The apparent frequency of a note, when a listener moves towards a stationary source, with velocity of \[40m/s\] is \[200Hz\]. When he moves away from the same source with the same speed, the apparent frequency of the same note is \[160Hz\]. The velocity of sound in air is \[(\]in \[m/s\]\[)\]
A. \[360\]
B. \[330\]
C. \[320\]
D. \[340\]

Hint:To solve this question, we have to know about the Doppler effect. This is also called Doppler shift. This is the change in frequency of a wave relation to an observer who is moving relative to the wave source. Doppler is an Austrian physicist who described this phenomenon in \[1842\].

Complete step by step answer:
Now, we are assuming the original frequency of the source is, \[{f_0}\].Doppler effect when an observer moves towards the stationary source. We can write, the apparent frequency is f’.
We know that, \[{f^1} = {f_0}(\dfrac{{{V_{sound}} + {V_0}}}{{{V_{sound}}}})\]
\[200 = {f_0}(\dfrac{{{V_{sound}} + 40}}{{{V_{sound}}}})\]
Doppler Effect when an observer moves away from the stationary source .
Now, we can say, the apparent frequency heard,
We know that,
\[160 = {f_0}(\dfrac{{{V_{sound}} - 40}}{{{V_{sound}}}})\]
Now, we are going to divide f’ by f”
So, \[200/160 = \dfrac{{{V_{sound}} + 40}}{{{V_{sound}} - 40}}\]
\[5{V_{sound}} - 200 = 4{V_{sound}} + 160\]
\[\therefore {V_{sound}} = 360m/s\]
This is the right answer.

So, the right option would be option A.

Note:Here we have to know about the sound wave properties and Doppler effect. We know that, Regular illustration of Doppler move is the difference in pitch heard when a vehicle sounding a horn draws near and subsides from a spectator. Contrasted with the discharged recurrence, the got recurrence is higher during the methodology, indistinguishable at the moment of cruising by, and lower during the downturn.