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A man is watching two trains, one leavin...

A man is watching two trains, one leaving and the other coming in with equal speed of `4 m//s`. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air is `320m//s` will be equal to

A

6

B

3

C

0

D

12

Text Solution

Verified by Experts

The correct Answer is:
A
Apparent frequency due to train which is coming in is
`n_1=(v)/(v-v_s)n`
Apparent frequency due to train which is leaving is
`n_2=(v)/(v+v_s)n`
So the number of beats is
`n_1-n_2=((1)/(316)-(1)/(324))320xx240impliesn_1-n_2=6`
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Knowledge Check

  • A man is watching two trains, one leaving and the other coming in with equal speeds of 4"m s"^(-1) . If they sound their whistles, each of frequency 240 Hz, the number of beats per second heard by the man is (velocity of sound in air =330"m s"^(-1) )

    A
    6
    B
    3
    C
    0
    D
    12

  • On a platform a man is watching two trains one leaving the station the other approaching the station, both with the same velocity 2 m/s and blowing their horn with a frequency 480 Hz. The number of beats heard by the man will be (Velocity of sound in air is 320 m/s)

    A
    6
    B
    3
    C
    2
    D
    4

  • A man is running with a fork of frequency 340 Hz towards a huge wall with a velocity 2m/s, then the number of beats heard per second is (Velocity of sound in air is 342 m/s)

    A
    2
    B
    6
    C
    4
    D
    8
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