0
$\begingroup$

Suppose we describe a wave travelling along at positive $x$ axis, using $y_1=a\sin(\omega t-kx).$

Suppose, at $x=0$, we have a rigid boundary, such that the wave is reflected. Let us write this reflected wave as $y_2=a'\sin(\omega t+kx)$.

The resultant wave is simply $y=y_1+y_2$.

Imposing the boundary condition, that at $x=0$, $y=0$, we have :

$$0=a\sin(\omega t)+a'\sin(\omega t)$$

This implies, $a'=-a$. Hence, our two waves can be summarized as :

$$y_1=a\sin(\omega t-kx)$$ $$y_2=-a\sin(\omega t+kx)$$

This shows that the reflected wave $y_2$ is inverted with respect to $y_1$. However, I'm still a little bit unclear about this. This inversion is quite apparent when we talk about a single pulse. For a wave, this concept of inversion is unclear. Does it simply mean that at the boundary, if the first wave is trying to cause an upward displacement, the second wave is trying to cause a downward displacement?

My actual problem has to do with the fact, that most texts mention a $\pi$ phase change at the reflection. I don't see where this phase difference is coming from in the first place. How do we compare the phase of two waves travelling in opposite directions in the first place?

One possible solution is that at $x=0$, we absorb the negative sign in front of $y_2$ into the argument. In that case :

$$y_1=a\sin(\omega t-kx)$$ $$y_2=-a\sin(\omega t+kx)=a\sin(\omega t+kx+\pi)$$

However, how does this mean that the reflected wave has a phase difference of $\pi$ relative to the incident wave? In the incident wave, the argument is $\omega t-kx$ and in the reflected wave, the argument is $kx+\omega t+\pi$. These are clearly two different functions of $t$ and $x$.So, I don't see how to 'define' a phase change of $\pi$ in this case.

I guess we could instead say something like if the boundary had not been rigid, then, we would have some reflected wave $y_3=a\sin(kx+\omega t)$. In this case, we can compare the phases because both of them are the same function of $t,x$. This reflected wave would be $\pi$ out of phase with the reflected wave in case of the rigid boundary.

So, instead of saying that the incident wave and reflected wave are out of phase by $\pi$, is it not more accurate to say that the reflected wave in case of a rigid boundary is $\pi$ out of phase with respect to the reflected wave in case of a free boundary ?

$\endgroup$

1 Answer 1

1
$\begingroup$

"My actual problem has to do with the fact, that most texts mention a 𝜋 phase change at the reflection. [...] How do we compare the phase of two waves travelling in opposite directions in the first place?"

I think that, to start with, you should think of the phase change just in the incident and reflected oscillations at $x=0$. As you are clearly aware, this can be represented by a phase angle of $±\pi$ added to the argument of the sine.

The reflected wave then travels back with its displacement at $-x$ being the same as that of the forward wave at $+x$ (had it continued forward beyond $x=0$), except for the additional phase angle of $±\pi$. Is this not the desired comparison?

"I don't see where this phase difference is coming from in the first place." Clearly if the boundary is rigid, the reflected wave must have a phase reversal to give zero displacement at the boundary. But you know that. A more physical explanation would depend on the type of wave. For example a transverse wave in a stretched string arriving at a rigid 'anchoring point' will exert, at any instant, a transverse force on the anchoring point, that will, according to Newton's third law, exert an equal and opposite force on the string – launching the reflected wave.

$\endgroup$
4
  • $\begingroup$ Yes, I think I get it a bit now, but from a purely mathematical perspective, I've normally seen the phase difference defined by subtracting the argument inside the sines of the two waves. Since the two waves here are traveling in opposite directions, even though the value of the reflected wave at $-x$ is the same as that for the incident wave at $x$ just shifted by $\pi$ phase, these two points are not the same point. $\endgroup$ Jan 26, 2022 at 19:53
  • $\begingroup$ Normally I'm used to dealing with phase changes with respect to interference, and in those cases, we try to find the difference of phase of two waves at some unique point on the screen. That is how phase difference is usually defined. $\endgroup$ Jan 26, 2022 at 19:54
  • $\begingroup$ In this case, we are talking about the difference in phase at two different points i.e. $+x$ for the first wave and $-x$ for the reflected wave. This definition for phase difference is something that I'm not used to. I don't think it makes much sense to talk about phase difference in two waves travelling in opposite directions in the first place. Is that correct ? $\endgroup$ Jan 26, 2022 at 19:57
  • 1
    $\begingroup$ As I've said, understanding the phase difference between the oscillations at $x=0$ presents no problems and is important in its own right. Seeing the phase shift as persisting as the wave propagates 'backwards' requires a little flexibility; as you've seen, subtracting the arguments of the sines doesn't work. I've suggested a strategy that does work. Keep thinking about it; in the meantime your equations are correct and you can work with them. And you might get another answer! $\endgroup$ Jan 26, 2022 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.