Modelling Forced Oscillations with a Single Trigonometric Function

Question

In a physics book I came across the following solution to a differential equation modeling forced, damped, oscillating motion: $$x(t)=A\cos(\omega t+\phi)$$ where $$A=\frac{F_0}{\sqrt{m^2\left(\omega^2-\omega_0^2\right)^2+b^2\omega^2}}$$ The differential equation which prompted this solution is $$-kx-b\dot{x}+F_0\sin(\omega t)=m\ddot{x}$$

This doesn't feel quite intuitively right, however. When the angular frequency $\omega$ of the thing which is causing the forced oscillating is equal to the natural angular frequency $\omega_0$ of the system it is causing an oscillation in, this seems to work intuitively. But when this isn't the case, it seems like $x(t)$ cannot be modeled as a single trig function, but rather a combination of multiple.

For example, in this video at 1:26, the frequency $\omega$ of the vibration generator is set to 2Hz, which is different than the natural frequency $\omega_0$ of the system (this turns out to be $\approx3.33$Hz) and the motion of the mass does not appear to resemble a single trig function, but rather multiple.

Here is my best attempt to graph the seeming motion at 1:26:

$$f(x)=\cos(x)-\cos(2x)$$

My question: Is the formula incorrect/a simplified version? If so, what is the correct one? If not, where am I going wrong here?

Answer 1

The solution of a forced and damped harmonic oscillator $$m\ddot{x}=-kx - b\dot{x} +F_0 \cos\omega_{\text{d}} t$$ is actually the sum of steady-state part $x_{\text{s}}$ and a transient part $x_{\text{t}}$ $$x = x_{\text{s}} + x_{\text{t}}$$ where $$x_{\text{s}} \sim A\cos(\omega t+\phi) \\ x_{\text{t}} \sim A' e^{-\gamma t} \sin (\omega' t+\phi')$$

The transient part depends on the initial conditions and both parts are present at the start. As time goes on, the transient part decays exponentially, leaving only the steady-state solution, which is the one you quoted. The full equations can be found here. Therefore, at the beginning, the motion will not be a single pure trig function but rather a sum of two trig functions.

Answer 2

When watching the video (beginning at around 0:35) carefully, we can see that the driving element (at the bottom of the device) does not move in a purely sine oscillation. It moves more like this:

The driving movement is still a periodic, but not a sine-like function.

You can decompose this oscillation (like every periodic motion) into a Fourier series of several pure sine oscillations.

In this case the Fourier decomposition is roughly like this: $$F(t)= F_0( \sin(\omega t) + 0.33\sin(3\omega t) + 0.2\sin(5\omega t) + ...)$$

You see, the movement contains not only the base frequency $\omega$, but also higher frequencies (which are multiples of the base frequency).

To convince you of the validity of the Fourier decomposition, here is a plot of $F(t)$ with only the first 3 components of the series being considered. When considering even more higher frequency components, then the series would approximate the function from the first image even better.

_{(image created by FooPlot)}

Then the solution of the differential equation $$-kx-b\dot{x}+F(t)=m\ddot{x}$$

will be $$\begin{align} x(t) &= A(\omega)\cos(\omega t+\phi(\omega)) \\ &+ 0.33 A(3\omega)\cos(3\omega t+\phi(3\omega)) \\ &+ 0.2 A(5\omega)\cos(5\omega t+\phi(5\omega)) \\ &+ ... \end{align}$$

where $A(\omega)$ is still the expression as given in your question and $\phi(\omega)$ is another expression which can also be calculated from the differential equation. It is no surprise, that the response $x(t)$ contains not only the base frequency $\omega$, but also the higher frequencies. Especially, when one of these higher frequencies ($n\omega$) happens to be near to the resonance frequency ($\omega_0$), then the component with $A(n\omega)$ will be quite large. This is exactly what you could observe in the video.