Question

The difference of sounds levels between two points is 40 dB. What is the ratio of pressure amplitudes between the two points?
${\text{A}}{\text{. 10}} \\\ {\text{B}}{\text{. 200}} \\\ {\text{C}}{\text{. 100}} \\\ {\text{D}}{\text{. 400}} \\\$

Answer 1

Hint: To find the ratio of pressure amplitudes between the two points, we calculate the ratio of intensities of sound at both points using the formula of decibel level. We substitute this ratio in the pressure amplitude formula to find the answer.

Complete step-by-step solution -

Given data,
Difference of sound = 40 dB

Formula used:
${\text{D = 10 log}}\left( {\dfrac{{\text{I}}}{{{{\text{I}}_{\text{o}}}}}} \right)$, where D is the decibel level, I is the intensity of sound at that point and ${{\text{I}}_{\text{o}}}$is the intensity of the least audible sound for humans.
${\text{P }}\alpha {\text{ }}\sqrt {\text{I}}$, where P is the pressure amplitude and I is the intensity of sound.

Let us consider two points with decibel levels${{\text{D}}_1},{{\text{D}}_2}$, intensities of sound${{\text{I}}_1},{{\text{I}}_2}$and pressure amplitudes ${{\text{P}}_1},{{\text{P}}_2}$respectively.
We know the sound or the decibel level at a point is given by the formula, ${\text{D = 10 log}}\left( {\dfrac{{\text{I}}}{{{{\text{I}}_{\text{o}}}}}} \right)$
Given the difference between the decibel levels at both points is 40 dB.
$\Rightarrow {{\text{D}}_1} - {{\text{D}}_2} = 40{\text{ dB}} \\\ \Rightarrow 40{\text{ = }}1{\text{0 log}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_{\text{o}}}}}} \right) - {\text{10 log}}\left( {\dfrac{{{{\text{I}}_2}}}{{{{\text{I}}_{\text{o}}}}}} \right) \\\$
$\Rightarrow 40{\text{ = }}10{\text{ log}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_{\text{o}}}}} - \dfrac{{{{\text{I}}_2}}}{{{{\text{I}}_{\text{o}}}}}} \right) \\\ \Rightarrow 40{\text{ = }}10{\text{ log}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_2}}}} \right){\text{ - - - - - }}\left( {\because \log {\text{a - logb = log}}\left( {\dfrac{{\text{a}}}{{\text{b}}}} \right)} \right) \\\ \Rightarrow 40{\text{ = }}10{\text{ log}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_2}}}} \right){\text{ }} \\\ \Rightarrow {\text{4 = lo}}{{\text{g}}_{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_2}}}} \right) \\\ \Rightarrow {10^4}{\text{ = }}\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_2}}}{\text{ - - - - (1)}} \\\$

Pressure amplitude at a given point is the maximum change is pressure at that point and is given by the formula ${\text{P }}\alpha {\text{ }}\sqrt {\text{I}}$
Now the ratio of pressure amplitudes between our two given points becomes –
$\dfrac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = \sqrt {\dfrac{{{{\text{I}}_1}}}{{{{\text{I}}_2}}}}$
Using equation (1) in the above we get
$\Rightarrow \dfrac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = \sqrt {{{10}^4}} \\\ \Rightarrow \dfrac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = {10^2} \\\ \Rightarrow \dfrac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = 100 \\\$
Hence the ratio of pressure amplitudes between the two points is 100. Option C is the correct answer.

Note – In order to answer this type of problem the key is to know the concepts of sound. Knowing that a common logarithm used in calculations has a base ‘10’ is one of the vital steps while solving this problem. We used an identity of logarithms in this problem that is, if a relation is of the form${\text{lo}}{{\text{g}}_{10}}\left( {\text{a}} \right) = {\text{b}}$ that implies${\text{a = 1}}{{\text{0}}^{\text{b}}}$.
Pressure amplitude is a physical property of sound waves and it has the unit Pascal shown as “Pa”. The ratio of pressure amplitude does not have any units as they are cancelled in the numerator and denominator.