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9.5 doppler


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doppler effect

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9.5 doppler

  1. 1. Wave Phenomena Topic 9.5 Doppler Effect
  2. 2. The Doppler Effect This effect is the change in the frequency of a wave received by an observer, compared to the frequency with which it was emitted. The effect takes place whenever there is motion between the emitter and receiver.
  3. 3. This is a phenomenon of everyday life. On a highway, an approaching car creates a high pitched sound. As it goes past us and recedes from us the frequency becomes lower.
  4. 4. In diagrams we can explain the Doppler effect as follows:
  5. 5. This diagram can be constructed accurately to show the pattern As can the pattern for a moving detector. Simple Doppler Effect Clips of various examples of Doppler Effect
  6. 6. The source moves towards observer B and away from observer A. The wavecrests are piling in front of the source and thus the crests reach B at time intervals which are shorter than those on emission. Thus the received period is smaller and hence the frequency is larger. On the other hand, the crests reach A at longer time intervals and thus the measured frequency is smaller.
  7. 7. The frequency of the sound emitted from the stationary source is f Observer A will hear a note of frequency fA where fA  f Observer B will hear a note of frequency fB where fB  f This shift in frequency is known as the Doppler effect
  8. 8. Which gives you the most chocolates?
  9. 9. Deriving the formulae (not required) Let us look at the simplest case in which the velocity of the source is in line with the observer In the diagram the observer 0 is at rest with respect to the medium and the source is moving with speed vs.
  10. 10. The source is emitting a note of constant frequency f that travels with speed v in the medium. S' shows the position of the source t later. In a time t the observer would receive ft waves and when the source is at rest these waves will occupy a distance vt .
  11. 11. The wavelength = distance occupied by the waves  the number of waves The wavelength = vt / ft = v/f Because of the motion of the source this number of waves will now occupy a distance vt - vst The ´new´wavelength = (vt - vst) / ft i.e. 1 = (v- vs) / f
  12. 12. If f1 is the new frequency, then 1 = v/ f1 = (v- vs) / f Rearranging f1 = v / (v- vs) * f Dividing throughout by v gives f1 = 1 f 1- (vs / v)
  13. 13. If the source was moving away from the observer then we have f1 = 1 f 1+ (vs / v)
  14. 14. And for moving observer Observer moving towards source Relative velocity = v +vO f1 = (V + VO)/  But  = v/f Therefore f1 = (V + VO)/ v/f Rearranging gives f1 = ((V + VO)/ v )f
  15. 15. If the observer is moving towards the source f1 = (1+ (vO / v)) f If the observer is moving away from the source f1 = (1- (vO / v)) f
  16. 16. Doppler effect for sound with moving observer
  17. 17. Doppler shift for SOUND source Be careful to apply the + or – the correct way.
  18. 18. Example of using Doppler to measure speed: 1. measuring blood flow
  19. 19. Example of using Doppler to measure speed: 2. measuring speed of vehicle
  20. 20. Caught speeding Take the case of the car and the radar speed trap. The emitted frequency of the police radar gun is f, the car acts like a moving observer receiving it at f'. It reflects it back at f', acting now as a moving source and the police receive it back shifted once again as f''. The total change in frequency (f'' - f) is therefore doubled.
  21. 21. It can be shown that the relative Doppler shift for electro- magnetic radiations like light, radio waves etc is given (approximately) by where c is the speed of light and v is the relative speed of source and observer. The speed of a car is being measured by a police-person using a "radar speed-measuring gun". The frequency of the transmitted signal is 5GHz. When "mixed" the transmitted and received signals beat with a frequency of 750Hz. If the speed limit for the road is 110kmh-1, should the driver be fined or not? Question