Question

Due to the Doppler effect, a moving aircraft sees the broadcast signals at 377 MHz from a stationary communications antenna that broadcasts signals at 356 MHz. Find the relative velocity between the aircraft and the broadcast station. Determine the direction of the moving aircraft.

Answer 1

Given that the frequency of broadcast signal that a moving aircraft sees is 377 MHz and the broadcast signal frequency of the stationary communications antenna is 356 MHz where the frequency identified by aircraft is generated.Here we have to find the relative velocity between aircraft and broadcast station.

so the difference in frequency is 377-356=21 MHz

For solving this we have to consider the velocity of air at 20 degree celcius equals to 343 m/s

frequency at aircraft=frequency of antenna*(velocity of air +/- initial velocity / velocity of air +/- final velocity)

377=356*(343 +/- 0 / 343 +/- final velocity)

377=356*(343 / 343+final velocity)

377=122108 / 356(343+final velocity)

interchanging numerator and denominator on both sides

1/377=356(343+final velocity) / 122108

122108/377=356(343+final velocity)

323.8938=356(343+final velocity)

323.8938/356=343+final velocity

0.9098-343=final velocity

final velocity=-342 m/s

The final velocity we obtained is the relative velocity between the aircraft and broadcast station.

The aircraft is moving opposite to the direction of broadcast station.