Question

In the case of sound waves, wind is blowing from source to receiver with speed ${U_w}$. Both source and receiver are stationary. If ${\lambda _0}$ is the original wavelength with no wind and V is the speed of sound in air, then find the wavelength as received by the receiver.
A. ${\lambda _0}$
B. $\left( {\dfrac{{V + {U_w}}}{V}} \right){\lambda _0}$
C. $\left( {\dfrac{{V - {U_w}}}{V}} \right){\lambda _0}$
D. $\left( {\dfrac{V}{{V + {U_w}}}} \right){\lambda _0}$

Answer 1

The above problem can be resolved using the mathematical relation for the two cases. One case is that when the sound travels in such a way that is no wind blowing at the moment, and the second case is by taking that the wing is blowing with some magnitude of velocity, then solve accordingly to obtain the desired result.

Complete step by step answer:
Given:
Wind is blowing from source to receiver with speed ${U_w}$.
The original wavelength is, ${\lambda _0}$.
The speed of sound in air is, $V$.
The speed of the sound, when there is no wind present. The mathematical relation is given as,
$V = f{\lambda _0}............................\left( 1 \right)$
Here, f denotes the frequency of the sound wave.
The speed of sound, when the velocity of wind is, ${U_w}$. The mathematic expression for the condition is given as,
$V + {U_w} = f{\lambda _1}..................................\left( 2 \right)$
Here, ${\lambda _1}$ is the wavelength of sound waves as received by the receiver.
Dividing the equation 1 and 2 as,

\dfrac{V}{{V + {U_w}}} = \dfrac{{f{\lambda _0}}}{{f{\lambda _1}}}\\\ {\lambda _1} = \left( {\dfrac{V}{{V + {U_w}}}} \right) \times {\lambda _0} \end{array}$$ Therefore, the wavelength of sound waves as received by the receiver is $$\left( {\dfrac{V}{{V + {U_w}}}} \right) \times {\lambda _0}$$ **So, the correct answer is “Option D”.** **Note:** In order to resolve the cases as given above, it is important to analyze the situation for the sound waves as compared to the wind; there can be two possible cases. The first case will be, either the wind will blow with some magnitude of velocity or there will be the magnitude of sound waves alone in the atmosphere. This will create a significant result as per the requirement.