Question

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. $340$

Answer 1

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.