wave_mechanics dopplers_effect Question 42
Question: A source of sound of frequency 90 vibrations/ sec is approaching a stationary observer with a speed equal to 1/10 the speed of sound. What will be the frequency heard by the observer
[MP PMT 2000]
Options:
A) 80 vibrations/sec
B) 90 vibrations/sec
C) 100 vibrations/sec
D) 120 vibrations/sec
Show Answer
Answer:
Correct Answer: C
Solution:
$ n’=n( \frac{v}{v-v _{s}} )=90( \frac{v}{v-\frac{v}{10}} ) $ = $ 100\frac{Vibration}{\sec } $
BETA
