is 0.1 of a 'B', the deci means 10.
    - we just use deciB rather than milliB or centiB for convenience and historical reasons.

    So 40db is 4 powers of 10 less - or 10,000
     
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