### Home > CCA2 > Chapter 11 > Lesson 11.2.1 > Problem11-49

11-49.

Logarithms are used to measure the “loudness” of sound. Decibels (dB) are logarithmic units used to describe a ratio of two levels of intensity or pressure. The difference between two levels of sound pressure ( $P_1$ and $P_2$ ) is defined as $10$ log($\frac { P _ { 1 } } { P _ { 2 } }$)dB. Usually, when decibels are used to describe just one sound, it is assumed that that sound is being compared to a reference level of $20$ micropascals.

1. How many decibels correspond to doubling the pressure of a sound?

$10$ log$(2) ≈ 3.0$

2. What is the sound pressure of a noise described as $60$ dB?

$10\cdot$ log $(\frac{P}{20})=60$

log$(\frac{P}{20})=6$

$10^6=\frac{P}{20}$

$P = 20 · 10^6 = 2 · 10^7$

3. What does $0$ decibels mean?

The sound ratio is $1$.

4. How many times more pressure is in a sound of $40$ dB than of $20$ dB?

$100$ times more pressure.