Question

Wavelengths of two notes in air are $ 5m $ and $ 30\text{ m} $ . Each of these notes produce 4 beats per seconds separately with a third note of fixed frequency find velocity of sound in air in $ \text{m/s} $ is:

Answer 1

Hint: The distance between one crest to another crust or we can say distance between one trough to another trough is known as wavelength of the wave. Its symbol is represented by $ \text{ }\!\!\lambda\!\!\text{ } $ . Basically, it is inversely proportional to frequency.
Frequency: It is the number of cycles or number of waves which passes through the fixed place in a given time that is known as frequency of the wave. Its symbol is represented by “f” or “v”
Frequency and wavelength are inversely proportional to each other.
Mathematically, it is given as:
 $ \text{ }\!\!\lambda\!\!\text{ }=\dfrac{\text{V}}{\text{F}} $
Using this formula, we will solve our problems.

Complete step by step solution
Let the speed of sound in air V and fixed frequency of third note of V.
Wavelength of first note $ \left( {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}} \right)=5\text{ m} $
Wavelength of second note $ \left( {{\text{ }\!\!\lambda\!\!\text{ }}_{2}} \right)=30\text{ m} $
We know that
$ \text{V}=\text{F}\times \text{ }\!\!\lambda\!\!\text{ } $
$ \text{ }\!\!\lambda\!\!\text{ } $ = wavelength
F = frequency
V = velocity of sound in air
Now, to find out $ {{\text{F}}_{1}} $
V = c = for air
$ \Rightarrow \text{V}={{\text{F}}_{1}}\times {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}} $
$ {{\text{F}}_{1}}=\dfrac{\text{V}}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}}} $
$ {{\text{F}}_{1}}=\dfrac{\text{V}}{\text{5}} $
Similarly for $ {{\text{F}}_{2}} $
$ \begin{align}
& \text{V}={{\text{F}}_{2}}\times {{\text{ }\!\!\lambda\!\!\text{ }}_{2}} \\
& {{\text{F}}_{2}}=\dfrac{\text{V}}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}}} \\
& \Rightarrow \dfrac{\text{V}}{30} \\
\end{align} $
Each of these notes produces = 4 beat/second
Two notes produces = $ 4\times 2=8 $ beat/second
 $ \therefore $ Expression will become
 $ {{F}_{1}}-{{F}_{2}}=8 $
 $ \Rightarrow \dfrac{V}{5}-\dfrac{V}{30}=8 $
 $ \Rightarrow \dfrac{6V-V}{30}=8 $
 $ \Rightarrow \dfrac{5V}{30}=8 $
 $ \Rightarrow 5V=30\times 8 $
 $ \Rightarrow V=\dfrac{240}{5}=48m/s $

Note
Remember that in air, the speed of light is faster than the speed of sound. If we want to compare the speed of sound and speed of light in air, we will found that the speed of sound is $ 343\text{ m/s} $ and the speed of light is $ 3\times {{10}^{10}}\text{ m/s} $ . In other words, we can say that a light wave travels 186 thousand miles in 1 second, while the sound wave travels 1 mile in 5 seconds.