Question

The human ear can detect continuous sounds in the frequency range from \[20{\rm{ Hz}}\] to \[20,000{\rm{ Hz}}\]. Assuming that the speed of sound in air is \[330{\rm{ m}}{{\rm{s}}^{ - 1}}\] for all frequencies, calculate the wavelengths corresponding to the given extreme frequencies of the audible range.

Answer 1

Hint: The wavelength of a sound wave is the distance between the adjacent crest and the trough of that wave. The expression of the wavelength of a sound wave is equal to the ratio of sound's speed in the wave's air to the wave frequency.

Complete step by step answer:
Given:
The speed of sound in air is \[v = 330{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\].
The minimum audible frequency for the human ear is \[{\nu _1} = 20{\rm{ Hz}}\].
The maximum audible frequency for the human ear is \[{\nu _2} = 20,000{\rm{ Hz}}\].
Assume:
The wavelength corresponding to minimum audible frequency is \[{\lambda _1}\].
The wavelength corresponding to maximum audible frequency is \[{\lambda _2}\].
We are required to calculate the value of \[{\lambda _1}\] and \[{\lambda _2}\].
Let us write the expression for minimum audible frequency for the human ear.
\[{\lambda _1} = \dfrac{v}{{{\nu _1}}}\]
Substitute \[330{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\] for v and \[20{\rm{ Hz}}\] for \[{\nu _1}\] in the above expression.
\[
{\lambda _1} = \dfrac{{330{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}}}{{20{\rm{ Hz}}}}\\
\Rightarrow{\lambda _1} = \dfrac{{330{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}}}{{20{\rm{ Hz}} \times \left( {\dfrac{{{1 {\left/
 {\vphantom {1 {\rm{s}}}} \right.
} {\rm{s}}}}}{{{\rm{Hz}}}}} \right)}}\\
\Rightarrow{\lambda _1} = 16.5{\rm{ m}}
\]
Write the expression for maximum audible frequency for the human ear.
\[{\lambda _2} = \dfrac{v}{{{\nu _2}}}\]

Substitute \[330{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\] for v and \[20,000{\rm{ Hz}}\] for \[{\nu _2}\] in the above expression.
\[
\Rightarrow{\lambda _2} = \dfrac{{330{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}}}{{20,000{\rm{ Hz}}}}\\
\Rightarrow{\lambda _2} = \dfrac{{330{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}}}{{20,000{\rm{ Hz}} \times \left( {\dfrac{{{1 {\left/
 {\vphantom {1 {\rm{s}}}} \right.
} {\rm{s}}}}}{{{\rm{Hz}}}}} \right)}}\\
\therefore {\lambda _2} = 0.0165{\rm{ m}}\]

Therefore, the wavelength corresponding to the minimum and maximum audible frequency for the human ear are \[16.5{\rm{ m}}\] and \[0.0165{\rm{ m}}\] respectively.

Note: Do not forget to convert the given unit of frequency into its base unit. We can also remember that the unit of time that is second is the inverse of hertz or vice-versa. It is given that the human ear can detect continuous sounds in the frequency range from \[20{\rm{ Hz}}\] to \[20,000{\rm{ Hz}}\], this range reduces with the age of the person. But it is found that the sound of frequency lower than \[250{\rm{ Hz}}\] is hard to detect. People are not able to recognize the sound below, \[250{\rm{ Hz}}\] although it is in the audible frequency range. Also, a long duration of exposure to loud noise within the range of audible frequency can cause hearing loss.