## 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 ? :/

1. Michele_Laino

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

ok! :)

3. anonymous

what do we plug in from there?

4. Michele_Laino

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

whoah! okay! so then we get 40?

6. Michele_Laino

no, since: $\large 10 \times {\log _{10}}\left( 4 \right) = 10 \times 0.60206 = ...$

7. anonymous

ohh so it is 6.0206 so our answer is +6 dB?

8. Michele_Laino

yes! that's right!

9. anonymous

yay!! thanks!!:)

10. Michele_Laino

:)