I have simple but a frustating doubt. I have seen that equation of a wave on a string can be given by $y=Asin(wt-Kx)$ or by $y=Asin(Kx-wt)$, where $w$ represents angular velocity of wave, $A$ represents amplitude, $K$ ($K=\frac{2\pi}{\lambda}$) represents wave number and $x$ represents displacement of wave at any time $t$. I was confused when I saw a sentence "Let's take a equation of a wave as $y=Asin(wt-Kx)$" in my textbook and a another statement "Let's take a wave equation as $y=Asin(Kx-wt)$" in another book. But how can both these different equation represents a wave tavelling in the same direction?. Because one of the wave equation represents wave travelling in $+ve$ y-axis and another represents the wave travelling in $-ve$ y-axis, from mean position at $t=0$

Being simple I would just ask that, If I am asked to write a wave equation which is traveling with a angular velocity $w$, with some amplitude $A$ starting from mean position at $t=0$ what would be my answer? Is it be $y=Asin(wt-Kx)$ OR $y=Asin(Kx-wt)$? I get much confused while dealing with word problem questions as using these two equation gives two different answers. So I don't know which one to use when.

  • $\begingroup$ For a travelling wave, a general express is $$y(t,\vec{x})=Asin(\omega t-\vec{k}\dot{}\vec{x}+\phi)$$ It involves two variables t,x, so if you want to decide the direction the wave moves along $x$, you must fix the variable $t$. So, the phase doesn't get changed with time $t$, or $\frac{d(\omega t-\vec{k}\dot{}\vec{x}+\phi)}{dt}=0$ Thus, $\omega=\vec{k}\frac{\vec{x}}{dt}$, If wave travels along x, we have $\vec{v_x}=\frac{\omega}{k}\hat{x}$ with wave eqution $y(t,\vec{x})=Asin(\omega t-kx+\phi)$, and if along -x, then $\vec{v_x}=-\frac{\omega}{k}\hat{x}$and $y=Asin(\omega t+kx+\phi)$ . $\endgroup$
    – Feynman
    Apr 1 '17 at 3:11

These two expression are just two instances of the general wave equation $$ y=Asin(\omega t - kx+ \phi) $$ where $ \phi $ is the initial phase. This initial phase depends on the initial conditions. How the wave looks like at t=0, for example. But for any initial conditions the wave propagates in the same direction. Your two example are just special cases, for $\phi =0$ and $\phi =\pi$
[$ y=sin(\omega t - kx)=-sin(kx- \omega t)$ and $sin(x+\pi)=-sin(x) ] $

If you don't know anything about initial condition, picking one or the other is quite arbitrary, like when you pick the positive side of an axis.


I have a trick for these questions, but by no means it is a formal method :P

See, fix the position, $x$.

Now, just observe the firs wave equation $y=Asin(wt−Kx)$ at a constant $x$, if time t increases, y will increase as well since $wt-kx$ increases (I am doing this for the first one-fourth part of the sine wave).

This must happen if the wave is traveling in +ve x direction!

That's pretty much how I remember which is the direction of wave propagation.

The method surely 'looks' longer than it really is, but at least you don't have to mug stuff up.

  • $\begingroup$ But both of the equations represents wave travelling in +ve x direction, only they differ in y directions. A wave travelling in -ve x direction is given as $y=Asin(wt+kx)$. $\endgroup$
    – Avi
    Mar 31 '17 at 18:03
  • $\begingroup$ Also how can you keep the x constant while varying the time (t) both at a same time? Because If x not changes with time, that means the wave is not moving. $\endgroup$
    – Avi
    Mar 31 '17 at 18:04
  • $\begingroup$ "...A wave travelling in -ve x direction is given as y=Asin(wt+kx)" exactly, i got the wave travelling in -ve direction for the other equation, didn't i? $\endgroup$ Apr 1 '17 at 4:00
  • $\begingroup$ x constant and time varies is just like taking a picture of the wave at two different instants and then comparing one single point on both those pictures! $\endgroup$ Apr 1 '17 at 4:01

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