Consider $E(x,t)=A\sin(kx-\omega t)$ where $k=2\pi / \lambda$, with $\lambda$ the wave length and $A$ its amplitude. We have $$E(x,t)=A\sin(k(x-vt))$$ so this wave is going to the right. Know, if I want to make it going to the left, I just have to change the sign of $v$ which leads : $$E(x,t)=A\sin(k(x+vt)).$$ Is that right? But, in my course it is written that changing the sign of $k$ can change the direction (left or right) of the propagation of the wave, but I don't understand why. Any help would be appreciated,

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    $\begingroup$ What happens if you change the sign of $k$ in $kx - \omega t$?, keeping the sign of $\omega$ fixed? The same thing as what happens when you change the sign of $v$ in $k(x-vt)$. Basically, changing the sign of $\omega / k = v$ changes a left moving to a right moving wave. $\endgroup$ – insomniac Mar 19 '20 at 10:04
  • $\begingroup$ Do you have a program at hand, which allows you to plot these functions? If yes, please start at $t=0$ and increase the time in "small" steps. Use the two functions and see how the wave travels to the left or to the right. If you don't have such a program use Mathematica online, e.g. type "Plot[sin(1*(x - 1*0.2 )), {x, 0, 2*pi}]" and increase the time $t=0.2$. $\endgroup$ – Semoi Mar 19 '20 at 11:46

The Plane progressive harminic wave of the form $$E(x,t)=Asin(kx-\omega t)$$ where $\omega =kv$ , represents the wave with speed $v$ travelling in $+x$ direction. While $$E(x,t)=Asin(kx+\omega t)$$ represent the wave with speed $v$ travelling in $-x$ direction. Now Let I change the sign of $k$ form first equation. $$E(x,t)=Asin(-kx-\omega t)=-Asin(kx+\omega t)$$ that is wave moving in $-x$ direction. Similearly changing sign in second equation $$E(x,t)=Asin(-kx+\omega t)=-Asin(kx-\omega t)$$ that is wave moving in $+x$ direction. So Either of these two can be done both are equivalent.


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