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anonymous

  • one year ago

Jim halves the distance between himself and a sound source. What is the change in decibels of the sound he hears? -6 dB, -2 dB, +2 dB, or +6 dB ? :/

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  1. Michele_Laino
    • one year ago
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    we start with an intensity I_1, after Jim halves the distance, the new intensity will be I_2= 4*I_1 so we can write: \[\large \begin{gathered} {N_1} = {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) \hfill \\ \hfill \\ {N_2} = {\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right) = {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) \hfill \\ \end{gathered} \] where I_0 is a reference intensity

  2. anonymous
    • one year ago
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    ok! :)

  3. anonymous
    • one year ago
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    what do we plug in from there?

  4. Michele_Laino
    • one year ago
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    sorry the right formulas are: \[\large \begin{gathered} {N_1} = 10 \times {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) \hfill \\ \hfill \\ {N_2} = 10 \times {\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right) = 10 \times {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) \hfill \\ \end{gathered} \] now the requested change is: \[\large \begin{gathered} \Delta N = {N_2} - {N_1} = 10 \times {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) - 10 \times {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) = \hfill \\ \hfill \\ = 10 \times {\log _{10}}\left( 4 \right) \hfill \\ \end{gathered} \]

  5. anonymous
    • one year ago
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    whoah! okay! so then we get 40?

  6. Michele_Laino
    • one year ago
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    no, since: \[\large 10 \times {\log _{10}}\left( 4 \right) = 10 \times 0.60206 = ...\]

  7. anonymous
    • one year ago
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    ohh so it is 6.0206 so our answer is +6 dB?

  8. Michele_Laino
    • one year ago
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    yes! that's right!

  9. anonymous
    • one year ago
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    yay!! thanks!!:)

  10. Michele_Laino
    • one year ago
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    :)

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