Change Intensity Levels of a Sound: What Happens to the Decibel Level?

Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of logarithms.
Solution
(1) Identify knowns:

The ratio of the two intensities is 2 to 1, or: $\frac{I_{2}}{I_{1}} = 2.00 .$

We want to show that the difference in sound levels is about 3 dB. That is, we want to show

$β_{2} - β_{1} = 3 dB.$
14.8

Note that

$\log_{10} b - \log_{10} a {= log}_{10} (\frac{b}{a}) .$
14.9

(2) Use the definition of β to get

$β_{2} - β_{1} = 10 {log}_{10} (\frac{I_{2}}{I_{1}}) = 10 {log}_{10} 2.00 = 10 (0.301) dB.$
14.10

Therefore, $β_{2} - β_{1} = 3.01 dB.$

0 of 2 steps open

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