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In a MIT OCW video lecture, a professor discusses a wave on a string having some density moving towards another string. Both of which are joined but have different densities.

He goes on to analyse it, but assumes a reflected wave and a transmitted wave to be present as the wave strikes the boundary. He further assumes that the frequency of the reflected and transmitted wave are the same as the incident one. He doesn't give an explanation, but how can one assume these?

What if we get back some arbitrary wave with different frequency than the incident one bouncing back and nothing as a transmitted wave?

What justifies these assumptions which the professor made?

I'd be glad if someone pointed out.

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I think it’s boundary conditions—an assumption that the wave is continuous: If something is wiggling at a frequency when it hits an interface, it wiggles at that frequency on “both sides” of that interface.

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  • $\begingroup$ Yes it makes sense thank you. What about when the incident wave is upside down, any intuitive explanation for that? $\endgroup$
    – Kashmiri
    Commented Dec 30, 2020 at 5:01
  • $\begingroup$ I think this is addressed in answers here: physics.stackexchange.com/questions/32122/… $\endgroup$ Commented Dec 30, 2020 at 5:13
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What justifies is that people did it, and found that the reflected signal was at the same frequency, every time (almost). Science is, at the heart, an empirical process.

What we find is that the way sound operates and interacts with mediums is defined by a wave equation: $\frac{\partial^2u}{\partial t^2}=c^2\frac{\partial^2u}{\partial x^2} $ Whether you know calculus or not, its an equation which predicts the behavior of sound and other "waves" astonishingly well. So well, in fact, that we turn it around and say that anything which obeys that equation, for any reason, is a "wave".

Now if you have enough calculus and differential equations under your belt to explore this, you find that the time evolution operator for waves is linear -- the way they change over time is a "linear operator." One of the side effects of a linear operator is that they cannot change frequencies. If you have a frequency on the input, it may only be turned into the same frequency on the output. You need a non-linear operator to change frequencies.

But all of this is just math. The real reason is that we went out and analyzed a bunch of physical experiments, and found that sound waves always reflected at the same frequency. We worked all of the math on the back end, trying to make sense of the results we see. For example, when we notice that the frequency always comes back the same, that's a very strong indicator that the operator is probably linear, so we explored linear operators until we found one that worked! The why in science is almost always stemming from experiments. Very rarely does the why stem from the math, and when it does, we quickly work to identify experiments to back it up!

And note I've said almost. You can get some non-linearity here with a moving object. The effect is one we have heard many times in our life: a Doppler shift. Experimentally, we found that if you bounce a sound wave off of something that is moving it can change the frequency.

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  • $\begingroup$ Thank you for your view, I do believe science is empirical but after the experiment we do develop some theory for the explanation. I was looking for that explanation. After studying some more I'm suspicious as is said by the other answer that it's the boundary conditions that makes it do what it does. I'd be glad if you would write some more on that. Thank you :) $\endgroup$
    – Kashmiri
    Commented Dec 30, 2020 at 5:00
  • $\begingroup$ @YasirSadiq The answers are really the same, just phrased differently. You can think of it as a boundary condition, where one aspect of the boundary (position) is perfectly and completely fixed, or you can think of it as a larger system with an "impedance mismatch," where it is simply a sound wave operating inside a larger system. Regardless of how you get there, eventually it is the linearity of the system which shows that you don't get any frequency changes. $\endgroup$
    – Cort Ammon
    Commented Dec 30, 2020 at 17:37

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