
What is the effect of humidity on sound waves when humidity increases.
A. Speed of sound waves increases
B. Speed of sound waves decreases
C. Speed of sound waves remains same
D. Speed of sound waves becomes zero
Answer
135k+ views
Hint: The density of humid air is less than the density of dry air. So, when the humidity of air increases then the density of air decreases.
Formula used:
\[v = \sqrt {\dfrac{\beta }{\rho }} \]
where v is the velocity of sound in a medium with bulk modulus \[\beta \]and density \[\rho \].
Complete step by step solution:
Let the density of the dry air is \[\rho \] and bulk modulus is \[\beta \]. Then the velocity of sound in dry air will be,
\[{v_d} = \sqrt {\dfrac{\beta }{{{\rho _d}}}} \]
Let the density of the humid air is \[{\rho _H}\] and bulk modulus is \[\beta \]. Then the velocity of sound in humid air will be,
\[{v_H} = \sqrt {\dfrac{\beta }{{{\rho _H}}}} \]
Taking the ratio of the velocity of sound waves in dry air and humid air, we get
\[\dfrac{{{v_d}}}{{{v_H}}} = \dfrac{{\sqrt {\dfrac{\beta }{{{\rho _d}}}} }}{{\sqrt {\dfrac{\beta }{{{\rho _H}}}} }} \\ \]
\[\Rightarrow \dfrac{{{v_d}}}{{{v_H}}} = \sqrt {\dfrac{{{\rho _H}}}{{{\rho _d}}}} \]
The density of humid air is less than the density of dry air because the suspended water molecules in air increase the effective volume occupied by the air mass. So, when the humidity of air increases then the density of air decreases.
As the density of humid air is less than the density of the dry air, so
\[{\rho _d} > {\rho _H} \\ \]
\[\Rightarrow \sqrt {\dfrac{{{\rho _H}}}{{{\rho _d}}}} < 1 \\ \]
So, the relation of the velocity of the sound waves in two different medium is,
\[\dfrac{{{v_d}}}{{{v_H}}} < 1 \\ \]
\[\therefore {v_d} < {v_H}\]
So, the velocity of the sound wave in humid air is greater than the velocity of sound waves in dry air. Hence, the velocity of sound increases when the humidity of the air increases.
Therefore, the correct option is A.
Note: As the ratio is pure constant, so ignore the relation for which the square root of the ratio of densities of dry air and humid air is less than 1.
Formula used:
\[v = \sqrt {\dfrac{\beta }{\rho }} \]
where v is the velocity of sound in a medium with bulk modulus \[\beta \]and density \[\rho \].
Complete step by step solution:
Let the density of the dry air is \[\rho \] and bulk modulus is \[\beta \]. Then the velocity of sound in dry air will be,
\[{v_d} = \sqrt {\dfrac{\beta }{{{\rho _d}}}} \]
Let the density of the humid air is \[{\rho _H}\] and bulk modulus is \[\beta \]. Then the velocity of sound in humid air will be,
\[{v_H} = \sqrt {\dfrac{\beta }{{{\rho _H}}}} \]
Taking the ratio of the velocity of sound waves in dry air and humid air, we get
\[\dfrac{{{v_d}}}{{{v_H}}} = \dfrac{{\sqrt {\dfrac{\beta }{{{\rho _d}}}} }}{{\sqrt {\dfrac{\beta }{{{\rho _H}}}} }} \\ \]
\[\Rightarrow \dfrac{{{v_d}}}{{{v_H}}} = \sqrt {\dfrac{{{\rho _H}}}{{{\rho _d}}}} \]
The density of humid air is less than the density of dry air because the suspended water molecules in air increase the effective volume occupied by the air mass. So, when the humidity of air increases then the density of air decreases.
As the density of humid air is less than the density of the dry air, so
\[{\rho _d} > {\rho _H} \\ \]
\[\Rightarrow \sqrt {\dfrac{{{\rho _H}}}{{{\rho _d}}}} < 1 \\ \]
So, the relation of the velocity of the sound waves in two different medium is,
\[\dfrac{{{v_d}}}{{{v_H}}} < 1 \\ \]
\[\therefore {v_d} < {v_H}\]
So, the velocity of the sound wave in humid air is greater than the velocity of sound waves in dry air. Hence, the velocity of sound increases when the humidity of the air increases.
Therefore, the correct option is A.
Note: As the ratio is pure constant, so ignore the relation for which the square root of the ratio of densities of dry air and humid air is less than 1.
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