Question

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 ? :/

Answer 1

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

Answer 2

ok! :)

Answer 3

what do we plug in from there?

Answer 4

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:
\[\large \begin{gathered}
\Delta N = {N_2} - {N_1} = 10 \times {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) - 10 \times {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) = \hfill \\
\hfill \\
= 10 \times {\log _{10}}\left( 4 \right) \hfill \\
\end{gathered} \]

Answer 5

whoah! okay! so then we get 40?

Answer 6

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

Answer 7

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

Answer 8

yes! that's right!

Answer 9

yay!! thanks!!:)

Answer 10

:)