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On the same path, the source and observer are moving in such a way that the distance between these two increases with the time. The speeds of source and observer are same and equal to 10 ms^-1 with respect to the ground while no wind is blowing. The apparent frequency received by observer is 1950 Hz, then the original frequency must be (Take, the speed of sound in present medium is 340 ms^-1)

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2068 Hz

2100 Hz

1903 Hz

602 Hz

answer is A.

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Detailed Solution

Let the original frequency be f, then apparent frequency,

$f' = \frac{v - u_{o}}{v + u_{s}} \times f$

where, u₀ = speed of observer,
u_s = speed of source
and v = speed of sound wave.

$\Rightarrow 1950 = \frac{340 - 10}{340 + 10} f$

$\Rightarrow Frequency, f = \frac{35}{33} \times 1950 = 2068.18 Hz \approx 2068 Hz$

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