Solveeit Logo
Login

Question

Question: About how many times more intense will the normal ear perceive a sound of \({10^{ - 6}}W/{m^2}\) tha...

About how many times more intense will the normal ear perceive a sound of 10−6W/m2{10^{ - 6}}W/{m^2} than one of 10−9W/m2{10^{ - 9}}W/{m^2}?

Explanation

Solution

Loudness is the subjective perception of sound pressure. To find the loudness in sound or the change in the sound, use ΔL=10×log⁡10I2I1\Delta L = 10 \times {\log _{10}}\dfrac{{{I_2}}}{{{I_1}}}, where I1,I2{I_{1,}}{I_2} are the given value of sounds. Substitute those intensities to find the more intensity perceived by the ear.

Complete step by step answer:
We are given to find how many times more intense will the normal ear perceive a sound of 10−6W/m2{10^{ - 6}}W/{m^2} than one of 10−9W/m2{10^{ - 9}}W/{m^2}
Intensities are given. We have to calculate the loudness.
Loudness is the human perception of sound intensity. It is frequently measured in decibels which is a scale based on the human threshold of hearing.
Loudness is given by the formula
ΔL=10×log⁡10I2I1\Delta L = 10 \times {\log _{10}}\dfrac{{{I_2}}}{{{I_1}}}
Where ΔL=L2−L1\Delta L = {L_2} - {L_1} which is also equal to ΔL=10×log⁡10I2I1\Delta L = 10 \times {\log _{10}}\dfrac{{{I_2}}}{{{I_1}}}
Loudness is the difference of the sounds.
ΔL=10×log⁡10I2I1\Delta L = 10 \times {\log _{10}}\dfrac{{{I_2}}}{{{I_1}}}
Substitute the values of the intensities given in the question.
ΔL=10×log⁡10I2I1 I1=10−9W/m2,I2=10−6W/m2 ⇒ΔL=10×log⁡1010−610−9 ⇒ΔL=10×log⁡10103 (∵aman=am−n) ⇒ΔL=10×3 (∵log⁡aan=nlog⁡aa=n) ⇒ΔL=30  \Delta L = 10 \times {\log _{10}}\dfrac{{{I_2}}}{{{I_1}}} \\\ {I_1} = {10^{ - 9}}W/{m^2},{I_2} = {10^{ - 6}}W/{m^2} \\\ \Rightarrow \Delta L = 10 \times {\log _{10}}\dfrac{{{{10}^{ - 6}}}}{{{{10}^{ - 9}}}} \\\ \Rightarrow \Delta L = 10 \times {\log _{10}}{10^3} \\\ \left( {\because \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right) \\\ \Rightarrow \Delta L = 10 \times 3 \\\ \left( {\because {{\log }_a}{a^n} = n{{\log }_a}a = n} \right) \\\ \Rightarrow \Delta L = 30 \\\
The loudness is 30 decibels.
Therefore, 30 times more intense the normal ear will perceive a sound of 10−6W/m2{10^{ - 6}}W/{m^2} than one of 10−9W/m2{10^{ - 9}}W/{m^2}

Additional Information: Sound is a vibration that propagates as an acoustic wave, through a transmission medium such as a gas, liquid or solid. In human psychology, sound is the reception of such waves and their perception by the brain.

Note: Sound intensity is defined as the sound power per unit area. The most common method to sound intensity measurement is to use the decibel scale and it is measured in decibels. Sound intensity can be measured in units of energy or work or in units of power, as watts per square metre.