On the Harmonic Oscillator

We’re in the case of a \(1\)-dimensional harmonic oscillator, depicted below as a mass \(m\) set on a frictionless ground attached to a fixed element to the left by a horizontally positioned spring of constant \(k\).

Harmonic oscillators play a major role in physics, as explained by Feynman in Chapter \(21\) of the first volume of his Lectures on Physics:

The harmonic oscillator, which we are about to study, has close analogs in many other fields; although we start with a mechanical example of a weight on a spring, or a pendulum with a small swing, or certain other mechanical devices, we are really studying a certain differential equation. This equation appears again and again in physics and in other sciences, and in fact it is a part of so many phenomena that its close study is well worth our while.

Some of the phenomena involving this equation are the oscillations of a mass on a spring; the oscillations of charge flowing back and forth in an electrical circuit; the vibrations of a tuning fork which is generating sound waves; the analogous vibrations of the electrons in an atom, which generate light waves; the equations for the operation of a servosystem, such as a thermostat trying to adjust a temperature; complicated interactions in chemical reactions; the growth of a colony of bacteria in interaction with the food supply and the poisons the bacteria produce; foxes eating rabbits eating grass, and so on; all these phenomena follow equations which are very similar to one another, and this is the reason why we study the mechanical oscillator in such detail.

Thus, we’ll take the time to analyze it more precisely than what is expected here: this is the occasion to present some aspects not covered, or not covered as explicitly in the book.
Despite everything happening in one-dimension, we’ll use a vector position \(\pmb{x}\), which is a function of time. Hence, the vectors for speed and acceleration will respectively be \(\pmb{v}=\dot{\pmb{x}} = \dfrac{d}{dt}\pmb{x}(t)\) and \(\pmb{a}=\ddot{\pmb{x}} = \dfrac{d^2}{dt^2}\pmb{x}(t)\). Note that we’re representing vectors with bold font instead of arrows; bold font is rather common in physics. This should be the only additional thing needed to follow through for someone having read the book up to this stage.

Contact force, normal force, friction force
The contact force is the force resulting from the contact of two objects. It is generally decomposed into vertical and horizontal components. The horizontal component perhaps is the most intuitive: it corresponds to friction forces. For instance, after having given a gentle push to an object an on ordinary table, it will only move up to a certain point: it is progressively stopped by the friction between the object and the table. You may think that this friction is caused at least in part because at a microscopic scale, both the object and the surface are far, far from being perfectly plane, and even more so at an atomic scale.
But this is false. As a counter-example, consider gauge blocks, manufactured to be extremely flat: their surfaces, far from being frictionless to each other, cling!
Because a rigorous understanding of friction forces requires advanced mechanics knowledge, we can satisfy ourselves with Feynman’s vulgarization, from Chapter \(12\) of the first volume of his Lectures on Physics:

There is another kind of friction, called dry friction or sliding friction, which occurs when one solid body slides on another. In this case a force is needed to maintain motion. This is called a frictional force, and its origin, also, is a very complicated matter. Both surfaces of contact are irregular, on an atomic level. There are many points of contact where the atoms seem to cling together, and then, as the sliding body is pulled along, the atoms snap apart and vibration ensues; something like that has to happen. Formerly the mechanism of this friction was thought to be very simple, that the surfaces were merely full of irregularities and the friction originated in lifting the slider over the bumps; but this cannot be, for there is no loss of energy in that process, whereas power is in fact consumed. The mechanism of power loss is that as the slider snaps over the bumps, the bumps deform and then generate waves and atomic motions and, after a while, heat, in the two bodies.
[...]
It was pointed out above that attempts to measure \(\mu\) by sliding pure substances such as copper on copper will lead to spurious results, because the surfaces in contact are not pure copper, but are mixtures of oxides and other impurities. If we try to get absolutely pure copper, if we clean and polish the surfaces, outgas the materials in a vacuum, and take every conceivable precaution, we still do not get \(\mu\). For if we tilt the apparatus even to a vertical position, the slider will not fall off—the two pieces of copper stick together! The coefficient \(\mu\), which is ordinarily less than unity for reasonably hard surfaces, becomes several times unity! The reason for this unexpected behavior is that when the atoms in contact are all of the same kind, there is no way for the atoms to “know” that they are in different pieces of copper. When there are other atoms, in the oxides and greases and more complicated thin surface layers of contaminants in between, the atoms “know” when they are not on the same part. When we consider that it is forces between atoms that hold the copper together as a solid, it should become clear that it is impossible to get the right coefficient of friction for pure metals.

