Reading the content in the harmonic oscillator solution


Any physical system will oscillate if it has a stable point, develops restoring forces that push it back towards the stable point when it moves away, and has inertia that carries it through the stable point if it moves toward the stable point. For small oscillations, this oscillation will be described by the same equations that govern the harmonic oscillator. This makes the equation that describes the solution to the motion of a mass attached to a spring incredibly important. Like many of our equations in physics, there is a lot of conceptual information coded into a complicated looking equation. Let's pull it apart to see what it's telling us.

For a system moving with harmonic motion (no damping, restoring force proportional to the displacement from equilibrium) the solution for the motion as a function of time looks like this:

$$x(t) = A \cos{(\omega_0 t - \phi)}$$

Although this is a fairly short equation, there are 7 distinct symbols, each telling us something different about the solution. Let's go through them one at a time.

1.$x$ — The dependent variable: The symbol on the left represents what the equation is calculating: the displacement from equilibrium. Note that this is NOT necessarily the position of an object. For our mass on a spring example it was convenient to choose it as such, but for our hanging mass on a spring we might have preferred to choose the position coordinate as the distance of the mass from the ground (and sonic ranger detector). Then, the "$x$" here would be the position shifted to make it 0 at the equilibrium point.

A second tricky issue is that "$x$" doesn't have to represent a distance at all. In describing the small oscillations of the pendulum we could describe it by a position coordinate — the distance deviated left and right from the hanging-straight-down line — or by the angle. In the latter case, "$x$" would actually represent the angle "$θ$". In other examples, such as oscillations in an electrical network, it might not represent a distance at all, but an electric current, $I$.

2. $t$ — The independent variable: This symbol represents the time, expressing the fact that we are considering the oscillation as occurring in time. (When we get to waves, we'll also find oscillations that occur in space.)

3. (..) — Parentheses: It might seem trivial to call out the parentheses in this formula as a symbol carrying conceptual information. But parentheses in a math equation can have a variety of very distinct meanings. It can simply mean grouping, as in the basic algebraic equation a(b+c) = ab + bc, but in our case here it represents functional dependence, as in the math equation $y = f(x)$. That equation says that the dependent variable y is a function of an independent variable $x$. In the harmonic oscillator solution, $x$ is our dependent variable and $t$ is our independent variable. This therefore says that time runs, we can choose any value of the time, and when we do, our equation tells us where the object will be at that time (what the value is of $x(t)$).*

4. $A$ — The amplitude: Our solution is proportional to a cosine function. But while this is oscillatory, and gives us the oscillation, a cosine is a unitless ratio. Therefore, to get the right dimensions in our equation — whatever dimensions "$x$" has — we have to multiply by something with the dimensions of $x$. Since we know that cosine oscillates between the values of +1 and -1, this parameter tells us the magnitude of the oscillation. As cosine goes between +1 and -1, $x$ goes between $+A$ and $-A$.

5. cos — The oscillating factor: The cosine function is the trig function that provides the oscillation. One way to see this is to think about a point moving on the unit circle in the x-y plane. Its coordinates are $(x,y) = (\cos θ, \sin θ)$. If $θ$ grows at a uniform rate, the point will move around the circle. Its x coordinate will go back and forth between +1 and -1 like $\cos θ$. We could also have chosen sin instead of cos or a linear combination of both. (Our inclusion of a phase shift allows us to transform our equation into either sin or expand it out.) The choice of cos is convenient if you start with a maximum displacement. (Then the phase shift is 0.)

6. $ω_0$ — The angular frequency: The cosine is a function of an angle — a dimensionless (though not a unitless) quantity. This means that we can't just take "$\cos{(t)}$" since different folks evaluating it might get different answers if they chose different units for $t$ (like seconds or hours). We fix this by multiplying t by a constant with dimensions of [$ω_0$] = 1/T. This makes $ω_0t$ dimensionless. Any evaluator plugging in numbers will get the same result (as long as they use consistent units).

This parameter combines the physics of the restoring force that brings the oscillator back towards the equilibrium point and the inertia that keeps it moving through that point even though there is no force there. For our mass on a spring, we get

$$\omega^{\mathrm{mass\;on\;a\;spring}}_0 = \sqrt{\frac{k}{m}}$$

In the pendulum case, we get

$$\omega^{\mathrm{pendulum}}_0 = \sqrt{\frac{g}{L}}$$

The ratio under the square root is also really a force-per-unit-length parameter, $mg/L$, (like $k$), divided by a mass parameter $m$. The fact that the mass cancels is special to the case of gravity.

7. φ — The phase shift: The function cos has its maximum value when its argument is 0. This might not be the case for our physical system. For example, if we want to start a pendulum swinging, we can pull it back to some starting angle and then release it. If we set the 0 of our time when we release it, then it will have a maximum at that time and cos(ωt) works just fine. But if instead of pulling it back, we give it a quick tap, while it's at the bottom, then our starting condition will be that it starts at position 0 but with a non-zero velocity. The function cos doesn't work for that (though sin does). What the parameter $φ$ is doing is essentially shifting the function cos so it has a different starting time than what was chosen for $t= 0$. You can see this by writing $φ = ω_0t_0$. Then we see that the argument of cos looks like $\cos{(ω_0t- ω_0t_0)} = \cos{(ω_0(t-t_0))}$ showing us that choosing to put in a $φ$ is like saying that the maximum of the oscillation is occurring at a time $t_0 = φ/ω$.

When does this equation hold?

The equation at the top of this page is the solution to the simple harmonic oscillator equation —

$$\frac{d^2x}{dt^2} = -\omega_0^2 x$$

where $x$ is our dependent variable (a function of time), $t$ is our independent variable, and $ω_0$ is a constant (parameter). Our solution is the most general solution to this equation. Any of the possible solutions (corresponding to different starting conditions) can be gotten by appropriate choices of $A$ and $φ$. We get an equation like this out of Newton's second law for a mass experiencing a net force that

  • has a stable equilibrium point (where the force is zero),
  • has a restoring net force (pushes back towards the equilibrium point when displaced away from it),
  • the net force can be approximated by Hooke's law (proportional to the displacement), and
  • resistive forces can be ignored. 

Of course this situation is an idealized model that never truly holds perfectly, but it is very good for almost any oscillating system for distances where the displacement is small so the force can be approximated by a linear function (straight line) and for times that are short enough that any resistive forces can be ignored. This turns out to be true for a lot of important situations! More than that, any system that can be modeled by the second derivative of a function is proportional to to negative of the value of the function (and there are many such cases) will have a solution that looks exactly like the one we have discussed here. (If the math is the same, the analogy is good!)

* This will become of particular importance when we go on to study waves since then, for example, for waves on a spring in one dimension, $x$ becomes an independent variable (which bit of spring are we choosing to look at) and our dependent variable, $y$, depends on both $x$ and $t$.

Joe Redish 4/6/16

Article 666
Last Modified: May 16, 2019