I'm facing much confusion about a rather silly problem recently. Suppose I have two functions $\sin(x+\phi_1)$ and $\sin(x+\phi_2)$. Defining $\theta_1=x+\phi_1$ as the phase of the first and $\theta_2=x+\phi_2$ as the phase of the second function.

The phase difference, by definition comes out to be simply:

$$\Delta\theta=\theta_1-\theta_2=x+\phi_1-x-\phi_2=\phi_1-\phi_2$$ Hence, $$\Delta\theta=\phi_1-\phi_2$$

So, there is a constant phase difference between these two waves.

However, what if I have two functions $\sin(x+\phi_1)$ and $\sin(\phi_2-x)$ instead ? If I follow the same definition above, then I have $\theta_1=x+\phi_1$ and $\theta_2=\phi_2-x$.

In this case, the phase difference comes out to be $$\Delta \theta=2x+(\phi_1-\phi_2)$$

As $x$ increases, the phase difference seems to increase here. However, this doesn't seem correct to me.

For example, I know $\cos(x)=\sin(\frac{\pi}{2}-x)=\sin(\frac{\pi}{2}+x)$. This automatically implies, that both $\sin(\frac{\pi}{2}-x)$ and $\sin(\frac{\pi}{2}+x)$ have the same phase, since they are both equal to $\cos(x)$.

However if I use the definiton as above, then $\theta_1=\frac{\pi}{2}-x$ and $\theta_2=\frac{\pi}{2}+x$. In this case however,

$$\theta_1-\theta_2=2x=2\omega t \,\,\,\,\,\,(x=\omega t)$$

However, this implies that the phase difference increases over time. When the difference is an even multiple of $\pi$, the waves are in phase, and apart from that, they are out of phase. However, in this example, since both are essentially the same function, we should have a obtained a constant phase difference that was always $0$.

What is wrong with my definition then, and how would I find the phase difference between two waves in general.

  • $\begingroup$ Forget your problem for a second. I feel you made a wrong judgement when you said "This automatically implies, that both $\sin(\pi/2−x)$ and $\sin(\pi/2+x)$ have the same phase, since they are both equal to $\cos(x)$." Think about $\sin(x)$ only. It becomes 0 at $x=0,\pi,2\pi,\ldots$. Does that mean the phase is the same?! $\endgroup$
    – Newbie
    Jan 21, 2022 at 2:03
  • $\begingroup$ @Newbie should I have instead said that the phase difference between them is a constant and an even multiple of $\pi$, since they represent the same function ? $\endgroup$
    – RayPalmer
    Jan 21, 2022 at 2:06
  • $\begingroup$ I was trying to guide you towards a correct judgement for your own problem by providing a simple example of why your judgement may not be true. The other way to look at it is whatever happens to $x$, the opposite happens to $-x$ so their difference is definitely going to increase as $x$ increases which is the case you already have with $2x+\ldots$ so I don't know what the problem is here. $\endgroup$
    – Newbie
    Jan 21, 2022 at 2:09
  • $\begingroup$ @Newbie my problem is suppose, I have $sin(x)$ and $sin(-x)$. I know that they are out of phase by $\pi$. However, their arguments are $x$ and $-x$ respectively. If I define phase difference to be the difference between the arguments of the two, I get $\phi=x-(-x)=2x$ instead of $\pi$ which I know is the correct phase difference. Is there something wrong with my definition then ? $\endgroup$
    – RayPalmer
    Jan 21, 2022 at 2:16
  • $\begingroup$ I think your analysis has this implicit definition that both functions should be the same, i.e., $\sin$ and the sign of $x$ should be the same as well. So for instance, in your second example in the problem you should have $\sin(\phi_{2}-x)=-\sin(x-\phi_{2})=\sin(x-\phi_{2}+\pi)$. I'm not sure however, why such a definition of phase difference exists but I think your phase difference is constant now. $\endgroup$
    – Newbie
    Jan 21, 2022 at 2:25

1 Answer 1


In calculating the phase difference between two waves, it seems necessary that both waves have the same functionality as a function of the independent variable and that the frequency of the two waves are the same, i.e., their period is the same. In that case the phase difference may be interpreted as the horizontal shift that is necessary to make the two waves fall right on top of each other (assuming they have the same amplitude). It is this fact that necessitates the same period of the two waves. Once the periods are the same, the shift necessary to make the two waves identical is definitely less than the equal wave period. With that, one can proceed to each of your examples to calculate the phase difference. In the first case the two waves are $\sin(x+\phi_{1})$ and $\sin(x+\phi_{2})$. Note that the two waves have the same exact frequency and both are the same periodic function. Thus the magnitude of the phase difference is $$|(x+\phi_{1})-(x+\phi_{2})|=|\phi_{1}-\phi_{2}|$$

In the second case the two waves are $\sin(x+\phi_{1})$ and $\sin(\phi_{2}-x)$. While both waves are $\sin$ functions, they do not have the exact same frequency (though the absolute value is the same). Thus, as mentioned in the comments $$\sin(\phi_{2}-x)=-\sin(x-\phi_{2})=\sin(x-\phi_{2}+\pi)$$ Subsequently, the magnitude of the phase difference in this case is $$|(x+\phi_{1})-(x-\phi_{2}+\pi)|=|\phi_{1}+\phi_{2}-\pi|$$ Note that as mentioned in your problem, the implication of the above equation for $\cos(x)=\sin(\pi/2-x)=\sin(\pi/2+x)$ is that the phase difference between $\sin(\pi/2-x)$ and $\sin(\pi/2+x)$ is 0. I think this answer provides general guidelines for finding the phase difference between two waves through specific examples mentioned in the question. Please let me know if you have any questions or find the answer incorrect.

  • $\begingroup$ Thanks. I had another question : how would one modify this reasoning for waves travelling in two different directions - $\sin(\omega t-kx)$ and $\sin(\omega t +kx)$. Should I fix the sign in front of $\omega t$ or $kx$ ? Depending on that, my phase difference seems to be $2kx$ or $2\omega t$. How do we know which one to choose ? $\endgroup$
    – RayPalmer
    Jan 21, 2022 at 10:08

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