Question

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?

Answer 1

`f_(beat) = 1.64\ Hz`

Step-by-step 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((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* ((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