From my last three posts, I think we are ready to model the motion of a mass on a spring:

So, just like I did with the tennis ball, I will show you equations that describe this motion.

Now again, to develop these equations requires calculus, so I will just provide the final result. But before I do, just how does one begin modelling a physical process like this?

What is usually done, is to write down known equations that describe the forces acting on the mass. If you think about it, there are two: a force due to gravity and a force from the spring. There are other forces as well like resistance from the air, but as before, we will assume these to be zero to simplify the development.

Let’s first set up the picture. We have a weight on a spring. The weight has mass *m*. Now weight is different than mass, but on earth, the units are the same. So on earth, a 1 kg weight has a mass of 1 kg. But on the moon, the mass is still 1 kg but its weight is 0.165 kg because gravity is weaker there. We have a spring with a spring constant of *k* which is a measure of how stiff the spring is. The higher the value of *k*, the stiffer the spring.

To start the mass moving, we have stretched it *A* centimeters down from its resting position, then let go. We set up a one dimensional coordinate system where the rest position of the mass is 0 and up is positive.

The force due to gravity is –*mg* where *m* is the mass and *g* is the acceleration due to gravity. From my post on the tennis ball, remember that *g* is 9.8 m/s². It’s negative because the force is acting in the down direction. This comes from Isaac Newton’s second law that says that force is equal to mass times acceleration. That is, *F* = *ma*. The force due to the spring comes from something called Hooke’s Law: *F* = *kx* where *k* is the spring constant and *x* is the amount that the spring is stretched (negative) or compressed (positive) from the resting position.

So the force equation for this setup is:

*F* = *ma* = *kx* – *mg*

This is the equation engineers start with before they do calculus on it. So now in this post, this is the part where a miracle happens, and I’ll give you the final result.

So the equation that shows where the mass is at a certain time is below where *t* is time in seconds. It is assumed that time starts (that is *t* = 0) when the mass is travelling upwards and is at the 0 position:

%Translator MathMagic Personal Edition Mac v9.41, LaTeX converter, 2019.1.28 09:06

{x}\hspace{0.33em}{=}\hspace{0.33em}{A}\hspace{0.33em}\sin\left({\frac{180t}{\mathit{\pi}}\sqrt{\frac{k}{m}}}\right)

\]

In my next post, I’ll dissect this a bit and put some actual numbers in it and plot the results.