Let's continue the example of a sawtooth driving force for a damped, driven oscillator. As written down last time, the particular solution is
\[ \begin{aligned} x_p(t) = \sum_{n=0}^\infty B_n \sin(n \omega t - \delta_n) \end{aligned} \]
where
\[ \begin{aligned} B_n = \frac{2(-1)^{n+1} F_0/m}{\pi n\sqrt{(\omega_0^2 - n^2 \omega^2)^2 + 4\beta^2 n^2 \omega^2}}, \\ \delta_n = \tan^{-1} \left( \frac{2\beta n\omega}{\omega_0^2 - n^2 \omega^2} \right). \end{aligned} \]
If we want the full solution, we add in whatever the corresponding particular solution is. For example, if the system is underdamped (\( \beta < \omega_0 \)), then
\[ \begin{aligned} x(t) = A e^{-\beta t} \cos (\sqrt{\omega_0^2 - \beta^2} t - \delta) + x_p(t) \end{aligned} \]
with \( A \) and \( \delta \) determined by our initial conditions. This solution is nice and easy to write down, but very difficult to work with by hand, particularly if we want to keep more than the first couple of terms in the Fourier series! To make further progress, let's pick some numerical values and use Mathematica to make some plots. Let's keep it underdamped so we can think about resonance; I'll take \( \omega_0 = 1 \) to start and \( \beta = 0.2 \). Let's also set \( F_0/m = 1 \) for simplicity. Finally, we'll consider two values of the driving frequency \( \omega \). Let's begin with a higher driving frequency than the natural frequency, \( \omega = 3 \):
(Note that this is only the particular solution \( x_p(t) \), so this is the behavior at some amount of time after \( t=0 \) when the transient solution is gone.) Here I'm keeping \( M=20 \) terms in the Fourier series. This looks surprisingly simple given the complex shape of the driving force! Why is that? We can find a clue if we look in "frequency space" once again, i.e. if we plot the ampltiudes \( B_n \) of each of the Fourier coefficients:
So the reason that the long-time response is so simple-looking is because there is a single Fourier component which is completely dominant; the first component, \( B_1 \). This comes in part from the Fourier series for the sawtooth wave force itself; we know the coefficients \( b_n \) of the force die off proportional to \( 1/n \). But there's another effect which is just as important. Remember that the Fourier series for the response is a sum of contributions at frequencies \( n \omega \), i.e. at integer multiples of the fundamental driving frequency. Going back to our discussion of resonance, let's think about where the first few frequencies lie on the resonance curve \( A(\omega) \):
The black curve shows the resonance curve that we've considered before for simple sine-wave inputs. But now from the sawtooth force, we have multiple terms at \( \omega, 2\omega, 3\omega... = 3, 6, 9... \) that all contribute. But since we start at \( \omega \) much larger than \( \omega_0 \), all of the contributions from this sawtooth force are out on the tail of the resonance curve, which gives an additional suppression of the higher-frequency terms. In fact, if we go back to the formula we found for \( B_n \), we see that \( B_n \propto 1/n^2 \) for \( n \omega \gg \omega_0 \).
This is a nice picture, and it should make you suspect that we will get a dramatically different result if we pick a driving frequency which is lower than \( \omega_0 \). Here's our solution once again, with all of the same physical inputs except that \( \omega = 0.26 \)/s:
Now we can see some remnant of the sawtooth shape here in the response, although it's far from perfect - we see especially complicated behavior at the ends of the cycle, where the force direction changes suddenly. As you can probably guess already, there are many more significant Fourier modes in the particular solution here. Here's a plot of \( B_n \) for this case:
This time, several low-lying modes are important - in fact, the first few modes after \( B_1 \) are larger than they are in the sawtooth wave itself! Once again, we can easily understand what is happening in terms of the resonance curve:
Since the fundamental driving force \( \omega < \omega_0 \) this time, we see that the first few multiples actually have their amplitudes enhanced by the resonance at \( \omega \approx \omega_0 \). This causes them to die off more slowly than the \( 1/n \) coming from the input force itself. Eventually, the higher frequencies move past the resonance, and we find the same \( 1/n^2 \) dependence as in the previous case. But if we had an even lower driving frequency, we might see a large number of Fourier modes contributing significantly.
