A man is standing on the platform and one train is approaching and another train is going away with the speed of 4 m s^(-1), frequency of sound…

Question

A man is standing on the platform and one train is approaching and another train is going away with the speed of 4 m s^(-1), frequency of sound produced by train is 240 Hz. What will be the number of beats heard by him per second? (velocity of sound = 300 m s^(-1))

MARKS Content Team · Accepted Answer

Correct answer: (C) 6. The apparent frequency heard by the man is given by n_app=n(v±v_o)/(v±v_s) On our case,v_o=0 ∴n_app=nv/(v±v_s) Given, v=300 m s^(-1), v_s=4 m s^(-1), n=240 Hz In the first case, the train is approaching the man, so frequency heard n_1=nv/(v±v_s)=240300/(300-4) =(240×300)/296=243.24 Hz In the second case, the train is going away from the man, so frequency heard n_2=nv/(v+v_s)=240 300/(300+4) =(240×300)/304=236.84 Hz Hence, the number of beats heard by the man per second =n_1-n_2 =243.24-236.84=6.4≈6

A man is standing on the platform and one train is approaching and another train is going away with the speed of $4 m s^{- 1}$ , frequency of sound produced by train is $240 Hz$ . What will be the number of beats heard by him per second? (velocity of sound = $300 m s^{- 1}$ )

Solution

More Waves and Sound Questions

A man is standing on the platform and one train is approaching and another train is going away with the speed of 4 m s-1, frequency of sound produced by train is 240 Hz. What will be the number of beats heard by him per second? (velocity of sound = 300 m s-1)

Solution

More Waves and Sound Questions

A man is standing on the platform and one train is approaching and another train is going away with the speed of $4 m s^{- 1}$ , frequency of sound produced by train is $240 Hz$ . What will be the number of beats heard by him per second? (velocity of sound = $300 m s^{- 1}$ )