wave_mechanics dopplers_effect Question 63
Question: A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5m/s. The speed of sound is 330 m/s. If the observer is between the wall and the source, then beats per second heard will be
[UPSEAT 2002]
Options:
A) 7.8 Hz
B) 7.7 Hz
C) 3.9 Hz
D) Zero
Show Answer
Answer:
Correct Answer: A
Solution:
The observer will hear two sound, one directly from source and other from reflected image of sound Hence number of beats heard per second =$ ( \frac{v}{v-v _{S}} )n-( \frac{v}{v+v _{S}} )n $ = $ \frac{2nvv _{S}}{v^{2}-v _{S}^{2}}=\frac{2\times 256\times 330\times 5}{335\times 325} $ = 7.8$ Hz $
BETA
