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My textbook says

"Consider an open organ pipe of length $L$ lying along $x$-axis with its ends at $x=0$ and $x=L$. The sound wave travelling along the pipe can be represented as $\Delta P=\Delta P^0\sin(kx-wt)$."

But I've always used a "$\sin(wt-kx)$" type wavefunction to represent a wave travelling in $+x$-direction. As I know that when the coefficient of $x$ in the wavefunction is negative, it represents a wave travelling towards $+x$-direction and the opposite when there's a — in the coefficient of $x$.

So my doubt is:

Whether both the aforementioned wavefunction can be used to represent a wave travelling towards $+x$-direction or not. If yes then please explain how.

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As $$ \sin(kx-\omega t)=- \sin(\omega t-kx) $$ the only difference between the waves is the sign of the amplitude coefficient. Both travel to the right if $k$ and $\omega$ are positive numbers. In general the wave velocity (counted as positive if to the right) is $v_{\rm phase}= \omega/k$

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  • $\begingroup$ I guess k and w are always positive as wavelength and time period are always positive. So v should always be positive. And if what you said is correct then how shall we represent a wave travelling in negative x-direction. $\endgroup$ – user50973 Dec 26 '17 at 16:30
  • $\begingroup$ @user50973. You make the wavenumber $k$ negative to get a leftgoing wave. It's not true that $k$ has to be a positive number. $\endgroup$ – mike stone Dec 26 '17 at 17:30

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