### Re: How do you calculate displacement in the y-axis wrt x-axis in wave motion?

Area: Physics
Posted By: Dan Goldner, grad student, MIT/Woods Hole Joint Program
Date: Mon Jul 21 09:52:57 1997
Area of science: Physics
ID: 868554022.Ph
Message:

You've asked many good questions! Before I talk specifically about them, I will give some general information about waves. You can read quickly through the parts you already understand, just to learn how I'm using the notation.

### Waves and math

In its simplest form, a progressive wave is just a repeating, measurable pattern that moves through space. An example is when you quickly move one end of a rope up and down. A pulse moves down the rope, as in this figure, showing the rope at three different times:

It's important to clarify what moves in what direction. Individual points on the rope ('particles'--like the red dot in the figure) move up and down, and we can measure their height. The pattern of high-low-high-low moves horizontally, so we can watch, for example, the "high spots" (circled) move along the rope. Of course, the movement of the pattern is related to the movement of the particles, since the pattern we see is made up of different particle heights.

Now for the math. An equation for a travelling wave must relate (at least) three variables:

y
the signal or property being measured, such as air pressure in a sound wave, or rope height in our example,
x
the distance travelled by the pattern,
t
time.
For the rope wave, the equation is

y(x,t)=a sin (kx+wt).

(This isn't quite the equation in your question--we'll get to that.) Notice that the height of the rope depends on where on the rope you are (x) and also on when you look at it (t) . In addtion to the variables, there are three parameters in the equation which determine the character of the wave:

a
the amplitude. This is the maximum value of the signal; in our case, the maximum vertical displacement of the string.
k
the wavenumber is 2*pi divided by the wavelength, which is the spatial length of the pattern. For the rope, the wavelength is the distance between high spots. Writing it out, kx = 2*pi*x/wavelength -- in other words distance is now measured by wavelengths, instead of by meters or centimeters or whatever. The 2*pi (pi=3.14125...) is thrown in because the sine function completes a cycle every 2*pi units, not every unit.
w
(often written as the greek letter omega) is defined as -2*pi divided by the period, which is the amount of time taken for one cycle of the pattern to occur or to pass a particular spot. This is exactly the same idea as was true for kx: time is now being measured in periods, instead of, say, seconds. Making omega negative is for later convenience; we could work through the problem with it positive if we wanted to.

As you remarked, w is sometimes called the angular velocity, which you noticed was strange since there is no circular motion in the wave we are studying. In some applications, it really does mean angular velocity! For instance, the height h of an object as it travels in a vertical circle is given by h = r sin wt, where r is the radius, and w is the actual angular velocity of the object. (If w is negative, is the object moving clockwise or counter-clockwise?) But the term can make sense in other uses, too, if you consider that anything moving in a circle repeats itself regularly, and represents a cycle. So some people use "angular velocity" for any cyclical process, in an analogy to circular motion. Other people avoid this usage, because as you know, it can cause confusion.

Let's examine the wave equation from both points of view: studying the cycle of displacement at one spot, and looking at the motion of the pattern. We'll use the figure above, which was plotted with a = k = 1 and w = -1.

### The bead on the rope

To think about how a rope "particle" is displaced in y, imagine the rope you are wiggling has a bead on it (the red dot in the figure). We will say the horizontal position of the bead is x=0. The x-position of the bead does not change, but of course, y does: from the equation, with x=0, we find

y=a sin(wt).

If we were to put the bead somewhere else, say x=3, the equation would be y=a sin(k*3+wt). (In that case k*3 would take the place of d in the equation from your question. That is a hint about what d is -- more below.)

### The speed of the wave

The rope "particles" don't move in x, but we definitely see waves moving in x. What's going on?

What we see moving is the pattern. You can see the "high points" on the rope moving along as the wave progresses. This is because one particle is highest one moment, then another particle, in another location, is highest the next. So the location of the highest particle has changed. Then the next particle is highest, and so on. The high point moves smoothly down the rope. Same for the low point, or points at the central height, or any other fixed height.

Any specific point in the cycle is called the phase. If we ask, "what the phase?", we mean, at what point in the high-low-high cycle are we? Mathematically (and more specfically), the phase is the argument of the sine function: kx+wt. If at some place and time, the phase is equal to pi/2, then the rope is at its maximum height, and there's a high spot at that place and that time (leftmost open circle in the figure). A short time later, t is larger. The phase (kx+wt) of this particular high point always equals pi/2. If t is larger then wt is more negative (remember w is negative!). For the phase to equal pi/2, x must be larger. So the high point must have moved to larger x (center open circle in the figure). A short time later, t and x have increased again, and the high point where kx+wt = pi/2 has moved again (rightmost open circle). (Notice that this wave is moving from left to right. What would you have to change in the equation to make it move from right to left?)

How fast does the wave move? We can make this question specific by asking, how fast does a particular phase, like a high spot, move? We can compare its postition at time t1 to its position at a later time t2. For the circled high spot in the figure, kx+wt always equals pi/2. That means kx1+wt1= pi/2 = kx2+wt2, which leads to

(x2- x1)/(t2- t1) = -w/k.

This has the units of a speed (length/time), and it is the speed at which the pattern moves down the rope. It is called the phase speed.

### The relationship of displacements in x and y

If I understand your question, you want to know how far a particle moves in y when the pulse, or pattern, moves a certain distance in x. We can now answer this. Here's how:

1. First we use the phase speed to calculate how long it takes for the pulse to move the distance in x.
2. Then we use the equation describing the wave to determine how far a particle moved in y during that time.

Say the wave pattern moves a distance dx. We know that the time it takes to do that, dt, is dx divided by the phase speed, so dt = -k*dx/w. (Remember w is negative, so dt is positive.)

How far does a particle move in the interval dt? It depends which particle it is! Say the particle we are interested in is at x=x0. At time t0, the particle's height is

y1=a sin(kx0+wt0).

After dt, the particle's height is

y2=a sin[kx0+w(t0+dt)] = a sin[kx0+w(t0-k*dx/w)] = a sin[k(x0-dx)+wt0].

In other words, the height at (x0, t0+dt) equals the height at (x0-dx, t0)!

The displacement dy = y2-y1 is the vertical displacement of the particle. Using trigonometric identities, you can show that this reduces to

dy = -2a cos[ k(x0-dx/2)+wt0 ]sin[ k*dx/2 ].

I hope I understood your question correctly! If I didn't, then I hope the information I've given helps you find the answer. Otherwise, please feel free either to rephrase and resubmit your question, or to e-mail me directly.