# I Question on Sinusoidal Wave

1. Jan 5, 2016

### toesockshoe

My textbook states the following:

The wave disturbance travels from x=0 to some point x to the right of the origin in an amount of time given by x/v, where v is the wave speed. So the motion of point x at time t is the same as the motion x=0 at the earlier time t-(x/v). Hence we find the displacement of point x at time t by simply replacing t in the equation: y(x=0,t)=Acos(wt) by (t-(x/v)) to get the following: y(x,t)=Acos(w(t-x/v)) .

My book is talking about sinusoidal waves and how to give them a function.

I dont understand the t-(x/v) part.... it says that "motion of point x at time t is the same as the motion x=0 at the earlier time t-(x/v)".... what does it mean by motion? Does it mean that its at the same phase of the sin wave? This wouldnt make sense because if the x displacement is not a factor of 2pi, then it would be in a different phase.... Can someone please clarify this part?

2. Jan 6, 2016

### Merlin3189

I think this author causes some confusion by using x to label his axis, represent a variable distance along it and to name the point which is a distance x along the x axis!
Perhaps we could have two points, A at the origin where x=0 and B some point to the right where x=b. (Assuming the x axis is from left to right.)
Each point of the wave moves, oscillates, about a fixed point. A oscillates about x=0 and B oscillates about x=b, but both stay centred on those positions. The movement (motion) could be transverse, say up and down in the y axis (I'd stick with that idea for now), or longitudinal where the point moves a little forward and backward along the x axis.

So they are saying that the movement (motion) of point B looks (is) exactly the same as the point A was doing at some previous time.
At this moment A could be going up and B going down. A quarter of a cycle later, A could be stationary at the top of its movement and B stationary at the bottom of its movement. Then a further quarter cycle later, A is back on the axis going down and B is back on the axis going up. Another quarter cycle later A is at its lower extreme and B is at its top extreme. Etc. B does exactly the same as A, but a bit later - in this case half a cycle later, for simplicity, but it could be any delay. If that delay were exactly one cycle, then they would be in step, but most points will be out of step.

They are saying that when the distance from A to B is x, the time difference (or delay) is = x/v (time = dist /speed)
So what B is doing now, is what A was doing x/v seconds before.
So if A's motion is described by y= A cos(w t)
then B's motion is what A was doing when the time was x/v earlier, ie. t - x/v , so for B, y = A cos( w (t - x/v))

3. Jan 6, 2016

### toesockshoe

ahh thank you. that makes perfect sense.

4. Jan 8, 2016

### toumaza

there is a difference between the distance x you stated and the displacement. i mean as far as the wave is traveling in the same medium the speed remains the same(provided no external force act upon my force mean any external condition)and the t-(x/v) is the phase difference. i also think you omitted the angular speed w(omega) unless the author omitted to write it. t has a dimension meanwhile angle is dimensionless so you can add the w to the equation and you would get the equation you are used to. sorry for my english

5. Jan 8, 2016

### sophiecentaur

It would have been easier to understand if the author had talked in terms of displacement, rather than motion. The "motion" he talks of is the motion of the medium and not the motion of the wave. The negative sign just means that, as you go along the x axis, you are seeing a displacement that was at the origin x/v seconds before so it corresponds to an earlier time i.e. subtracting delay time.

6. Jun 5, 2016