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A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz, it bounces off the moving car (it is traveling toward or away from the signal emitted) and the radar gun measures the frequency of the signal received. For one car suspected of speeding the received frequency was 1100 Hz. The speed of sound is approximately 350 m/s (note our radar gun uses sound waves rather than electromagnetic waves)

a. How fast was the car moving?
b. Was it moving towards or away from the police car?
c. The police car follows it by moving towards the suspicious car with the speed Upolice-50m/s. What is the frequency the radar gun is measuring now?

Respuesta :

Answer:

a) The car was moving at a speed of [tex]29.167\ m.s^{-1}[/tex]

b) The negative sign of [tex]v_o[/tex] denotes that the observer is coming towards the police car which is the source of the sound.

c) [tex]f_o=1283.33\ Hz[/tex]

Explanation:

Given:

  • original frequency of the source, [tex]f=1200\ Hz[/tex]
  • speed of the source, [tex]v_s=0\ m.s^{-1}[/tex]
  • velocity of the obstacle car be, [tex]v_o[/tex]
  • speed of sound, [tex]s=350\ m.s^{-1}[/tex]
  • observed frequency, [tex]f_o=1100\ Hz[/tex]

Using the equation from the Doppler's effect:

[tex]\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}[/tex]

[tex]\frac{1100}{1200} =\frac{(350+v_o)}{350-0}[/tex]

[tex]v_o=-29.167\ m.s^{-1}[/tex]

a)

The car was moving at a speed of [tex]29.167\ m.s^{-1}[/tex]

b)

The negative sign of [tex]v_o[/tex] denotes that the observer is coming towards the police car which is the source of the sound.

c)

Now when, [tex]v_s=50\ m.s^{-1}[/tex]

Then, [tex]f_o=?[/tex]

Using the Doppler's eq.:

[tex]\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}[/tex]

[tex]f_o=1283.33\ Hz[/tex]

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