wave_mechanics dopplers_effect Question 55
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
[CPMT 2003]
Options:
A) 320 s?1
B) 360 sec?1
C) 390 sec?1
D) 420 sec?1
Show Answer
Answer:
Correct Answer: A
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=321.12se{{c}^{-1}}\approx 320sec^{1} $
BETA
