GRAPHING OSCILLATING FUNCTIONS |
During this tutorial you will be asked
to look at a graphs and you will encounter a number of numerical calculations.
You should have your calculator handy to check these calculations. It's
important to actually do this in order to get the maximum benefit.
A class of functions which occur frequently in all sciences are those which are oscillatory, and in particular, those which are characterized by a sine or cosine. These functions arise in many ways in science but your first introduction to them will probably be in the study of waves.
Waves can be realized in many ways and in many media, but here we will examine transverse waves on a string because, in this case, the wave on the string is a picture of the graph we want to be able to draw. However, this example will be similar to many other waves (for example, water waves, sound waves, light waves, etc.). The purpose of this tutorial is not to teach you the physics of waves. It is assumed that in your other studies, you have learned the equations of travelling and standing waves.
Panel 1 |
You must realize that you can't draw a graph of this whole function because it's a function of two variables, x and t. You can do two things. You can imagine taking a snapshot of the string and showing y vs x at a given time or you can choose a particle on the string at some given position x and see how its displacement varies with time. But normally, you can't do both at once. (Three-dimensional graphs are hard to draw, but some computer graphics programs can now do them easily.)
Panel 1 |
The quantity "k" is called the wave vector and is equal to 2p/l, where "l," is the wavelength.
The quantities "A" and "l," are shown in
the figure. The wave travels with the speed "v" given by
v = fl, . Now, anyone can sketch a sine
curve like this and label it but the real problems in this graph have been
avoided.
How does it intersect the y axis and at what positions from the origin does it cross the x axis? We know that these crossings are l, apart but how far is any one of them from the origin? The way to show you how to do this is best done by using a specific example.
Let's take the case of y = 0.5 sin(3t - 3p x). First, what do we know about this wave? Let's make a list.
Panel 2 |
Panel 3 |
Panel 3 |
How is the function going through these blue dots? Is it rising, or falling? Well, let's imagine increasing x by a little bit. |
Now you know everything and can sketch in the whole wave.
Here's another situation to consider.
Suppose the original wave function had been y = -0.5 sin(3t - 3p
x). Notice it's the same as before except for a minus sign in front. One
way to handle this is to do the analysis exactly as before but everything
gets reversed in the y direction by the minus sign.
At t = 3p seconds, for example, the crossing
points are as before but instead of the wave going down and to the right
through them, they will go up and to the right. All other considerations
are the same and the graph can be sketched easily as shown in Panel 4.
Make sure you understand this.
Panel 4
Let's look briefly at another case.
Suppose a graph of our travelling wave is wanted at t = 3.2p
seconds.
Now our equation is y = 0.5 sin(9.6p - 3p
x). You can now see that the "zero-crossing" occurs at a point given by
(9.6p - 3px) = 0
or x = 3.2 metres. We also have "zero-crossings" every 2/3 of a metre on
either side of this. These are indicated in blue. Again, the crossings
half-way between go the other way and the graph can be sketched.
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At t = 3.2 p s
y = 0.5 sin(9.6 p - 3 p x) y = 0 when (9.6 p - 3 p x) = 0 So, x = 3.2 m At x = 0,
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Now, the value of the function is obviously not 0 at x = 0, but of course the value is easy to find. At x = 0, y = 0.5 sin(9.6p). Remember the 9.6p is in radians and since the sine function repeats itself every 2p radians, we can take 8p radians away from it, leaving sin(1.6p).
Recall that p radians is 180°, so that 1.6p is 288°. Sin(288°) = -sin(72°) = -0.95. So y at x = 0 is negative and is, in fact, -0.475 metres. Again, the sketch of the graph is completely determined. Study panel 5 carefully.
Let's now look at the problem of graphing
a standing wave. A general equation of a standing wave is
y = (2 A cos wt) sin kx. All the symbols
are the same as we defined previously. The 2a cos wt
is put in parentheses to emphasize that what we have is the sine of function
of x with an amplitude of 2a cos wt which varies
periodically in time. Again, let's look at a specific example.
Suppose we're asked to sketch the standing wave y = 0.5 cos (7p
t) sin(3p x). First let's list all the things
we know.
w = 7p radians/second,
so f = 7p/2p = 7/2
cycles/second. k = 3p metres-1 so
l
= 2p/3p = 2/3 metre.
Now, again we can only plot the standing wave if we're told the time at which the graph is wanted, that is, the time at which the snapshot picture is to be taken. For example, what's the graph at time t = 0? Then, cos (7p t) = 1 and the function is y = 0.5 sin(3p x). As before, this has a maximum value of 0.5 metres.
Panel 6
Where are the crossing points? Well, y = 0 when (3p x) = 0, which means x = 0, so the function is a positive sine curve with wavelength = 2/3 metre. Make sure you understand the graph in panel 6.
Panel 7 |
But just to see that it is so, let's work it out in detail. Substituting t = 1/14 second into our standing wave equation gives for the amplitude 0.5 cos(7p 1/14) = 0.5 cos(p/2). Since cos(p/2) = 0, then y is 0 everywhere.
Finally, what's a graph of the standing wave at t = 1/10 second? Now
our equation is
y = 0.5 cos[7p (1/10)] sin(3p
x). Looking at the amplitude term, this is the cos(0.7p)
or
cos(126
). Cos(126
)
= -cos(54
) = -0.588. This multiplied by
0.5 is -0.294. The graph is thus the negative sine curve with an amplitude
of 0.294 metres, shown in Panel 8.
Panel 8
This is the end of this tutorial on graphing oscillating functions.