How would I find the direction in which these waves are travelling?

mrh124

1. The problem statement, all variables and given/known data
These are two problems, but what I'm supposed to do is the same. The general case is that i'm given two Y[m] vs. t[s], 2 Y[m] vs. x[m], or a Y[m] vs. x[m] and Y[m] vs. t[s] graph. It asks me to explain how the y(t) graphs tell you "whether this wave is moving left or right." How would it been done in case? I just need a general method that I can apply because the way i've had it explained before was really unclear

Note these are for two different problems

Graph set #1: two Y[m] vs. t[s]

Graph set #2: a y[m] vs. x[m] and a Y[m] vs. t[s]


2. Relevant equations

Y=Asin(2∏/T*t±2∏/λ*x+phi)

3. The attempt at a solution

For graph set #1 (the first problem):
I looked at the graphs and noticed that second one's position is .1ms away from the first and noticed that if you shift the first graph over to the left it looks like the second graph. Does that mean it's going left?

For graph set #2 (the second problem):
I looked at the y(t) vs. t graph and noticed that it's going down originally and realized that if i shifted the graph to the right (y(x) vs. x(t)) at point x=0 I would get the same graph, so it must be going left? I dont understand this at all.

Also what if I had two y(x) vs. x[s] graphs. What would happen then?

Simon Bridge

Have you tried constructing their wave equations? [tex]y(x,t)=A\sin \left [{k(x-vt)+\phi}\right ][/tex] ... solve for v.

Of course what you are expected to do is understand waves :) to get that understanding, you should play around with them. Make up a bunch of travelling sine waves and work out what they look like at different times and places. After a while you'll get a feel for it.

eg in the first one - look at the shape of the wave with time: at x=1m, the wave hits zero at 0.3s, but only 10cm to the right of it the wave has already crossed zero a second ago. Presumably the wave will cross zero 10cm to the left at t=0.4s ... so which direction is the wave moving?

mrh124

no, i have not. my class uses a different wave equation. what would v tell us and would this would work in all the cases?

Simon Bridge

In general, if y(x)=f(x) and the shape is travelling in the +x direction at speed v, then at some time t, y(x,t)=f(x-vt).

This equation is derived by considering that at t=0 [itex]y(x,0)=A\sin{(kx+\phi)}:k=2 \pi / \lambda[/itex] ... at t > 0, then x [itex]\rightarrow[/itex] x-vt ... you should be able to derive the equation your class uses from that.

v is the wave speed - if it is positive, then the wave is moving in the +x direction.
it should work for all waves which can be described by that equation.

mrh124

Okay thank you. How would lamba found with two displacement vs. time graphs? the wavelength is like the period except for displacement Y[m] vs. position X[m].

mrh124

uhm.. hmm it looks like it's going left since the second graph looks like a version of the first graph that has been shifted over to the left? im not sure..

Simon Bridge

This is where understanding the waves comes in - you'll have to play around with the equations and the graphs until you get it.

That's actually the point of giving you this kind of exercise - forces you to experiment with the math and so gain understanding: which is the prize here. Go for it.

Take care about what the graphs are actually telling you though - just looking at the first pair - the horizontal axis is in time. The second is shifted to the left wrt the first one means that the displacement is happening earlier in the different place. If the second graph were for x=0.9m instead of 1.1m the direction of travel would be opposite... but it would be the same graph.

mrh124

thank you =). I figured out both problems with your help =)

Simon Bridge

No worries :) and well done.

I was bracing myself to mention what happens when the wave is fast enough for almost whole wavelength to go by ... which can trick you into thinking the wave is travelling backwards. This is what happens to wagon wheels in old movies. But it sounds like you've managed to figure it out.