The normal force is the vertical component of the contact force. An object on a table is affected by the Earth gravity, but naturally resists going through the table, unless of course, the object is really massive and/or the table very weak. There’s a bunch of complicated interactions at the atomic level from which this situation occurs: at a macroscopic scale, we simplify things and wrap this complexity by abstracting it as a (macroscopic) force.

Newton’s laws
Let us start by recalling Newton’s laws of motion:

1. Principle of inertia: Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it;

2. "\(\pmb{F} = m\pmb{a}\)": The change of motion [momentum] of an object is proportional to the force impressed; and is made in the direction of the straight line in which the force is impressed;

3. "action \(\Rightarrow\) reaction": To every action there is always opposed an equal reaction; or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.

Hooke’s law
Hooke’s law, as Newton’s laws, is an empirical law, that is, which has been found by repeated experimentation. Essentially, it states that the force needed to extend or compress a spring by a given distance varies linearly with the distance during which the spring is extended/compressed. That linear factor \(k\) is the spring’s constant. This force is thus a function of that distance/displacement \(\pmb{\delta}\): \[\pmb{F}_s(\pmb{\delta}) = k \pmb{\delta}\] The equation could be read as something like:

• If I don’t compress/stretch the spring (\(\pmb{\delta}=\pmb{0}\)), no force is exerted;

• The more I compress/stretch the spring, the greater the force, where the relation between both varies linearly.

Forces diagram
The goal of a force diagram, or a free body diagram, is to identify the forces acting on an object, with the intent of applying Newton’s second law, so as to later determine an equation of movement for that object: we want to identify the sum of all the forces acting on the object, \(\pmb{F}_\text{net}=\sum_i\pmb{F}_i\), who are likely to contribute to the shape of its trajectory. Let’s complete the diagram of our horizontal spring setup with:

\(\pmb{F}_{e,o}\)

the gravitational force exerted by the earth on our object. Note that we haven’t talked about gravitation yet, and as you’ll see in a moment, we don’t need to, but this will be necessary for the vertical spring setup;

\(\pmb{F}_{t,o}\)

the normal force exerted by the table on our object. Because the surface is frictionless, the normal force is the contact force;

\(\pmb{F}_{s,o}\)

the force exerted by the spring on our object, which can be described by Hooke’s law.

Placement of the origin of the \(\vec{x}\)-axis As we’ve already observed before, it can be interesting to identify the origin of our coordinate system to the position where the system will be at equilibrium/rest.
Intensities and signs Thanks the previous convention, as discussed in Remark \(\ref{rem:L03E04:origin}\), Hooke’s law can be used to determine the force exerted by the spring on the object, and can be written as \(\pmb{F}_{s,o} = -k\pmb{x}\).
We know, from Newton’s second law, that the contact/normal force exerted by the table on the object is equal to \(-\pmb{F}_{o,t}\), where \(\pmb{F}_{o,t}\) is the force exerted by the object on the table, which itself results from the gravitational force \(\pmb{F}_{e,o}\), i.e. \[\pmb{F}_{t,o} = -\pmb{F}_{o,t} = -\pmb{F}_{e,o}\]

As we’ll see quickly, we don’t actually need to compute anything more, in this situation. We’ll nevertheless show how to compute the gravitational force in the next section, when considering the similar case of a mass attached to a vertical spring.
Applying Newton’s second law We can now use Newton’s second law, \(\sum_i \pmb{F}_i = m\pmb{a} = m\ddot{\pmb{x}}\) to deduce an equation of motion:

\[\begin{aligned} ~ && \pmb{F}_{e,o} + \pmb{F}_{t,o} + \pmb{F}_{s,o} &= m\ddot{\pmb{x}} \\ \Leftrightarrow && \pmb{F}_{e,o} - \pmb{F}_{e,o} + \pmb{F}_{s,o} &= m\ddot{\pmb{x}} \\ \Leftrightarrow && \pmb{F}_{s,o} &= m\ddot{\pmb{x}} \\ \Leftrightarrow && -k\pmb{x} &= m\ddot{\pmb{x}} \,\boxed{\,} \end{aligned}\]

Which is a second order, linear differential equation, homogeneous, with constant coefficients. It’s often rewritten, as is the case in the book, by defining \(\omega^2 = \dfrac{k}{m}\):

\[\boxed{\ddot{\pmb{x}} = -\omega^2\pmb{x}}\]

