I have an equation $y(x,y) = y_0 \sin (\omega t \pm kx \pm \phi)$ and there are two graphs. One is $y(x,t=0)$ in units $mm(mm)$ and the other is $y(t,x=0)$ in units $mm(s)$.

Now I have to find all of the constants inside of the equation by looking at graphs and it goes well until I am to determine phase shift $\phi$. If I calculate it by looking at $y(x=0,t=0)$ it works out well and I get right solution, but when I try to read $\phi$ from graph $y(x,t=0)$ or $y(t,x=0)$ I get two wrong solutions.

The idea for reading $\phi$ from graphs was to express some distance $x'$ or time $t'$ in radians but it looks like I can't get it right. The sign in front of $\phi$ is not a problem - if my sinus (in red) is shifted right it is minus otherwise it is plus.

Can anyone clear things out for me and explain why I get wrong results which also differ from one-another?

enter image description here

  • $\begingroup$ Your result using $y(x, t=0)$ appears correct, it agrees to within measurement error. Your result for $\phi$ from $y(x=0, t)$ should be replaced with $\pi - \phi$, since there is a relative minus sign between $\omega t$ and $kx$ in this wave. Upon this replacement, you also get the right answer. $\endgroup$
    – knzhou
    Jul 2, 2016 at 23:29

2 Answers 2


Formal note: you use "," in the picture but here the numbers look better with "." so I will use ".".

I think your $-0.628$ is close enough to $-0.73$. The difference may be because of $x'$ error. Notice that the upper plot crosses the $x$ axis at $0.0061$ and $-0.0039$ rather than $0.006$ and $-0.004$ respectively. It may be even $0.00615$ and $-0.00385$, so

$$x' \approx 0.00115$$ $$\phi \approx -0.72 \approx -0.73$$

The calculated value of $-0.73$ is uncertain to its own degree.

So far so good.

The problem is with $-2.5$ value. To explain that I need to get rid of "$\pm$" in your equation and rewrite it as:

$$y(x,t)= y_0 \sin(\omega t+kx+\phi)$$

The negative factor (if any) should be included in $k$ and/or $\phi$ instead of ambiguous "$\pm$"; these values may be negative, as well as $\omega$ may be.

From the slopes of respective plots:

$$k y_0 \cos(\phi) = \frac d {dx} y(x,0) \rvert _{x=0} > 0$$ $$\omega y_0 \cos(\phi) = \frac d {dt} y(0,t) \rvert _{t=0} < 0$$

Conclusion: $k$ and $\omega$ have opposite signs.

Another thing, for any $\alpha$:

$$\sin(\alpha) = \sin(-\alpha - \pi)$$


$$y(x,t)= y_0 \sin(\omega t+kx+\phi) = y_0 \sin(-\omega t-kx-\phi-\pi) = y_0 \sin(\Omega t+Kx+\Phi)$$


$$\Omega = -\omega\\ K=-k\\ \Phi=-\phi - \pi$$

The forms with $\phi$ and $\Phi$ describe the same wave with two different (yet connected) sets tuples of parameters. Note that you can change $\phi$ by $2n \pi$ addend ($n \in Z $) and the wave will remain the same; this is true also for $\Phi$.

Now let's get back to your $\phi = -0.73$.

$$\Phi= -\phi - \pi \approx 0.73 - 3.14 = -2.41 \approx -2.5$$

I think you would get coherent results if $k$ and $\omega$ had the same signs. You measure $x'$ and $t'$ going from $0$ to the right. In your case, when there are different signs of $k$ and $\omega$, on the lower plot you should go to the left (try it!) and get approximately $-0.73$; xor go to the left on the upper plot and get approximately $-2.41$.

Any of these two numbers might be your $\phi$ calculated by looking at $y(0,0)$, because they both satisfy:

$$\sin(\phi) \approx - \frac 2 3$$

Can anyone (…) explain why I get wrong results which also differ from one-another?

Every result you've got is (approximately) good on its own; none is wrong. They differ because there are different $( y_0, k, \omega, \phi )$ tuples that describe the same wave. With different ways you've got the values from different tuples.

  • $\begingroup$ One more question. How can an $w$ be negative? What does that mean? $\endgroup$
    – 71GA
    Jul 3, 2016 at 7:33
  • 1
    $\begingroup$ @71GA It is purely mathematical thing, means nothing on its own. It's the sign of the $\frac \omega k$ (phase velocity) that tells if the wave propagates towards $x+$ or $x-$ as the time goes. You can demand (choose) positive $\omega$ only and allow $k$ to be negative, or write the equation with $\pm$, as you originally did. Yet pure math allows $\omega$ to be negative. I had to consider a general case to show how you can get different values for $\phi$. $\endgroup$ Jul 3, 2016 at 8:09
  • $\begingroup$ Thank you! I really couldn't explain to myself the negative $\omega$, but if it is just a mathematical thing I can live with it. =) $\endgroup$
    – 71GA
    Jul 3, 2016 at 18:09

Your results are not wrong, all of them are totally valid (with a small rounding error, see below). They just seem to "not fit together", since you dropped results you shouldn't have dropped.

Lets take your first graph: You are stating that $k = \frac{2\pi}{\lambda} = 628 \frac{1}{m}$. Why not $k=-628 \frac{1}{m}$?
In your lower calculation you basically assume $k x' + \phi = 0$ leading to $\phi = -\frac{x'}{\lambda}2\pi$. But the equation would actually be $k x' + \phi = n\pi$ leading to $\phi =n\pi -\frac{x'}{\lambda}2\pi$ which actually leads to $\phi = -0.628, 2.5, 5.65, ...$ If you do the same calculation, but use the point $(-0.0039|0)$ instead (notice it is not exactly 0.004), your calculation will lead you to $\phi = -2.45$.

If you keep the possibilities for a different sign of $k$ and $\omega$ and do the same calculations you marked as wrong for the other points where your function becomes zero, you will see that there is two possible values for $\phi$. Then take into account, that sine can become $-\frac{2}{3}$ on more than one point and that your values are not exact and you are done.
Well, not completely, you still have some choices when it comes to signs and multiples of $2\pi$, but thats just how it is, so find a combination that fits.


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