Today, we start by asking a very simple question: when we attach a box to a spring and pull the box backwards, can we find a function that gives us the displacement of the box from its initial position?
To make things harder, we're going to actually include the effects of friction; a situation almost unheard of in almost every physics question. While this may seem simple enough, its solution takes us through the realms of Newtonian physics, ordinary differential equations, complex analysis, and some heavy duty maths. GCSE Physics, A Level Maths, Further Maths, and a healthy knowledge of extracurricular mathematics is recommended.
To clarify, what we're looking for is a function, x(t) that tells us the displacement of the box as time changes. To find this, we first need to consider the forces acting on the box itself as it moves. The first force is easy: the force exerted on the box by the string, known as the restoring force of the string. From previous (GCSE) knowledge in physics, we know this is equal to the spring constant, k, multiplied by the extension, x. However, since the acceleration of the box will obviously be in the opposite direction to the extension, the restoring force is actually -kx.
The second force is friction. There are a few different models for friction, but the most common one is "wet" friction, which acts in the opposite direction to velocity and is directly proportional to it. In the animation, I use b as this constant of proportionality, and the notation x with a dot above it to signify the first derivative of displacement, or velocity.
By Newton's second law, the resultant force on the box is equal to its mass multiplied by acceleration. The resultant force on the box is equal to the restoring force add the friction force, as we can add together force vectors to find the resultant force vector. The mass of the block is constant, and acceleration is the second derivative of displacement, represented by x with two dots above it. Rearranging this gives us the second order differential equation governing the motion of all springs undergoing this type of motion, known as standard harmonic oscillation.
Taking a step back and looking at what we've done shows us that we've found an equation which, when we plug our function, x, into, MUST equal 0. This is interesting because the equation actually looks very similar to a quadratic, except instead of powers we have derivatives. If we could put this equation in the form of a quadratic, this could give us solutions. Secondly, we see that adding together the function x, and scalar multiples of its first and second derivative will give us 0. This means the first and second derivative of x must be proportional to x itself. There is only one function which satisfies this requirement, and that is the exponential function, e^t. Therefore, we can assume that x(t) = Ae^(alpha * t), where A and alpha are constants, which we then solve for.
It's important to note here that this process of stepping back and analysing like this is a cornerstone method for all maths. It's not just blindly applying knowledge wherever you see even the slightest hint of its use, but to really think about and analyse your results to gain some key insight.
Once we find the first and second derivative using the chain rule, plug it back into the differential equation and factorise, we find that either A must equal 0, or a quadratic in terms of alpha must equal 0. If A equals 0, it would solve this differential equation, but there would also be no movement at all, as x(t) would also equal 0. This is kind of boring, and it is is known as a trivial solution.
A more interesting solution (and the one we really care about) is found when the quadratic in alpha equals 0.
We can use the dreaded quadratic formula to solve this equation for alpha. As we're doing so, we take another step back and notice something; alpha is in the exponent in the function Ae^(alpha * t). What's confusing is that Ae^(alpha * t) is supposed to give us some sort of oscillation/wave movement, but exponential graphs don't exactly represent that type of movement at all. However, one very well known, very famous identity in maths is Euler's formula, e^(it) = cos(t) + isin(t), where i is the imaginary constant. Here, we see an exponential function defined in terms of two oscillating function, cos and sin. This is exactly what we want!
Armed with this knowledge, we factorise out sqrt(-4k/m), as this gets rid of the 2 in the denominator and introduces i in alpha, which allows us to use Euler's formula. This step is highlighted in red in the animation, as it is pivotal in solving this ordinary differential equation.
Finally, there's a bunch of messy stuff inside a square root next to i. It only contains constants, such as k, m and b, so we can nicely simplify it to a single constant called omega prime, which is a fancy w with a dash next to it.
Although this equation actually has two solutions for alpha, I'm only going to demonstrate the positive solution and explain why the negative solution produces the same equation for x(t) later. We now have alpha, which we can substitute back into the function x(t). This produces a messy concoction of a formula; A, which is any arbitrary complex number (displayed as a white dot and text in the animation below), e^(-bt/2m), a solely real function of time (displayed as a red dot and text in the animation below), and e^(i*omega prime * t), a complex function of t (displayed as a blue dot and blue text in the animation below).
To introduce a bit of order in this equation, we're going to convert A into its polar form, where |A| is the real distance from the origin to A, and phi (circle with a line through it) is the real angle off the origin in radians. This lets us separate x(t) into a real part, |A|e^(-bt/2m), multiplied by an imaginary part, e^(i(phi + omega prime * t)). According to all our maths so far, this should be a complete solution for x(t) (displayed as a green dot and text in the animation below), so let's see it's movement over time in the complex plane!
Clearly something is wrong here. The green dot isn't supposed to move vertically - it represents displacement of the box over time, and the box only moves horizontally as the spring only oscillates it horizontally. So what are we missing?
Well, the only part of this equation producing motion in the imaginary direction is the e^i(phi + omega prime * t) part of this equation. But remember: e^i(phi + omega prime *t) = cos(phi + omega prime *t) + isin(phi + omega prime *t). This means that, actually, the only purely imaginary component in this entire equation is the isin(phi + omega prime *t) part. Since vertical displacement is entirely independent of horizontal displacement, we can completely ignore the imaginary part of this equation, leaving us with our final solution for the function x(t):
Fantastic! We found an equation, x(t) which describes, the oscillation of a spring on a surface with friction. Although I haven't worked through how this would pan out with the other solution for alpha, we find that it really doesn't matter, as it only causes a difference in the sign of the imaginary part of the solution, which we ignore anyway. We could plug this back into our initial ordinary differential equation to show that it actually does equal 0, but that's left as an exercise to the reader.
A perhaps more interesting and exercise is to play around with this final function of x and what it looks like visually. What happens if b, the coefficient of friction, is 0? What if its 10? What happens when we change the spring constant, or stiffness, k, of the spring, so that the restoring force is very large for a given extension? What if the mass of the system is really large, so the force exerted by the spring causes very little acceleration? And above all, are all the changes in these values intuitive?
There are two good takeaways from this whole exercise. The first is that simple questions can have really complicated solutions. Just because a question seems straightforward doesn't mean it's very easy. The second is that sometimes, the way to solve a difficult equation isn't to enter a maths-induced stupor, constantly fumbling around with variables until you see some semblance of an answer. It's to take a step back and really think about what you're seeing, relating it to other instances of similar maths, and finding the best way to proceed from there.
Further Reading:
Brown, R. G., 2013. (Week 9 - Oscillations) Duke University. [Online] Available at: https://webhome.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1.pdf [Accessed 20 January 2020].
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