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Q17.3-24PE

Expert-verified
Found in: Page 629

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by$$40.0\;{\rm{dB}}$$?

The sound intensity level goes up by a factor of $${10^4}$$.

See the step by step solution

## Given Data

The intensity of the sound is $$d = 40.0\;{\rm{dB}}$$ more than the reference level.

## Sound intensity and Loudness

The watt per square meter is the SI measure of intensity which depends on the wave's amplitude. The loudness is the effect of sound intensity in the human ear when it exceeds the threshold.

## Calculation of the intensity

Use the intensity of the sound,

$$\begin{array}{c}{d_1} = 10\log \frac{{{I_1}}}{{{{10}^{ - 12}}}}\\{10^{\frac{{{d_1}}}{{10}}}} \times {10^{ - 12}} = {I_1}\end{array}$$

Now

$$\begin{array}{c}{d_1} + 40.0 = 10\log \frac{{{I_2}}}{{{{10}^{ - 12}}}}\\{I_2} = {10^{\frac{{{d_1} + 40.0}}{{10}}}} \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{array}$$

The ratio of the intensities is,

$$\begin{array}{c}\frac{{{I_2}}}{{{I_1}}} = \frac{{{{10}^{\frac{{{d_1} + 40.0}}{{10}}}} \times {{10}^{ - 12}}}}{{{{10}^{\frac{{{d_1}}}{{10}}}} \times {{10}^{ - 12}}}}\\\frac{{{I_2}}}{{{I_1}}} = {10^4}\end{array}$$

Therefore, the factor is $${10^4}$$ when the amplitude of a sound wave increase if the sound intensity level goes up by$$40.0\;{\rm{dB}}$$