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I have seen the equations that show the coefficient of reflection etc.

But I'm searching for an intuitive rather than solely mathematical explanation for why waves change phase by π when reflected (eg- from a solid wall)?

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    $\begingroup$ Does this answer your question? Phase shift of 180 degrees of transversal wave on reflection from denser medium $\endgroup$ – FakeMod Apr 13 at 11:55
  • $\begingroup$ After being reflected, the reflected wave combines with the incident wave to produce a standing wave pattern. If the medium is fixed at the boundary, there must be a standing wave node (zero amplitude) at the boundary. That requires that the displacement associated with the reflected wave at that point be equal and opposite to that of the incident wave. $\endgroup$ – R.W. Bird Apr 13 at 14:59
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When it is reflected, the wave switches direction of travel , hence it should be the same wave but with a negative sign. If a sine is modelling it , a negative sign factor outside the sine is equivalent to adding 'pi' into the argument.

Key: stuff in double square braces is an identity used i.e.: $$\sin(x+\pi)=\sin(\pi-(-x))$$

$$[[\sin(\pi-z)=\sin(z)]]$$

$$\sin(x+\pi)=\sin(-x)$$ $$[[\sin(-x)=-\sin(x)]]$$ $$\sin(x+\pi)=-\sin(x)$$

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  • $\begingroup$ Re "When it is reflected, the wave switches direction of travel , hence it should be the same wave but with a negative sign" How does this apply to a single positive pulse? $\endgroup$ – user45664 Apr 13 at 17:03
  • $\begingroup$ you get a negative pulse back when it is reflected $\endgroup$ – DDD4C4U Apr 13 at 17:07
  • $\begingroup$ But switching direction of travel in general does not change the sign, eg. in general $f(x-ct)$ is not equal to $-f(x+ct)$ $\endgroup$ – user45664 Apr 13 at 17:19

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