Question

jim halves the distance between himself and a sound source. What is the change of decibels of the sound he hears?

Answer 1

Answer:

6.02dB

Explanation:

The sound level intensity can be calculated by using the formula:

$I(dB)=10log_{10}(\frac{I}{I_o})$

where Io is the threshold of human hearing, and  I is the intensity of the sound in at a certain distance. I is given by

$I=\frac{P}{4\pi r^2}$

where P is the power of the source and r is the distance to the source. By replacing I in the expression for I(dB) we obtain

$I(dB)=10log(\frac{P}{4\pi r^2I_o})=10log(\frac{P}{\pi r^2I_o})-10log(4)$

when jim halves the distance we have r'=1/2r:

$I'(dB)=10log(\frac{P}{4\pi (\frac{1}{2}r)^2I_o})=10log(\frac{P}{\pi r^2I_o})$

Hence, the change o decibels is:

$I'-I=10log(4)=6.02dB$

hope this helps!!