Why does my wave equation appear to move backwards?

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The discussion revolves around a wave equation of the form y=10sin(18.35x-6283t) that initially appears to move backwards when graphed. The confusion arises from the interpretation of the phase component in the equation, which affects the wave's direction. It is clarified that waves can move in either direction, but the observed backward movement was due to a misunderstanding of the equation's parameters. After reevaluating the phase component, it is concluded that the wave is actually a forward traveling wave. Understanding the phase is crucial for accurately determining wave direction.
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I have a wave eqn that is of the standard form (y=sin(kx-wt)) and have a positive constant velocity. But when I graph it, it appears to be moving backwards with a greater t producing a lesser x for each specific y. How can this be?

The eqn is y=10sin(18.35x-6283t)
 
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It could be a reflected wave.

The wave direction is either + or - with respect to the coordinate system. In one dimension, waves can move in one of two directions - forwards or backwards.
 
How can it be a reflected wave when there is nothing to reflect it with? The time I subed in was tiny and it still showed a backward moving wave. So the wave is going backwards immediately after t=0.
 
I know where I made my mistake. I didn't fully understand how the phase component worked. The wave turns our to be a forward traveling wave after all.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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