Gravitational force
To handle the case of a vertical spring, we won’t be able to brush the gravitational force away as we did previously, and we’ll need to compute it.
There are a few different ways to reach a formula for the gravitational force, but perhaps the simplest one, in the context of studying a mass attached to a vertical spring, would be to empirically study the fall of an object of mass \(m\) in Earth’s gravitational field.
One could proceed for instance by video-taping a vertically falling object of known mass \(m\). The video can then be discretized into a series of images, on which we could measure the evolution of the position of the mass over time. From there, we could compute the speed, as the variation of position, and the acceleration, as the variation of speed.
Then, neglecting the air resistance (well, we could perform the experiment in a vacuum), we can pretend that there’s a single force acting on the mass, which is our gravitational force.
The experiment could obviously be performed multiple times to improve the accuracy. In the end, physicists have found that Earth’s gravity is an acceleration vector \(\pmb{g}\) of intensity \(9.81\,m.s^{-2}\), directed towards the center of the Earth, which experimentally gives a gravitational force of: \[\pmb{F}_{e,o} = m\pmb{g}\]

Vertical spring
We know have all we need to study a variant of our previous setup, slightly more complicated mathematically, where a mass \(m\) is attached to the end side of a vertical spring of constant \(k\), itself firmly attached to a fixed element at its top:

We’ll choose the same convention regarding the origin of our coordinate system (see Remark \(\ref{rem:L03E04:origin}\)). Contrary to the previous case, there’s no other force to counterbalance the gravitational force \(\pmb{F}_{e,o}\); applying Newton’s second law gives: \[\begin{aligned} ~ && \sum_i \pmb{F}_i = \pmb{F}_{e,o} + \pmb{F}_{s,o} &= m\ddot{\pmb{x}} \\ \Leftrightarrow && m\pmb{g} - k\pmb{x} &= m\ddot{\pmb{x}} \\ \Leftrightarrow && \ddot{\pmb{x}} &= -\omega^2\pmb{x}+\pmb{g} \end{aligned}\] Again, with \(\omega^2 = \dfrac{k}{m}\); so the only difference is that we get an additional constant \(\pmb{g}\), but this still is a second-order linear differential equation with constant coefficients.

Verifying the solution
Solving differential equations, mathematically, can be rather involved. For such simple equations, mathematicians themselves may find it sufficient to make an educated guess by tweaking an exponential-based function, despite having explored the subject with much more depth.
We won’t discuss the details here, and will pragmatically satisfy ourselves with verifying that the proposed solution actually solves the equation: \[x(t) = A\cos\omega t + B\sin\omega t\]

And indeed, the proposed solution does solve the equation for the horizontal spring setup. \[\begin{aligned} \ddot x(t) &= -A\omega^2\cos\omega t - B\omega^2\sin\omega t \\ ~ &= -\omega^2 (A\cos\omega t + B\sin\omega t) \\ ~ &= -\omega^2 x(t) \,\boxed{\,} \end{aligned}\]

But it doesn’t solve the vertical setup; we’ll get there quickly, this isn’t much more complicated. You may want to try by yourself to see if you can tweak the solution of the homogeneous equation to solve the non-homogeneous one.

Initial conditions
Let’s start by computing the expression of the velocity: \[\dot x(t) = -A\omega\sin\omega t + B\omega\cos\omega t\] Then, at \(t=0\), we have: \[\begin{aligned} x(0) = && A\cos(0) + B\sin(0) &= \boxed{B} \\ \dot x(0) = && -A\omega\sin(0) + B\omega\cos(0) &= \boxed{A\omega}\,\boxed{\,} \end{aligned}\]

Equations and initial conditions for the vertical setup
There are general mathematical methods for reaching a solution to a non-homogeneous linear second-order differential equation with constant coefficient from a homogeneous, but again, suffice for us to verify that the following would work: \[x(t) = A\cos\omega t + B\sin\omega t + \frac{g}{\omega^2}\] Indeed: \[\begin{aligned} \ddot x(t) &= -A\omega^2\cos\omega t - B\omega^2\sin\omega t \\ ~ &= -A\omega^2\cos\omega t - B\omega^2\sin\omega t -\frac{\omega^2}{\omega^2}g + g\\ ~ &= -\omega^2 (A\cos\omega t + B\sin\omega t + \frac{g}{\omega^2}) + g \\ ~ &= -\omega^2 x(t) +g\,\boxed{\,} \end{aligned}\] From there, we can again express the initial position and speed as functions of \(A\) and \(B\), first by computing the expression of the speed: \[\dot x(t) = -A\omega\sin\omega t + B\omega\cos\omega t\] And then by looking at what’s happening at \(t=0\): \[\begin{aligned} x(0) = && A\cos(0) + B\sin(0)+\frac{g}{\omega^2} &= \boxed{B+\frac{g}{\omega^2}} \\ \dot x(0) = && -A\omega\sin(0) + B\omega\cos(0) &= \boxed{A\omega}\,\boxed{\,} \end{aligned}\]