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3 years ago

9. a) What is the Doppler effect? b) How does it work? c) If 12 points

1 answer:

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a) Doppler effect is an apparent change in the frequency of a wave due to the relative motion between the source and the observer

b) It is given by the equation $f'=\frac{v\pm v_o}{v\pm v_s}f$

c) The star is moving towards us

Explanation:

The Doppler effect is a phenomenon that occurs whenever there is a source of a wave in relative motion to an observer. When such situation occurs, the apparent frequency of the sound as perceived by the observe is different from the proper frequency of the wave emitted by the source.

A typical example of this situation is when an ambulance is approaching you. The sound of the siren is perceived as having a higher pitch (higher frequency) as the ambulance moves towards you, and then is perceived as having a lower pitch (lower frequency) when the ambulance moves away from you.

The same phenomenon occurs not only with sound waves, but also with light waves and other types of waves.

Mathematically, the Doppler effect can be summarized by the following equation:

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where:

f is the proper frequency of the wave emitted by the source

f' is the apparent frequency, as perceived by the observer

v is the speed of the wave

$v_o$ is the velocity of the observer, which is positive if the observer is moving towards the source and negative if the observer is moving away from the source

$v_s$ is the velocity of the source, which is positive if the source is moving away from the observer and negative if the source is moving towards the observer

Applied to the example of the ambulance, we have that:

$v_o = 0$ , assuming that the observer is at rest

- when the ambulance is moving towards the observer, $v_s$ is negative, and therefore the fraction is larger than 1, therefore $f'>f$ and the apparent frequency is higher than the real frequency

- when the ambulance is moving away from the observer, $v_s$ is positive, and therefore the fraction is smaller than 1, therefore $f'$ and the apparent frequency is lower than the real frequency

As we mentioned earlier, the Doppler effect also occurs with light waves. This is particularly relevant for stars or galaxies moving towards or away from us, since the light coming from these objects will have a frequency (and also a wavelength) "shifted" due to the Doppler effect.

In particular, we have two possible cases:

- For a star moving away from us, the frequency of the light emitted by the star will appear lower than the real frequency --> this means that its wavelength will appear longer than the real wavelength (because wavelength is inversely proportional to the frequency), and this means that the light will appear shifted towards longer wavelengths (so, towards the red end of the visible spectrum)

- For a star moving away towards us, the frequency of the light emitted by the star will appear higher than the real frequency --> this means that its wavelength will appear shorter than the real wavelength, and this means that the light will appear shifted towards shorter wavelengths (so, towards the blue end of the visible spectrum)

Therefore, if a star looks bluer to us than it should, the star is moving towards us.

Learn more about waves:

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