[Notes on Diffy Qs home] [PDF version] [Buy paperback on Amazon]

[next] [prev] [prev-tail] [tail] [up]

Note: 2 lectures, §9.4 in [EP], not in [BD]

Let us return to the forced oscillations. Consider a mass-spring system as before, where we have a mass on a spring with spring constant , with damping , and a force applied to the mass. Suppose the forcing function is -periodic for some . We have already seen this problem in chapter 2 with a simple . The equation that governs this particular setup is

(4.9) |

The general solution consists of the complementary solution , which solves the associated homogeneous equation , and a particular solution of (4.9) we call . For , the complementary solution will decay as time goes by. Therefore, we are mostly interested in a particular solution that does not decay and is periodic with the same period as . We call this particular solution the steady periodic solution and we write it as as before. What will be new in this section is that we consider an arbitrary forcing function instead of a simple cosine.

For simplicity, let us suppose that . The problem with is very similar. The equation

has the general solution

where . Any solution to is of the form . The steady periodic solution has the same period as .

In the spirit of the last section and the idea of undetermined coeﬃcients we ﬁrst write

Then we write a proposed steady periodic solution as

where and are unknowns. We plug into the diﬀerential equation and solve for and in terms of and . This process is perhaps best understood by example.

Example 4.5.1: Suppose that , and . The units are again the mks units (meters-kilograms-seconds). There is a jetpack strapped to the mass, which ﬁres with a force of 1 newton for 1 second and then is oﬀ for 1 second, and so on. We want to ﬁnd the steady periodic solution.

The equation is, therefore,

where is the step function

extended periodically. We write

We compute

SoWe want to try

Once we plug into the diﬀerential equation , it is clear that for as there are no corresponding terms in the series for . Similarly for even. Hence we try

We plug into the diﬀerential equation and obtain

So , for even , and for odd we get

The steady periodic solution has the Fourier series

We know this is the steady periodic solution as it contains no terms of the complementary solution and it is periodic with the same period as itself. See Figure 4.12 for the plot of this solution.

Just like when the forcing function was a simple cosine, resonance could still happen. Let us assume and we will discuss only pure resonance. Again, take the equation

When we expand and ﬁnd that some of its terms coincide with the complementary solution to , we cannot use those terms in the guess. Just like before, they will disappear when we plug into the left hand side and we will get a contradictory equation (such as ). That is, suppose

where for some positive integer . In this case we have to modify our guess and try

In other words, we multiply the oﬀending term by . From then on, we proceed as before.

Of course, the solution will not be a Fourier series (it will not even be periodic) since it contains these terms multiplied by . Further, the terms will eventually dominate and lead to wild oscillations. As before, this behavior is called pure resonance or just resonance.

Note that there now may be inﬁnitely many resonance frequencies to hit. That is, as we change the frequency of (we change ), diﬀerent terms from the Fourier series of may interfere with the complementary solution and will cause resonance. However, we should note that since everything is an approximation and in particular is never actually zero but something very close to zero, only the ﬁrst few resonance frequencies will matter.

Example 4.5.2: We want to solve the equation

where

extended periodically. We note that

The solution must look like

for some particular solution .

We note that if we just tried a Fourier series with as usual, we would get duplication when . Therefore, we pull out that term and multiply by . We also have to add a cosine term to get everything right. That is, we must try

Let us compute the second derivative.

This series has to equal to the series for . We equate the coeﬃcients and solve for and .

That is,

When , you will not have to worry about pure resonance. That is, there will never be any conﬂicts and you do not need to multiply any terms by . There is a corresponding concept of practical resonance and it is very similar to the ideas we already explored in chapter 2. We will not go into details here.

Exercise 4.5.2: Let . Find the steady periodic solution to . Express your solution as a Fourier series.

Exercise 4.5.3: Let . Find the steady periodic solution to . Express your solution as a Fourier series.

Exercise 4.5.4: Let . Find the steady periodic solution to . Express your solution as a Fourier series.

Exercise 4.5.5: Let for and extended periodically. Find the steady periodic solution to . Express your solution as a series.

Exercise 4.5.6: Let for and extended periodically. Find the steady periodic solution to . Express your solution as a series.

Exercise 4.5.101: Let . Find the steady periodic solution to . Express your solution as a Fourier series.

Exercise 4.5.102: Let . Find the steady periodic solution to . Express your solution as a Fourier series.