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Question: On the same path, the source and observer are moving such a way that the distance between these two ...

On the same path, the source and observer are moving such a way that the distance between these two increases with the time. The speeds of source and observer are the same and equal to 10m/s10m/s with respect to the ground while no wind is blowing. The apparent frequency received by observer is 1950 Hz1950{\text{ }}Hz , then the original frequency must be (the speed of sound in present medium is 340 m/s340{\text{ }}m/s )
(A) 2068Hz2068Hz
(B) 2100Hz2100Hz
(C) 1903Hz1903Hz
(D) 602Hz602Hz

Explanation

Solution

Hint
When both the source and observer are in state of motion, there is a Doppler effect observed. We can use the given values of speed and apparent frequency to calculate the original frequency.
fL=(v+vL)(v+vS)fS\Rightarrow {f_L} = \dfrac{{(v + {v_L})}}{{(v + {v_S})}}{f_S}

Complete step by step answer
As given in the question the distance between the source and observer increases with time, the frequency seems to be decreased to the observer. Since both have the same velocity but the distance is said to be increased, this means that the source and observer are moving in opposite directions.
Doppler effect is the apparent change observed in the frequency of source due to the relative motion of observer and source.
Apparent frequency in this case will be
fL=(vvL)(v+vS)fS\Rightarrow {f_L} = \dfrac{{(v - {v_L})}}{{(v + {v_S})}}{f_S}
Where,
fL\Rightarrow {f_L} refers to the frequency of sound heard by the observer
v is the speed of sound in the medium in which it propagates in this case air hence 340 m/s
vL{v_L} refers to the velocity of the observer
vS{v_S} is the velocity of the source
fS{f_S} refers to the frequency of sound source
Here vL=vS=10m/s{v_L} = {v_S} = 10m/s and fL=1950Hz{f_L} = 1950Hz v=340m/sv = 340m/s
Putting these values in the apparent frequency equation discussed above
1950=(34010)(340+10)fSfS=1950×350330=2068.18Hz\Rightarrow 1950 = \dfrac{{(340 - 10)}}{{(340 + 10)}}{f_S} \Rightarrow {f_S} = \dfrac{{1950 \times 350}}{{330}} = 2068.18Hz
Which is approximately equal to option (A), Hence, the correct answer is (A).

Additional Information
Dopplers effect is used in various day-to-day applications like velocity of submarines in water, velocity of airplanes in flying air and also in radar stations to measure velocity of objects like speeding vehicles. Doppler effect is also seen in light and is used to determine velocities of stars and galaxies.

Note
If there is wind blowing then it will not affect the frequency of source until and unless there is a relative motion between source and the observer.