Explanation of simple harmonic motion as an example of problem based learning

After exams were over this term, students doing further maths in Abingdon have the opportunity of doing a maths project. Given two weeks, we are asked to find a topic related to maths that interests us, and that we can give a presentation on afterwards. I decided to choose a simple example of a dynamical system, because they are often used to model biological phenomenon. As it happens, the topic I chose wasn't as simple as I thought, but here's how it works, from basic maths upwards:
(warning: huge post)



Simple harmonic motion is easiest to explain by using it as a way of modelling a mass on the end of a spring:

You'll have to excuse the scribbles.

when the spring is at rest, there is no force acting on the mass. When the mass is displaced from this position of equilibrium, the spring exerts a force on it in the opposite direction to the direction it moved (in a completely 1D system). This force is directly proportional to the displacement, according to Hooke's law. That is to say:

F = -|k|x

Where F is the force, x is the displacement and k is some constant that depends on the strength of the spring. k is negative because it is in the opposite direction of x.

Because there's a force and a mass, and F = ma, there is acceleration (=F/m) towards the point of equilibrium, where x = 0, so F = -|k|x = 0. However, the object still has velocity and mass when it reaches x=0, so therefore it has a momentum (=mass*velocity), so it continues past the point of equilibrium. 

Therefore x changes direction, so there is a force in the opposite direction to the movement, which makes it accelerate towards x = 0 again. With no energy loss, this continues forever, and is periodic:

It looks like you should be able to find relationships very easily between x and v, and therefore x and a. It is possible, but not as easy as it looks. 

Here, I was a bit stuck. This is a second order differential equation: I can't integrate it, so I have to find a general solution, which was something I'd never come across before. So for those people who've never done something like this either, this is how to do it:

Because all the 'a' numbers are known constants, it is possible to solve for λ. The general solution is then found using y = e^(λx), by substituting in the value of λ. It is also important to remember the constants, because we are integrating (sort of). Because this is an exponential function, the constants are manifested as coefficients of e. So in our specific example, it looks like this:

To find λ, we need to find the square root of a negative number. This hadn't been covered in my syllabus before either, but by sixth form most people are aware that the root of -1 is a different type of number, called i. You can read this brief explanation of imaginary numbers if you want, but it isnt really necessary: 

All numbers I had encountered so far could be represented on a 1-dimensional numberline, like a graph without a y axis. I was also aware that none of the numbers I knew could be multiplied by themselves to produce a negative number. This is because a minus times a minus is always a plus, and a plus times a plus is always a plus, and there was nothing on the numberline apart from minus and plus. 

However, sometimes it is necessary to take the square root of a negative number. Therefore something apart from either 'minus' or 'plus' has to be created. Because they are not representable on our number line, or by an actual quantity of physical things, these numbers are called imaginary. Imaginary numbers are shown as a y-axis on the numberline. This means that all the numbers I already knew had an imaginary number 'coordinate', but that it was 0. 

To find the square root of a negative number, first factorise out -1, then take the root of both factors. For example:

sqrt(-16) = sqrt[(-1)*(16)] = (i)*(±4)

=±4i

So, back to finding λ:

Someone else in the class had just done a very impressive explanation of Euler's formula, so I used that without proving it, but an explanation can be seen here.

This proves that the displacement is sinusoidal, which means that it is periodic, provided no energy is lost. However, more interestingly, it also means that:

v = x' = ωRcos(ωt +α)

and

a = x'' = -(ω^2)Rsin(ωt +α)

So displacement, velocity and acceleration are all periodic. 

Also, on a side note, substituting x into the equation for acceleration gives:

a = -(ω^2)x

Which is quite neat.

Applications

SHM is used in very many aspects of very many sciences; here's a few examples from the three main ones we study in school:

Physics

The mass-on-a-spring can be used as a model for circular motion with one dimension, distance from the starting point. This means that it's very easy to find the acceleration or velocity of the moving object at any one point. Pendulums also follow simple harmonic motion, and any other kind of oscillators too. I don't know, I'm not really interested in physics.

Chemistry

When two particles are connected together, AS chemists will know that they oscillate (which is what absorbs different frequencies of radiation and therefore gives us IR spectra etc.). The oscillation is basically a mass on a spring. SHM is useful here because you can use it to find the energy of the system:

Because k and R are both arbitrary constants that you can find out given starting conditions, this shows that the energy of the system is constant. Which is quite exciting.

Biology

We don't really know much about biology, which is one of the reasons I want to have a job in it. Nothing's really changed in non-quantum physics since the 30's, and by and large, we understand a lot of chemistry. With biology, there are so many possible factors to everything that it's very hard to make any kind of mathematical model of a biological system (it is still possible, but currently beyond my skills). Nevertheless, simple harmonic motion has been used as a model of neural oscillations, heartbeats, and the circadian cycle, among others.