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Len [333]
3 years ago
13

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?
Physics
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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