They say the frequency of a wave is its fundamental character, thus remain constant throughout its propagation regardless the medium through which it travels. Could anyone explain why frequency of wave is fundamental character but its wavelength isn't?
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3$\begingroup$ The frequency of a wave does not stay constant in general. See e.g. en.wikipedia.org/wiki/Nonlinear_optics. Hence the premise of the question is already wrong. $\endgroup$– WolpertingerCommented Jun 18, 2016 at 10:39
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/22385/2451 , physics.stackexchange.com/q/21336 and links therein. $\endgroup$– Qmechanic ♦Commented May 9, 2019 at 5:07
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6 Answers
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The frequency must remain constant to avoid a discontinuity at the boundary.
The easiest way to see this is to consider 2 ropes of different linear densities - e.g. a thin rope and a thick rope - joined in series.
If you shake one end at a frequency f, then (transverse) waves will travel along the joined ropes. The waves travel slower along the thicker rope than the thin rope.
At the junction between the ropes (and to either side of the junction) the frequency must still be f - it wasn't the rope would have to split due to adjacent points having different frequencies.
The same is true for any wave - you can't have a sudden jump in the electric field of an EM wave for example - the electric field can only vary continuously, with no discontinuities.
As a consequence of remaining constant, wavelength and speed change proportionately (e.g. if speed doubles, wavelength doubles).
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$\begingroup$ This is true when considering macroscopic observations, but "jumps" or "splits" are the norm at the quantum scale. $\endgroup$ Commented Jun 9, 2023 at 2:23
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The assumptions under the statement are that A. the oscillation count in a wave is conserved and B. the passage of time is universal and uniform. Since the frequency of a wave is the count of oscillations measured within a given time interval by a stationary observer, it remains the same anywhere the wave can reach. On the other hand, the wavelength is how far a wave travels from one oscillation to the next thus depends on how fast the wave is traveling in the medium. There are cases we can find either A or B is violated, for example, in a nonlinear medium or a gravitational field, respectively.
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A good intuitive way to think about this is to think about how a wave propagates. In general, a wave is what happens whenever you have a "hand-off" effect in the sense of that if you disturb some medium at one point, then it disturbs the adjacent - and only the adjacent - points in some other way, and then those, receiving that disturbance, go on to disturb their adjacent points, and so forth.
So imagine a string to be divided into a large number of small, evenly-sized pieces that are connected to each other and so can't move independently. Now start wiggling the piece at one end (the other is pegged and tensioned appropriately, of course). That piece of string, because it's firmly held by your hand, has no choice but to move up and down at the same rate as your hand, and so it moves up and down at the same frequency. Now consider the next piece of string a little bit over from that - the first one not held by your hand. Because it's connected to the first piece which is being moved up and down by your hand at the same frequency as your hand, that first piece will act as a "hand" of sorts upon it and move it up and down at its frequency which, in turn, is that of your hand. That is, we have
$$\mbox{Frequency of Motion of Piece 1} = \mbox{Frequency of Motion of Hand}$$ $$\mbox{Frequency of Motion of Piece 2} = \mbox{Frequency of Motion of Piece 1}$$
and we can continue this way, all the way down the string, and by transitivity we thus must have that all pieces of the string are moving with the same frequency. This applies even if they are made of different materials, because that doesn't change the reasoning which is that the frequency of any segment only depends on that of those immediately nearby it which, in turn, we have connected in a chain back to that of your hand.
answered Feb 28, 2020 at 3:40
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You can visualize the situation like this. Lets say there is a source and an observer. They are not moving relative to each other. The source is emitting 10 peaks of wave per second (i.e. frequency is 10 Hz) and observer is observing 8 (only) peaks per time. i.e. due to medium property the frequency is changed. In this scenario 10 peaks enter the medium per second and 8 leave. where are the two peaks gone?
The wave propagation states that peak will remain peak and valley will remain valley during propagation. Hence this scenario contradict with basic nature of the wave and may be thats why the frequency is constant.
Change in wavelength may be visualized as compression/rarefaction of pule trains/wave which seems entirely physical.
Please forgive me if the language is bit non-technical.
Hope this helps
Regards,
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a different way to look at it is conservation of energy, since the intensity of wave depends upon its frequency (at given velocity and wavwlenght), so the frequency mustnt change when the medium changes cause no energy is being added or removed....
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When source and receiver are moving away or towards each other, the observed frequency changes. This effect is called a "doppler shift". The frequency becomes lower ("red shift) when the source moves away. The frequency becomes higher ("blue shift") when the source moves towards the observer. So yes, one can change the frequency of light. With light this effect is noticible by looking at stars.
The same phenomenon can be observed / heard when an emergency car, with sirens on, is passing by.
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$\begingroup$ I think that the question was about the actual, not observed frequency. $\endgroup$– jng224Commented Jul 1, 2021 at 11:20