The simple take-away message is that, for an underdamped oscillator, the system "wants" to resonate at the natural frequency \( \omega_0 \). If we drive at a lower frequency than \( \omega_0 \), we will tend to get several Fourier modes that contribute in the long-term behavior, and an interesting and complicated response. If instead we try to drive at a higher frequency than \( \omega_0 \), the response dies off very rapidly as frequency increases, so we get a response dominantly at the driving frequency itself. (Notice also that this response is much smaller in amplitude than the one that picks up the resonance - compare the amplitudes on the two \( x(t) \) plots above!)
Non-periodic forces and the driven oscillator
Clearly, Fourier series are a very powerful method for dealing with a wide range of driving forces in the harmonic oscillator. However, we are still restricted to considering only periodic driving forces. This is useful for a number of physical systems, but in the real world there are a larger number of examples in which the driving force is not periodic at all.
In fact, a simple example of a non-periodic driving force that has many practical applications is the impulse force:
This force is zero everywhere, except for a small range of time extending from \( t_0 \) to \( t_0 + \tau \), where \( \tau \) is the duration of the step, over which it has magnitude \( F_0 \). This is a decent model for most forces that you would apply yourself, without any tools: kicking a ball, or pushing a rotor into motion.
The easiest way to solve this problem is to break it into parts. Let's suppose that we begin at rest, so \( x(0) = \dot{x}(0) = 0 \). Let's also assume the system is underdamped - as usual, this gives the most interesting motion, since overdamped systems tend to just die off exponentially no matter what. Let's analyze things based on the regions, marked I, II, III on the sketch. For \( t < t_0 \), we have the general solution
\[ \begin{aligned} x_I(t) = A e^{-\beta t} \cos (\omega_1 t - \delta) \end{aligned} \]
with \( \omega_1 = \sqrt{\omega_0^2 - \beta^2} \) as usual. Since the force is zero up until \( t=t_0 \), in this region there is no particular solution. Let's apply our initial conditions:
\[ \begin{aligned} x_I(0) = A \cos (-\delta) = 0 \\ \dot{x}_I(0) = -\beta A \cos(-\delta) - \omega_1 A \sin(-\delta) = 0 \end{aligned} \]
Now, to solve the first equation we either have \( A = 0 \) or \( \delta = \pi/2 \). But if we try to use \( \delta = \pi/2 \), the second equation becomes \( \omega_1 A = 0 \) anyway. So in this region we just have \( A = 0 \), and there is no motion at all - not surprising, since there is no applied force and the oscillator starts at rest!
