The audible frequency for a normal human being is 20 Hz to 20 KHz. Find the corresponding wavelength if the speed of sound in air is 320 m/s.
Answer
661.5k+ views
Hint: Sound waves that have frequencies within the range of 20Hz-20 KHz is called audible sound. Wavelength is inversely proportional to frequency. The relation between speed of sound, frequency of sound and its wavelength is given as, speed of sound is the product of wavelength and frequency of the sound.
Formula used:
$\lambda =\dfrac{v}{f}$, where ‘$\lambda $’ is the wavelength, ‘$v$’ is the velocity and ‘$f$’ is frequency.
Complete step by step answer:
The audible frequency of human beings is given in the question, 20 Hz to 20 KHz.
We have to find its wavelength when the speed of sound in air is 320 m/s.
Minimum audible frequency,
${{f}_{audible}}\left( \min \right)=20Hz$
Maximum audible frequency,
${{f}_{audible}}(\max )=20KHz$
Speed of sound in air,
$v=320m/s$
We have the equation for wavelength,
$\lambda =\dfrac{v}{f}$ , Where ‘$\lambda $’ is the wavelength of the sound, ‘$v$ ‘is velocity of sound and ‘$f$ ‘is the frequency of sound.
Let${{\lambda }_{1}}$ be the wavelength corresponding to minimum audible frequency.
Then,
$\begin{align}
& {{\lambda }_{1}}=\dfrac{v}{{{f}_{\min }}} \\
& {{\lambda }_{1}}=\dfrac{320}{20} \\
& {{\lambda }_{1}}=16m \\
\end{align}$
Now, let${{\lambda }_{2}}$ be the wavelength of maximum audible frequency
Then,
$\begin{align}
& {{\lambda }_{2}}=\dfrac{v}{{{f}_{\max }}} \\
& {{\lambda }_{2}}=\dfrac{320}{20KHz} \\
& {{\lambda }_{2}}=\dfrac{320}{20\times {{10}^{3}}} \\
& {{\lambda }_{2}}=16\times {{10}^{-3}}m \\
\end{align}$
Therefore the corresponding wavelength to minimum frequency $=16m$
Corresponding wavelength to maximum frequency, $=16\times {{10}^{-3}}m=16mm$.
Note:
From this we can see that, for a minimum frequency we get maximum wavelength. And for maximum frequency, the wavelength we get will be minimum. I.e. We verify that wavelength and frequency are inversely proportional to each other.
To convert Kilohertz to Hertz, we multiply Kilohertz by 1000
$1KHz=1000Hz$
To convert meter to millimeter, we multiply meter with 1000.
$1m=1000mm$
Formula used:
$\lambda =\dfrac{v}{f}$, where ‘$\lambda $’ is the wavelength, ‘$v$’ is the velocity and ‘$f$’ is frequency.
Complete step by step answer:
The audible frequency of human beings is given in the question, 20 Hz to 20 KHz.
We have to find its wavelength when the speed of sound in air is 320 m/s.
Minimum audible frequency,
${{f}_{audible}}\left( \min \right)=20Hz$
Maximum audible frequency,
${{f}_{audible}}(\max )=20KHz$
Speed of sound in air,
$v=320m/s$
We have the equation for wavelength,
$\lambda =\dfrac{v}{f}$ , Where ‘$\lambda $’ is the wavelength of the sound, ‘$v$ ‘is velocity of sound and ‘$f$ ‘is the frequency of sound.
Let${{\lambda }_{1}}$ be the wavelength corresponding to minimum audible frequency.
Then,
$\begin{align}
& {{\lambda }_{1}}=\dfrac{v}{{{f}_{\min }}} \\
& {{\lambda }_{1}}=\dfrac{320}{20} \\
& {{\lambda }_{1}}=16m \\
\end{align}$
Now, let${{\lambda }_{2}}$ be the wavelength of maximum audible frequency
Then,
$\begin{align}
& {{\lambda }_{2}}=\dfrac{v}{{{f}_{\max }}} \\
& {{\lambda }_{2}}=\dfrac{320}{20KHz} \\
& {{\lambda }_{2}}=\dfrac{320}{20\times {{10}^{3}}} \\
& {{\lambda }_{2}}=16\times {{10}^{-3}}m \\
\end{align}$
Therefore the corresponding wavelength to minimum frequency $=16m$
Corresponding wavelength to maximum frequency, $=16\times {{10}^{-3}}m=16mm$.
Note:
From this we can see that, for a minimum frequency we get maximum wavelength. And for maximum frequency, the wavelength we get will be minimum. I.e. We verify that wavelength and frequency are inversely proportional to each other.
To convert Kilohertz to Hertz, we multiply Kilohertz by 1000
$1KHz=1000Hz$
To convert meter to millimeter, we multiply meter with 1000.
$1m=1000mm$
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

