wave_mechanics critical_thinking Question 33
Question: A man is watching two trains, one leaving and the other coming in with equal speeds of 4 m/sec. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air = 320 m/sec) will be equal to
[NCERT 1984; CPMT 1997; MP PET 1999; RPMT 2000; BHU 2004, 05]
Options:
A) 6
B) 3
C) 0
D) 12
Show Answer
Answer:
Correct Answer: A
Solution:
Frequency of sound heard by the man from approaching train $ n _{a}=n( \frac{v}{v-v _{s}} )=240( \frac{320}{320-4} )=243Hz $ Frequency of sound heard by the man from receding train $ n _{r}=n( \frac{v}{v+v _{s}} )=240( \frac{320}{320+4} )=237Hz $ Hence, number of beats heard by man per sec $ =n _{a}-n _{r}=243-237=6 $ Short trick : Number of beats heard per sec $ =\frac{2nvv _{S}}{v^{2}-v _{S}^{2}}=\frac{2nvv _{S}}{(v-v _{S})(v+v _{S})}=\frac{2\times 240\times 320\times 4}{(320-4)(320+4)}=6 $
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