Let's see how our Fourier series solution to the damped, driven oscillator works by putting our previous example to work.
Example: sawtooth driving force
Suppose we have a driving force which is described well by a sawtooth wave, the same function that we found the Fourier series for above:
\[ \begin{aligned} F(t) = 2F_0 \frac{t}{\tau},\ -\frac{\tau}{2} \leq t \leq \frac{\tau}{2}. \end{aligned} \]
What is the response of a damped, driven oscillator to this force?
Our starting point is finding the Fourier series to describe \( F(t) \), but we already did that: we know that \( a_n = 0 \) and
\[ \begin{aligned} b_n = \frac{2F_0}{\pi n} (-1)^{n+1}. \end{aligned} \]
Thus, applying our solution for the damped driven oscillator, we have for the particular solution
\[ \begin{aligned} x_p(t) = \sum_{n=0}^\infty B_n \sin(n \omega t - \delta_n) \end{aligned} \]
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 complementary 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.
The Fourier transform
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 \( \Delta/2 \). It's straightforward to find the Fourier series for this force, but we don't have to because Taylor already worked it out as his example 5.4; the result is
\[ \begin{aligned} a_0 = \frac{F_0 \Delta}{\tau}, \\ a_n = \frac{2F_0}{n\pi} \sin \left( \frac{\pi n \Delta}{\tau} \right), \end{aligned} \]
and all of the \( b_n \) coefficients are zero by symmetry, since the force is even.
Now, to apply this to the single impulse, we want to take the limit where \( \tau \gg \Delta \). This is straightforward to do in the Fourier coefficients themselves: we have
\[ \begin{aligned} a_n \approx \frac{2F_0 \Delta}{\tau}. \end{aligned} \]
But now we run into a problem when we try to actually use the Fourier series, which takes the approximate form
\[ \begin{aligned} F(t) \approx \frac{F_0 \Delta}{\tau} \left[1 + 2 \sum_{n=1}^\infty \cos \left( \frac{2\pi n t}{\tau} \right) \right]. \end{aligned} \]
The problem is that the usual factor of \( 1/n \) cancelled out, so the size of our terms is not dying off as \( n \) increases. So our usual approach of truncation won't be useful at all!
There's something else that is happening at the same time as we try to take \( \tau \rightarrow \infty \), which is equivalent to \( \omega \rightarrow 0 \). Remember that we're summing up distinct Fourier components with frequency \( \omega, 2\omega, 3\omega... \) But as \( \omega \) goes to zero, these frequencies are getting closer and closer together - and our sum is getting closer and closer to resembling a continuous integral!
This suggests that we try to replace the sum with an integral. We want to think carefully about the limit that \( \tau \rightarrow \infty \). Let's define the quantity
\[ \begin{aligned} \omega_n = \frac{2\pi n}{\tau}. \end{aligned} \]
As \( \tau \) becomes larger and larger, the interval between \( \omega_n \) is becoming smaller and smaller, to the point that we can imagine it becoming an infinitesmal:
\[ \begin{aligned} d\omega = \frac{2\pi}{\tau} \Delta n \end{aligned} \]
where \( (\Delta n) = 1 \) by definition, since we sum over integers, but it will be useful notation for what comes next. Let's take this and rewrite the Fourier sum slightly:
\[ \begin{aligned} F(t) = \sum_{n=0}^\infty (\tau a_n) \cos \left( \omega_n t \right) \frac{\Delta n}{\tau} \end{aligned} \]
All I've done is multiplied by 1 twice: first to introduce \( \Delta n \), which is 1, and second to multiply and divide by \( \tau \). But the key is that this is now a sum over a bunch of steps which are infinitesmally small, so as \( \tau \rightarrow \infty \), this is a proper Riemann sum and it becomes an integral. Using \( \Delta n / \tau = d\omega / (2\pi) \), we can write it as an integral over \( \omega \):
\[ \begin{aligned} F(t) = \frac{1}{2\pi} \int_0^\infty G(\omega) \cos(\omega t) d\omega. \end{aligned} \]
The function \( G(\omega) \) comes from the Fourier coefficients, specifically from the product \( \tau a_n \):
\[ \begin{aligned} G(\omega) = \lim_{\tau \rightarrow \infty} \tau \left( \frac{2}{\tau} \int_{-\tau/2}^{\tau/2} f(t) \cos \left( \omega t \right) dt \right) \\ = 2 \int_{-\infty}^\infty f(t) \cos(\omega t) dt. \end{aligned} \]
The function \( G(\omega) \) is known as the Fourier transform of \( F(t) \). Once again, just like the Fourier series, this is a representation of the function. In this case, there's no questions about infinite series or truncation; we're trading one function \( F(t) \) for another function \( G(\omega) \). There is also no restriction about periodicity - we can use the Fourier transform for any function at all, periodic or not.
One note: there are several equivalent but slightly different-looking ways to define the Fourier transform! One simple difference to watch out for is where the factor of \( 2\pi \) goes - it can be partly or totally moved into the definition of \( G(\omega) \) instead of being kept in \( F(t) \). More importantly, the cosine version that we're using is actually not as commonly used, in part because it can only work on functions that are even; in general we need both sine and cosine. A more common definition of the Fourier transform is in terms of complex exponentials:
\[ \begin{aligned} F(t) = \int_{-\infty}^\infty G(\omega) e^{i\omega t} d\omega \\ G(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty F(t) e^{-i\omega t} dt. \end{aligned} \]
Notice that in this case, unlike the Fourier series, there's a certain symmetry to the definitions of \( F(t) \) and \( G(\omega) \). This is because both representations are functions here, instead of trying to match a function onto a sum. But the symmetry of the definitions suggests that it should be just as good to go backwards, thinking of \( F(t) \) as a representation obtained from \( G(\omega) \).
