# Physics Lesson 11.5.6 - Sound Level

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Welcome to our Physics lesson on Sound Level, this is the sixth lesson of our suite of physics lessons covering the topic of Sound Waves. Intensity and Sound Level, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

## Sound Level

Sound Intensity does not vary linearly with power because distance is raised in power two in the formula that relates sound intensity and power. Therefore, it is not that easy dealing with it in calculations. Hence, scientists prefer to use another physical quantity known as sound level to describe the strength of sound waves. Sound level, L is measured in decibel (dB) in honour to Alexander Graham Bell and it is related to the sound intensity according a logarithmic scale, i.e.

L(dB) = 10 × log(I/I0)

where I is the intensity of the actual sound and I0 is the threshold of hearing. You may have read in the parameters plate of air conditioners or refrigerators the noise in dB these devices produce during their operation. (Noise level is another quantity measured in decibels as it represents irregular sound waves).

Since intensity of sound is the power per unit area, we can also use the equation

L(dB) = 10 × log(P/P0)

to represent the noise level. We can take the threshold sound power as P0 = 10 - 12 W.

### Example 3

Sound power in heavy traffic is 0.001 W. What is the value of noise level in dB caused by the heavy traffic?

### Solution 3

We have P = 0.001 W = 10 - 3 W and P0 = 10 - 12 W. Thus,

L(dB) = 10 × log(P/P0 )
= 10 × log(10-3/10-12))
= 10 × log10-3-(-12)
= 10 × log109
= 9 × 10
= 90 dB

(Here the logarithm properties log ab = b × log a and log 10 = 1 are used).

The advantage of using a logarithmic scale for calculating the sound level consists on the small range of sound level values when intensity changes too much. Let's consider an example to illustrate this point.

### Example 4

A machinery in a factory produces a sound of intensity equal to 100 dB. What is the sound level produced by 40 such machineries if they are turned on simultaneously?

### Solution 4

The procedure to solve this problem is as follows: We must calculate the sound intensity produced by a single machinery first, and then multiply it by 40. Then we must calculate the new sound level in dB. Thus,

L1 = 10 × log(I1/I0)
100 = 10 × log(I1/10-12)
100/10 = log I1 - log10-12
10 = log I1 - (-12) × log10
10 = log I1 + 12 × 1
log I1 = -2
I1 = 10 - 2
= 0.01 W/m2

The sound intensity produced by 40 such machineries therefore is

I40 = 40 × I1
= 40 × 0.01 W/m2
= 0.4 W/m2
= 4 × 10-1 W/m2

Hence, the sound level produced by these 40 machineries when they turn on simultaneously is

L40 = 10 × log(I40/I0)
= 10 × log(4 × 10-1/10-12 )
= 10 × log4 × 1011
= 10 × (log 4 + log1011)
= 10 × (log 4 + 11 × log10)
= 10 × (0.6 + 11)
= 10 × 11.6
= 116 dB

You may notice that for a 40 times increase in sound intensity, the sound level increases only by 16 dB (from 100 dB to 116 dB). Therefore, don't neglect the slight changes in the values of sound level as they represent a very big change in sound intensity.

Below some typical values for sound intensity and sound level are given. You have reached the end of Physics lesson 11.5.6 Sound Level. There are 7 lessons in this physics tutorial covering Sound Waves. Intensity and Sound Level, you can access all the lessons from this tutorial below.

## More Sound Waves. Intensity and Sound Level Lessons and Learning Resources

Waves Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
11.5Sound Waves. Intensity and Sound Level
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
11.5.1Things You Already Know About Sound Waves
11.5.2Things You Need To Know About Sound Waves
11.5.3Limits of Audibility. Audible Sound. Infra and Ultrasound<
11.5.4Echo
11.5.5Intensity of Sound Waves
11.5.6Sound Level
11.5.7Loudness and Pitch

## Whats next?

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