First slide

Doppler's effect

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

Moderate

A

6
B

3
C

0
D

12

Solution

Frequency of sound heard by the man from approaching train
$n_{a} = n (\frac{v}{v - v_{s}}) = 240 (\frac{320}{320 - 4}) = 243 Hz$
Frequency of sound heard by the man from receding train $n_{r} = n (\frac{v}{v + v_{s}}) = 240 (\frac{320}{320 + 4}) = 237 Hz$
Hence, number of beats heard by man per sec
   $= n_{a} - n_{r} = 243 - 237 = 6$
Short trick : Number of beats heard per sec
   $= \frac{2 {nvv}_{S}}{v^{2} - v_{S}^{2}} = \frac{2 {nvv}_{S}}{(v - v_{S}) (v + v_{S})} = \frac{2 \times 240 \times 320 \times 4}{(320 - 4) (320 + 4)} = 6$

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