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3 years ago

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In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second

identical tuba played the same tone while at rest in the railway station. What beat frequency was heard in the station if the train car approached the station at a speed of 8.50 m/s?

1 answer:

wolverine [178]3 years ago

3 0

Answer:

$f_{beat} = 1.64\ Hz$

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

$f' = f(\dfrac{v}{v-v_s})$

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

$f' = 64\times (\dfrac{340}{340 - 8.50})$

f' = 65.64 Hz

now, beat frequency is equal to

$f_{beat} = f' - f$

$f_{beat} = 65.64 - 64$

$f_{beat} = 1.64\ Hz$

hence, beat frequency is equal to 1.64 Hz

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Complete Question

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