Doppler by Dimitris

Classical Mechanics Level 2

A train is directed to a tunnel and transmits audio frequency fs. The sound reflected from the vertical wall. The train is moving in the tunnel at a speed U = 90 m/sec U=90\text{ m/sec} , the observer 2 moves against the direction of the train speed U 2 = 10 m/sec U_2=10\text{ m/sec} . Αt the same time the other observers remains stationary next to the train line and in front of the tunnel.

Calculate the absolute difference of the reflected frequency to which the two observers.hear the train produces a sound frequency. That is, what is ∣ f 1 − f 2 ∣ | f_1 - f_2 | ?

Give your answer in hertz.

Take f s = 1000 Hz f_s=1000\text{ Hz} .

It is given that the speed of sound 340 m/sec 340\text{ m/sec} .


The answer is 40.8.

Dimitris Kouroupos by dimitris kouroupos , 23, Greece

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1 solution

  • First computed the frequency in which the reflectivity tunnel : f [ t u n n e l ] = u o u o − U f [tunnel]=\frac{uo}{uo-U} fs= 340 340 − 90 \frac{340}{340-90} 1000 = 340 250 \frac{340}{250} 1000=1,36 1000 =1360Hz
(f [tunnel]=1360Hz

-calculates the reflectance listening frequency by 1 observer : f 1 = u o − U 1 u o f1=\frac{uo-U1}{uo} f [tunnel] = 340 − 10 340 \frac{340-10}{340} 1360= 330 340 \frac{330}{340} 1360=0.97 1360=1319.2Hz

f2=1319.2Hz
f1=f [tunnel]
  • I f 1 − f 2 I = f 1 − f 2 = 1360 − 1319.2 = 40.8 H z If1-f2I=f1-f2=1360-1319.2=40.8Hz
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