Traveling Wave Equation $\sin(kx-wt)$ vs $\sin (wt-kx)$

In my textbook most if the times it uses $A\sin(wt-kx)$, but occasionally there is a problem using $A\sin(kx-wt)$

So i just changed it from $A\sin(kx-wt) \to -A\sin(wt-kx)$ but does the amplitude change to $-A$? Is the wave going downwards first?(as negative amplitude would imply)

• The solution of the differential equation depends on the inertial conditions. As you sayed in the setting where you change $A \rightarrow -A$ it goes at first 'downwards' (it has negative derivative in 0). – Alpha001 Aug 15 '17 at 9:38
• The forms you show are for traveling waves, not standing waves. Standing waves take on forms like $\sin(kx) \sin(\omega t)$. Also note that the angular velocity is usual a lowercase omega $\omega$, rather than a double-u $w$; use \omega to get the right glyph in LaTeX and MathJax. – dmckee Aug 15 '17 at 17:14

Using $\sin(kx-\omega t)$ or $\sin(\omega t-kx)$ does not make a difference: it is just a matter of convention. You can very legitimately change $A\sin(kx-\omega t)$ to $-A\sin(\omega t- kx)$ without causing any damage.

Note that the full solution to the wave equation is of the form $A\sin(kx-\omega t+\phi)$. Expanding: $$A\sin(kx-\omega t+\phi)= A\sin(kx-\omega t)\cos\phi+A\cos(kx-\omega t) \sin\phi$$ and choosing $\phi=\pi$ transform $$A\sin(kx-\omega t+\phi)\to -A\sin(kx-\omega t)=A\sin(\omega t -kx)$$ Hence, the solution $A\sin(\omega t -kx)$ is just "shifted" by one half of a cycle compared to $A\sin(kx-\omega t)$: this shift does not change the wave.

Finally, the amplitude $A$ is usually defined to be a positive number, and it is one half of the difference between the maximum and the minimum reached by a wave of the type $\sin(kx-\omega t)$. If the wave is "going down first" then it might be better to have an explicit negative sign in front of $A$.

• Something is wrong with your first equation (aside from a missing $A$ on the cosine term). If you put in $\phi=0$ then you get $\sin(kx-\omega t)=-\cos(kx-\omega t).$ Also, a shift of $\pi/2$ should take sine into cosine. A shift of $\pi$ is required to go from sine to negative sine. – Bill N Aug 15 '17 at 14:49
• @BillN Yes you are right. Somehow I copied the trig identity wrong and it tumbled from there. Hopefully fixed now. Thanks! – ZeroTheHero Aug 15 '17 at 14:55
• You are still off. $\sin(A+B)=\sin (A) \cos (B) + \cos (A) \sin (B)$. Change your negative between the terms. It doesn't affect the final outcome in this situation, but you still want to be careful with signs in an answer. – Bill N Aug 15 '17 at 15:17
• @BillN Gee... thanks again... I really need to get more coffee. – ZeroTheHero Aug 15 '17 at 15:19
• -A means the wave is traveling downwards first, and +kx means the wave is traveling to the left side. That is what my textbook says so when the equation is sin(kx-wt) i got really confused... – SVS Aug 15 '17 at 21:55

Is the wave going downwards first?

That all depends an what you mean by downwards and what type of change you're interested in.

Let's investigate "downwards": The wave solution describes the disturbance of some quantity from its equilibrium solution. It may be a displacement of a section of string from its rest position (say, in a violin or cello or bass). What is "downwards" for the bass? The string is almost vertical.

Or the wave could be a pressure wave in some medium. In that case, molecules are being condensed and rarified along the path of transmission. How is downward defined? Actually, downward is irrelevant.

I believe I understand why you say "downward" and that is because of a routine association between the variable $y$ and the local gravitational pull direction. That is an association that you need to break. The variable $y$ can represent anything, and for the standard wave solution it is merely the disturbance about an equilibrium position.

Now, on to the type of change. You could investigate how the wave disturbance from equilibrium, $y$, changes as you move along the direction of travel of the wave (the $x$ direction) for a given instant of time, $t$, OR you could investigate how $y$ changes for a particular point, $x$, as the time moves forward ("time keeps on slippin' into the future" according to Steve Miller Band).

Let's consider $y=A\sin(kx-\omega t)$ as our starting convention. If we take a time snapshot, start at $x=0$ and slide along the $+x$ direction, the disturbance initially becomes larger in the (pre)defined positive $y$ direction, then it turns around and goes back in the negative $y$ direction. On the other hand, if we choose a particular $x$ location and monitor what happens with time something different will happen. Consider watching $x=0$, our previous starting point. As time progresses, that point will move in the $-y$ direction, reach a maximum negative and turn around and move back toward equilibrium. Again, downward is irrelevant. And the choice of positive $y$ direction is arbitrary.

The importance of the sign of $y$ only shows up when one makes an association with a physical system.

Yes that is correct.

If we look at the addition formulae for sine functions we see that $$\sin (A - B) \equiv \sin A\cos B - \cos A\sin B$$ and that $$\sin(B-A) \equiv \sin B\cos A - \cos B\sin A$$We can easily see that the second equation is just $-1$ times the first equation. By adding amplitude all we are doing to the equations is multiplying both sides by a scalar as such $$X(\sin (A - B)) = X(\sin A\cos B - \cos A\sin B)$$ So we can say that $$X(\sin (A - B)) = -X(\sin (B - A))$$