
Given that sound travels in air at $340m{s^{ - 1}}$ , find the wavelength of the waves in air produced by a $20kHz$ sound source. If the same source is put in a water tank, what would be the wavelength of the sound waves in water? Speed of sound in water $ = 1,480m{s^{ - 1}}$
Answer
558.9k+ views
Hint: We are required to find the wavelengths of the sound in air and water. We are given with speed of sound and frequency of sound. Recall a formula which related the three: wavelength, speed and frequency of sound.
Complete step by step answer:
Lets first note down the given quantities from the question and find a relation between them and wavelength.
Speed of sound in air: $340m{s^{ - 1}}$
Frequency of sound: $20kHz$
Speed of sound in water: $1,480m{s^{ - 1}}$
Now, in order to find the wavelength, we will have to relate the given quantities with wavelength.
Remember that the frequency of sound remains the same, irrespective of the medium it travels through.
We have the relation:
$v = f \times \lambda $
Where $v$is the velocity of the sound in the medium
$f$ is the frequency of the sound wave
$\lambda $ is the wavelength of the sound in a medium.
Using this formula and the given values from the question, we can easily calculate the wavelength of sound. The wavelength for air can be calculated in the following manner:
$v = f \times \lambda $
$ \Rightarrow \lambda = \dfrac{v}{f}$
Substituting the values of velocity of sound in air and frequency of sound, we have:
$ \Rightarrow \lambda = \dfrac{{340}}{{20000}}$
$ \therefore \lambda = 0.017m$
Therefore, the wavelength of sound in air is: $0.017m$
Now the wavelength of sound in water will be given as:
$v = f \times \lambda $
$\lambda = \dfrac{v}{f}$
$ \Rightarrow \lambda = \dfrac{{1480}}{{20000}}$
$ \therefore \lambda = 0.074m$
The wavelength of sound in water is: $0.074m$
Therefore, the wavelength of the sound in air and water is $0.017m$ and $0.074m$ respectively.
Note:
Irrespective of the medium, the frequency of sound in any medium remains constant.
Convert all units to SI units to get the final answer correct.
The speed of sound given in the problem is the average speed, in reality the speed of sound is not constant rather changing continuously.
Complete step by step answer:
Lets first note down the given quantities from the question and find a relation between them and wavelength.
Speed of sound in air: $340m{s^{ - 1}}$
Frequency of sound: $20kHz$
Speed of sound in water: $1,480m{s^{ - 1}}$
Now, in order to find the wavelength, we will have to relate the given quantities with wavelength.
Remember that the frequency of sound remains the same, irrespective of the medium it travels through.
We have the relation:
$v = f \times \lambda $
Where $v$is the velocity of the sound in the medium
$f$ is the frequency of the sound wave
$\lambda $ is the wavelength of the sound in a medium.
Using this formula and the given values from the question, we can easily calculate the wavelength of sound. The wavelength for air can be calculated in the following manner:
$v = f \times \lambda $
$ \Rightarrow \lambda = \dfrac{v}{f}$
Substituting the values of velocity of sound in air and frequency of sound, we have:
$ \Rightarrow \lambda = \dfrac{{340}}{{20000}}$
$ \therefore \lambda = 0.017m$
Therefore, the wavelength of sound in air is: $0.017m$
Now the wavelength of sound in water will be given as:
$v = f \times \lambda $
$\lambda = \dfrac{v}{f}$
$ \Rightarrow \lambda = \dfrac{{1480}}{{20000}}$
$ \therefore \lambda = 0.074m$
The wavelength of sound in water is: $0.074m$
Therefore, the wavelength of the sound in air and water is $0.017m$ and $0.074m$ respectively.
Note:
Irrespective of the medium, the frequency of sound in any medium remains constant.
Convert all units to SI units to get the final answer correct.
The speed of sound given in the problem is the average speed, in reality the speed of sound is not constant rather changing continuously.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

