First slide

Doppler's effect

Question

A source and an observer approach each other with same velocity 50 m/s. If the apparent frequency is 435 sec^–1, then the real frequency is

Easy

A

320 sec^–1
B

360 sec^–1
C

390 sec^–1
D

420 sec^–1

Solution

$n' = n [\frac{v + v_{O}}{v - v_{S}}];$ Here v = 332 m/s and $v_{0} = v_{s} = 50 m / s$
$\Rightarrow 435 = n [\frac{332 + 50}{332 - 50}] \Rightarrow n = 321 .12 \sec^{- 1} \approx 320 \sec^{- 1}$

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