Next, let's consider what happens during the step, for \( t_0 < t < t_0 + \tau \). Here we have a constant driving force, which as we found before gives us a particular solution which is also constant, so we now have a slightly different general solution,
\[ \begin{aligned} x_{II}(t) = A' e^{-\beta t} \cos (\omega_1 t - \delta') + \frac{F_0}{m\omega_0^2}. \end{aligned} \]
Now, we can think of applying our initial conditions. Since our region starts at \( t=t_0 \), the initial condition is actually a boundary condition; we impose the requirement that the final position and speed of \( xI(t) \) have to match the initial position and speed of \( x{II}(t) \). Mathematically, we write this as
\[ \begin{aligned} x_I(t_0) = x_{II}(t_0) \\ \dot{x}_I(t_0) = \dot{x}_{II}(t_0) \end{aligned} \]
Now, since \( xI(t) \) is just zero everywhere, this turns out to be the same initial conditions that we started with. But I'm writing this out carefully now, because we'll need to match the boundary conditions again at the right-hand side to find \( x{III}(t) \). Let's apply the boundary conditions:
\[ \begin{aligned} x_{II}(t_0) = 0 = A' e^{-\beta t_0} \cos (\omega_1 t_0 -\delta') + \frac{F_0}{m\omega_0^2} \\ \dot{x}_{II}(t_0) = 0 = -\beta A' e^{-\beta t_0} \cos (\omega_1 t_0 - \delta') - \omega_1 A' e^{-\beta t_0} \sin(\omega_1 t_0 - \delta') \end{aligned} \]
This time, the presence of the force in the first equation means \( A' = 0 \) won't work. So taking the second equation and cancelling off the \( A' \), we have
\[ \begin{aligned} \tan(\omega_1 t_0 - \delta') = -\frac{\omega_1}{\beta} \end{aligned} \]
which gives us the phase shift \( \delta' \). To find the amplitude, we need the cosine of this angle to plug in to the first equation. Let's use the trick that I showed on the homework of drawing a "reference triangle":
from which we can immediately see that
\[ \begin{aligned} \cos (\omega_1 t_0 - \delta') = \frac{\beta}{\omega_0} \end{aligned} \]
and so the other boundary condition equation gives us
\[ \begin{aligned} 0 = A' e^{-\beta t_0} \frac{\beta}{\omega_0} + \frac{F_0}{m \omega_0^2} \\ A' = -\frac{F_0}{m \beta \omega_0} e^{\beta t_0} \end{aligned} \]
So, our general solution in region II looks like
\[ \begin{aligned} x_{II}(t) = \frac{F_0}{m \omega_0^2} \left[ 1 - \frac{\omega_0}{\beta} e^{-\beta(t-t_0)} \cos (\omega_1 t - \delta') \right]. \end{aligned} \]
If there was no region III, i.e. if the force just switched on and stayed on at \( t_0 \), we see that at long times the second term dies off, and we're just left with the first term, which is exactly the simple solution in the presence of a constant force. Now, let's confront the last boundary. In region III, after \( t=t_0 + \tau \), the force is gone once again and we have simply the underdamped solution,
\[ \begin{aligned} x_{III}(t) = A'' e^{-\beta t} \cos(\omega_1 t - \delta''). \end{aligned} \]
Matching solutions between regions II and III, we must have
\[ \begin{aligned} x_{II}(t_0 + \tau) = x_{III}(t_0 + \tau) \\ \dot{x}_{II}(t_0 + \tau) = \dot{x}_{III}(t_0 + \tau). \end{aligned} \]
If we just try to plug in directly, we'll get a really messy pair of equations to deal with! In fact, it's not really worth solving the fully general case here. Instead, we can make a simplifying assumption: let's suppose that \( \tau \) is very short compared to the other timescales in the problem, i.e. assume that \( \beta \tau \ll 1 \) and \( \omega_1 \tau \ll 1 \). In other words, we assume that the force is applied for a very brief window of time. We can then series expand at small \( \tau \) to simplify our equations greatly.
This will still lead to some very messy algebra, which I'll skip past; you know how to set it up at this point. If we slog through it all, we will find a relatively simple final result:
\[ \begin{aligned} x_{III}(t) \approx \frac{F_0 \tau}{m \omega_1} e^{-\beta(t-t_0)} \sin (\omega_1 (t-t_0)). \end{aligned} \]
for \( t > t_0 \). We might have guessed at the qualitative behavior: if we hit a damped oscillator, it will "ring" and start to oscillate, but the oscillations will die off quickly because of the damping.
You might notice that all of this was much more complicated than our Fourier series approach for periodic forces. Unfortunately, the impulse force isn't periodic, so there's no way we can use our Fourier series technique - it only happens once. But what if there was a way we could use the Fourier series anyway?
Consider the following periodic force, which resembles a repeated impulse force:
Within the repeating interval from \( -\tau/2 \) to \( \tau/2 \), we have a much shorter interval of constant force extending from \( -\Delta/2 \) to \( \Delta/2 \). Next time, we'll consider this example in detail.