Question

Question: (a) What is the intensity of a sound that has a level \(7.00\;{\rm{dB}}\) lower than a\(4.00 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)sound? (b) What is the intensity of a sound that is \(3.00\;{\rm{dB}}\)higher than a \(4.00 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)sound?

Short answer

1. The intensity is\(7.94 \times {10^{ - 10}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

b. The intensity is\(7.94 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)

Step-by-step solution

Given Data

The intensity of a sound has a level\(7.00\;{\rm{dB}}\)lower than a\(4.00 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

The intensity of another sound has a level \(3.00\;{\rm{dB}}\) higher than a\(4.00 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

Loudness

The loudness is the ratio of the sound intensity to the threshold level. It is in the unit of Bel or dB. The loudness refers to the effect of intensity.

Calculation of the intensity of the first sound

(a)

Use the intensity of the first sound as ,

\(\begin{align}d &= 10\log \frac{{4.00 \times {{10}^{ - 9}}}}{{{{10}^{ - 12}}}}\\d &= 36\;{\rm{dB}}\end{align}\)

Now the intensity level of the sound is,

\(\begin{align}I &= 36 - 7\\ &= 29\;{\rm{dB}}\end{align}\)

The intensity of the sound is,

\(\begin{align}29 &= 10\log \frac{I}{{{{10}^{ - 12}}}}\\2.9 &= \log \frac{I}{{{{10}^{ - 12}}}}\\I &= {10^{2.9}} \times {10^{ - 12}}\\I &= 7.94 \times {10^{ - 10}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

The intensity of a sound is \(7.94 \times {10^{ - 10}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)

Calculation of the intensity of the second sound

(b)

Now the intensity level of the sound is,

\(36 + 3 = 39\;{\rm{dB}}\)

The intensity of the sound is,

\(\begin{align}39 &= 10\log \frac{I}{{{{10}^{ - 12}}}}\\3.9 &= \log \frac{I}{{{{10}^{ - 12}}}}\\I &= {10^{3.9}} \times {10^{ - 12}}\\I &= 7.94 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

Therefore,the intensity of a sound is \(7.94 \times {10^{ - 9}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)