
Two waves of wavelength 2m and 2.02m respectively moving with the same velocity superimpose to produce 2 beats per second. The velocity of the wave is –
A) 400\[m{{s}^{-1}}\]
B) 402\[m{{s}^{-1}}\]
C) 404\[m{{s}^{-1}}\]
D) 406\[m{{s}^{-1}}\]
Answer
545.1k+ views
Hint: We need to understand the relation between the wave parameters such as the wavelength, the frequency and the velocity of the waves in the given medium with the beat formation due to superimposition to solve this problem easily.
Complete Step-by-Step Solution:
We are given two waves such that they have different wavelengths and frequencies but have the same velocity in a given medium. We know that the beats formed are a result of the difference in the frequencies of the two waves which superimpose.
We know that the wave properties wavelength, frequency and the velocity are inter related by the simple relation given by –
\[\nu =\dfrac{v}{\lambda }\]
Where, \[\nu \] is the frequency of the wave, v is the speed of the wave and \[\lambda \] is the wavelength of the wave.
We can use this equation two find the relation between the two waves and from that we can find the velocity of the wave which is the same for both the waves.
i.e.,
\[\begin{align}
& {{\nu }_{1}}=\dfrac{v}{{{\lambda }_{1}}} \\
& \text{and,} \\
& {{\nu }_{2}}=\dfrac{v}{{{\lambda }_{2}}} \\
\end{align}\]
Now, we can find the relation which gives the beat formed by the superposition of the two waves as –
\[\begin{align}
& \nu ={{\nu }_{1}}-{{\nu }_{2}} \\
& \Rightarrow \nu =\dfrac{v}{{{\lambda }_{1}}}-\dfrac{v}{{{\lambda }_{2}}} \\
& \Rightarrow 2=v(\dfrac{1}{2}-\dfrac{1}{2.02}) \\
& \Rightarrow v=2\times 202m{{s}^{-1}} \\
& \therefore v=404m{{s}^{-1}} \\
\end{align}\]
This gives the velocity of the waves that superimpose in order to give a beat of 2 beats per second. The waves should be travelling with a speed of 404 \[m{{s}^{-1}}\] in the medium under consideration.
This is the required solution.
The correct answer is option C.
Note:
The beats are used in creating musical effects when used appropriately. The superimposing of the sound waves in different frequencies and at different distances give rise to a wide variety of sounds and beats which is the basic idea of different musical instruments.
Complete Step-by-Step Solution:
We are given two waves such that they have different wavelengths and frequencies but have the same velocity in a given medium. We know that the beats formed are a result of the difference in the frequencies of the two waves which superimpose.
We know that the wave properties wavelength, frequency and the velocity are inter related by the simple relation given by –
\[\nu =\dfrac{v}{\lambda }\]
Where, \[\nu \] is the frequency of the wave, v is the speed of the wave and \[\lambda \] is the wavelength of the wave.
We can use this equation two find the relation between the two waves and from that we can find the velocity of the wave which is the same for both the waves.
i.e.,
\[\begin{align}
& {{\nu }_{1}}=\dfrac{v}{{{\lambda }_{1}}} \\
& \text{and,} \\
& {{\nu }_{2}}=\dfrac{v}{{{\lambda }_{2}}} \\
\end{align}\]
Now, we can find the relation which gives the beat formed by the superposition of the two waves as –
\[\begin{align}
& \nu ={{\nu }_{1}}-{{\nu }_{2}} \\
& \Rightarrow \nu =\dfrac{v}{{{\lambda }_{1}}}-\dfrac{v}{{{\lambda }_{2}}} \\
& \Rightarrow 2=v(\dfrac{1}{2}-\dfrac{1}{2.02}) \\
& \Rightarrow v=2\times 202m{{s}^{-1}} \\
& \therefore v=404m{{s}^{-1}} \\
\end{align}\]
This gives the velocity of the waves that superimpose in order to give a beat of 2 beats per second. The waves should be travelling with a speed of 404 \[m{{s}^{-1}}\] in the medium under consideration.
This is the required solution.
The correct answer is option C.
Note:
The beats are used in creating musical effects when used appropriately. The superimposing of the sound waves in different frequencies and at different distances give rise to a wide variety of sounds and beats which is the basic idea of different musical instruments.
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