Question

In one of the original Doppler experiments, one tuba was played on a moving flat train car at a frequency of 75 Hz, 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 approached the station at a speed of 5.0 m/s?

f(beat) = Hz

Answer 1

The beat frequency heard in the station is 73.91 Hz.

Step-by-step explanation:

To calculate the beat frequency heard in the station, we need to consider the Doppler effect. The beat frequency is the difference between the frequency of the moving tuba and the frequency of the stationary tuba.

Given that the frequency of the moving tuba is 75 Hz and the train is approaching the station at a speed of 5.0 m/s, we can use the Doppler effect equation:

f (beat) = f(stationary) - f(moving) = 75 Hz - (75 Hz * (5.0 m/s / 343.00 m/s)) = 75 Hz - 1.09 Hz = 73.91 Hz

Answer 2

1.109 Hz

Step-by-step explanation:

According to Doppler's effect, when there is relative motion between the source and the observer, an apparent frequency different from the actual one is measured.

Doppler effect:

`f'=f((v+v_o)/(v-v_s))\\\Rightarrow f'=75* ((343)/(343-5))=76.109Hz`

Beat frequency:

`f(beat)=f'-f\\f(beat)= 76.109-75 \\f(beat)=1.109 Hz`