Question

a) How much more intense is a sound that has a level \(17.0\;{\rm{dB}}\)higher than another? (b) If one sound has a level \(23.0\;{\rm{dB}}\) less than another, what is the ratio of their intensities?

Short answer

1. The sound is \(50.118\) times more intense than the first.
2. The ratio of the intensities is \(5.01 \times {10^{ - 3}}\).

Step-by-step solution

Given data:

The intensity of a soundhas a level \(17.0\;{\rm{dB}}\) higher than another.

The intensity of a sound has a level \(23.0\;{\rm{dB}}\) less than another.

Definition of sound intensity

Sound intensity is the power carried by sound waves per unit area in a direction perpendicular to a region. The intensity provides the sensual effect of loudness, and the unit of loudness is dB.

Calculation of the intensity of the first sound

(a)

Use the intensity of the first sound as,

\(\begin{align}d &= 10\log \frac{I}{{{{10}^{ - 12}}}}\\\frac{d}{{10}} &= \log \frac{I}{{{{10}^{ - 12}}}}\\{10^{\frac{d}{{10}}}} &= \frac{I}{{{{10}^{ - 12}}}}\\I &= {10^{\frac{d}{{10}}}} \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

The intensity of the second sound is,

\(\begin{align}d + 17 &= 10\log \frac{{{I_1}}}{{{{10}^{ - 12}}}}\\\frac{{d + 17}}{{10}} &= \log \frac{{{I_1}}}{{{{10}^{ - 12}}}}\\{10^{\frac{{d + 17}}{{10}}}} &= \frac{{{I_1}}}{{{{10}^{ - 12}}}}\\{I_1} &= {10^{\frac{{d + 17}}{{10}}}} \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

The ratio of the intensities is,

\(\begin{align}\frac{{{I_1}}}{I} &= \frac{{{{10}^{\frac{{d + 17}}{{10}}}} \times {{10}^{ - 12}}}}{{{{10}^{\frac{d}{{10}}}} \times {{10}^{ - 12}}}}\\{I_1} &= I \times {10^{\frac{{17}}{{10}}}}\\{I_1} &= I \times 50.118\end{align}\)

Therefore the sound is \(50.118\) times more intense than the first.

Calculation of the intensity of the second sound

(b)

The intensity of the second sound is,

\(\begin{align}{c}d - 23 &= 10\log \frac{{{I_2}}}{{{{10}^{ - 12}}}}\\\frac{{d - 23}}{{10}} &= \log \frac{{{I_2}}}{{{{10}^{ - 12}}}}\\{10^{\frac{{d - 23}}{{10}}}} &= \frac{{{I_2}}}{{{{10}^{ - 12}}}}\\{I_2} &= {10^{\frac{{d - 23}}{{10}}}} \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

The ratio of the intensities is,

\(\begin{align}\frac{{{I_2}}}{I} &= \frac{{{{10}^{\frac{{d - 23}}{{10}}}} \times {{10}^{ - 12}}}}{{{{10}^{\frac{d}{{10}}}} \times {{10}^{ - 12}}}}\\\frac{{{I_2}}}{I} &= {10^{ - \frac{{23}}{{10}}}}\\\frac{{{I_2}}}{I} &= 5.01 \times {10^{ - 3}}\end{align}\)

Therefore, \(5.01 \times {10^{ - 3}}\)is the ratio of the intensities.