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Doppler Shift

If you have ever been standing around where a train or car goes by, you are familiar with the eeeeeeeeeee–aaaaaaaaaah sound it makes as it passes. Why does this happen?

Figure 12-14–Figure 12-16 show this phenomenon. Figure 12-14 shows a train whistle making sound waves when it is still. The man hears these pressure waves, so that his ear records a certain frequency of pressure-maxima arrival times. In Figure 12-15, the train whistle is approaching, so the man perceives the pressure maxima coming more frequently. He perceives a higher frequency note. In Figure 12-16 the train is receding, each successive pressure maximum has a longer way to travel, and the man perceives pressure maxima less frequently.

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Figure 12-14


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Figure 12-15


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Figure 12-16


When the emitter of a wave and the detector are moving relative to each other, the detector detects a different frequency from the one emitted—a Doppler shift. The frequency is higher if they are coming together and lower if they are going apart.
There is one special case that deserves note. If a wave strikes a moving object and bounces back, then there are two Doppler shifts.
For example, when a police officer uses a radar device to detect the speed of an oncoming vehicle, there is one shift when the radar is intercepted by the car and another shift when the reflected signal is intercepted by the police device.



A police officer uses sonar to determine the speed of an approaching car. (Actually police use electromagnetic waves, but that is in the next chapter.) It emits a frequency of 60 kHz. The car is approaching at 38 m/s. The Doppler shift is


where fdet is the detected frequency, vs is the speed of the wave in the medium, vdet is the speed of the detector, vem is the speed of the emitter, and fem is the emitted frequency. The speed of sound is 343 m/s.

  1. What frequency would the car detect if it could detect the sonar?
  2. What frequency would the officer detect from the reflection?
  1. The frequency that the car would pick up if it could intercept the police sonar is

where f0 = 60 kHz and vcar = 38 m/s. We know to choose the positive sign because we know the result must be a higher frequency. Thus,


  1. The frequency that the police intercepts is given by

We choose the negative sign because, again, we know the result must be a higher frequency. Thus,



In this chapter we looked at sound as an example of waves. We especially noted resonating cavities of air which exhibit standing waves just like the waves on the guitar string in the previous chapter. The key to doing problems involving these cavities is drawing the pictures correctly.
The most important thing to remember about the Doppler shift is that the detected frequency is greater than the emitted frequency if the emitter and the detector are approaching each other, and less if they are receding.

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