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A person standing close to a railroad crossing hears the whistle of an approaching train. He notes that the pitch of the whistle drops as the train passes by and moves away from the crossing. The frequency of the approaching whistle is 516 Hz, and drops to 494 Hz after the train is well past the crossing. What is the speed of the train?

Respuesta :

Answer:

v_s = 7.47 m/s

Explanation:

given,

Varying frequency of the whistle = 516 Hz to 494 Hz

speed of sound = 343 m/s

speed of the train = ?

using equation of Doppler's

When the train is approaching

[tex]f_s = f_0(\dfrac{v-v_s}{v})[/tex]......(1)

Doppler's equation when train is moving away

[tex]f_s = f_1(\dfrac{v+v_s}{v})[/tex]...........(2)

equating both the equation

[tex] f_0(\dfrac{v-v_s}{v}) = f_1(\dfrac{v+v_s}{v})[/tex]

on simplifying the above equation we get

[tex]v_s = v \dfrac{f_0-f_1}{f_0 + f_1}[/tex]

f_0 = 516 Hz

f_1 = 494 Hz

now,

[tex]v_s = 343\times \dfrac{516 -494}{516+494}[/tex]

    v_s = 7.47 m/s

speed of the train is equal to v_s = 7.47 m/s

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