It makes intuitive sense to me for the apparent frequency of a sound as modified by the Doppler effect to be based entirely on the speed at which the observer and the source get closer or farther apart, regardless of whether the source or the observer is moving, based on the principle of relativity. However, the general equation I receive from every source $$f'=f((v±v_o)/(v∓v_s))$$ does not have this property. For example, if the speed of sound in the medium is 2 units and the observer and source are coming closer at a rate of 1 unit, then if we take the observer to be moving then we get $$f'=f(3/2)$$ but if we take the source to be moving then we get $$f'=f(2/1)$$ What is the difference between these two cases?
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2 Answers
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The difference is that the classical Doppler effect assumes a static background. In atmosphere, there is a marked difference between a moving observer and a stationary one - a gentle (or not so gentle) breeze. To exaggerate these effects, consider two jets flying above Mach 1. If the first jet is ahead of a second, the first jet will not hear any of the noise the second one makes, even if their relative speed is zero.
Note that with the relativistic Doppler effect, the situation is symmetric, and the doppler shift of light will depend only on relative measures of velocity/position between the two objects. This is because relativity does not assume a static/fixed background in the same sense as an atmosphere.
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The two cases are different. When the observer moves toward the source, the increase in frequency heard by the observer is due to the observer's speed being added to the speed of sound, while the wavelength of the sound stays the same. But when the source moves toward the observer, the increase in frequency heard by the observer is due to the wavelength being shortened, while the speed of the sound stays the same. When you derive the two changes, you get two different functions. They give results which are close for frequencies in the audible range, but not equal.
