wave_mechanics dopplers_effect Question 22
Question: A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be
[MP PET 1994]
Options:
A) 9 : 8
B) 8 : 9
C) 1 : 1
D) 9 : 10 (Speed of sound v = 340 m/s)
Show Answer
Answer:
Correct Answer: A
Solution:
When source is approaching the observer, the frequency heard $ n _{a}=( \frac{v}{v-v _{S}} )\times n=( \frac{340}{340-20} )\times 1000=1063Hz $ When source is receding, the frequency heard $ n _{r}=( \frac{v}{v+v _{S}} )\times n $ =$ \frac{340}{340+20}\times 1000=944 $
$ \Rightarrow n _{a}:n _{r}=9:8 $ Short tricks : $ \frac{n _{a}}{n _{r}}=\frac{v+v _{S}}{v-v _{S}}=\frac{340+20}{340-20}=\frac{9}{8}.
$
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