wave_mechanics dopplers_effect Question 43
Question: A whistle of frequency 500 Hz tied to the end of a string of length 1.2 m revolves at 400 rev/min. A listener standing some distance away in the plane of rotation of whistle hears frequencies in the range (speed of sound = 340 m/s)
[KCET 2000; AMU 1999; Pb. PET 2003]
Options:
A) 436 to 586
B) 426 to 574
C) 426 to 584
D) 436 to 674
Show Answer
Answer:
Correct Answer: A
Solution:
The linear velocity of Whistle $ v _{S}=r\omega =1.2\times 2\pi \frac{400}{60}=50m/s $ When Whistle approaches the listener, heard frequency will be maximum and when listener recedes away, heard frequency will be minimum So, $ {n _{\max }}=n( \frac{v}{v-v _{s}} )=500( \frac{340}{290} )=586Hz $
$ {n _{\min }}=n( \frac{v}{v+v _{s}} )=500( \frac{340}{390} )=436Hz $
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