The Dirac delta function
To understand inverting the Fourier transform, we can take these formulas and plug them in to each other. Let's use the complex exponential version:
\[ \begin{aligned} G(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty \left( \int_{-\infty}^\infty G(\alpha) e^{i\alpha t} d\alpha \right) e^{-i \omega t} dt \end{aligned} \]
(notice that I can't use \( \omega \) twice. In the definition of \( F(t) \), \( \omega \) is a dummy variable; we're integrating over it, so I can call it anything I want.) Now, I can do some rearranging of these integrals:
\[ \begin{aligned} G(\omega) = \int_{-\infty}^\infty G(\alpha) \left[ \frac{1}{2\pi} \int_{-\infty}^\infty e^{i(\alpha-\omega)t} dt \right] d\alpha \end{aligned} \]
Notice what is going on here; we have the same function \( G \) on both sides of the equation, but on the left it's at a single frequency \( \omega \), while on the right it's integrated over all possible frequencies. This is sort of familiar, actually; it resembles the orthogonality relation we had for the cosines and sines in the Fourier series, where the integral would be zero except for the special case where the two cosines match in frequency.
In line with the idea of orthogonality, the only way this equation can be true for all \( \omega \) and all \( G \) is if the object in square brackets is picking out only \( \alpha = \omega \) from under the integral, and discarding everything else! In other words, if we define
\[ \begin{aligned} \delta(\alpha - \omega) \equiv \frac{1}{2\pi} \int_{-\infty}^\infty e^{i(\alpha - \omega) t} dt, \end{aligned} \]
known as the Dirac delta function (although technically it's not a function - more on that shortly), then it has the property that
\[ \begin{aligned} \int_{-\infty}^\infty dx f(x) \delta(x-a) = f(a). \end{aligned} \]
The delta function resembles the Kronecker delta symbol, in that it "picks out" a certain value of \( x \) from an integral, which is what the Kronecker delta does to a sum. Note that we can put in any function we want, so if we use \( f(x) = 1 \), we get the identity
\[ \begin{aligned} \int_{-\infty}^\infty dx \delta(x) = 1. \end{aligned} \]
In other words, if we think of \( \delta(x) \) as a function on its own, the "area under the curve" is equal to 1. At the same time, we know that it is picking off a very specific value of \( x \) from any given function. Combining these two properties, if we try to plot \( \delta(x) \), it must look something like this:
or defining it as a "function",
\[ \begin{aligned} \delta(x) = \begin{cases} \infty, & x = 0; \\ 0, & x \neq 0. \end{cases} \end{aligned} \]
such that the integral over \( \delta(x) \) gives 1, as we found above. Once again, this isn't technically a function, so it's a little dangerous to try to interpret it unless it's safely inside of an integral. (If you're worried about mathematical rigor, it is possible to define the delta function as a limit of a regular function, for example a Gaussian curve, as its width goes to zero.)
From the plot, we can visually observe that it's important that the delta function spike is located within the integration limits, or else we won't pick it up and will just end up with zero. In other words,
\[ \begin{aligned} \int_{-1}^3 dx\ \delta(x-1) = 1 \end{aligned} \]
but
\[ \begin{aligned} \int_{-1}^3 dx\ \delta(x-4) = 0. \end{aligned} \]
There are some other useful properties of the delta function that we could derive (you can look them up in a number of math references, including on Wikipedia or Mathworld.) Let's prove a simple one, to get used to delta-function manipulations a bit. What is \( \delta(kt) \), where \( k \) is a constant real number? As always, to make sense of this we really need to put it under an integral:
\[ \begin{aligned} \int_{-\infty}^\infty f(t) \delta(kt) dt = ? \end{aligned} \]
We can do a simple \( u \)-substitution to rewrite:
\[ \begin{aligned} (...) = \int_{t=-\infty}^{t=\infty} f\left(\frac{u}{k} \right) \delta(u) \frac{du}{k} \\ = \frac{1}{k} \int_{-\infty}^\infty f\left( \frac{u}{k} \right) \delta(u) du \end{aligned} \]
assuming \( k>0 \) for the moment; we'll come back to the possibility of negative \( k \) shortly. Now, the fact that the argument of \( f \) has a \( 1/k \) in it doesn't matter, because when we do the integral we pick off \( f(0) \):
\[ \begin{aligned} (...) = \frac{f(0)}{k} = \int_{-\infty}^\infty f(t) \left[ \frac{1}{k} \delta(t) \right] dt. \end{aligned} \]
Now compare what we started with and what we ended with, and we get \( \delta(kt) = \delta(t)/k \). Again, if \( k<0 \) we get an extra sign flip, which would give us a minus sign - but \( -k \) if \( k \) is negative is just \( |k| \). So we have the full result
\[ \begin{aligned} \delta(kt) = \frac{1}{|k|} \delta(t). \end{aligned} \]
Clicker Question
What is the result of the integral
\[ \begin{aligned} \int_{-\infty}^\infty x^2 \delta(2-x) dx = ? \end{aligned} \]
A. 2
B. -2
C. 4
D. -4
E. \( \infty \)
Answer: C
The careful way to do this is once again a \( u \)-substitution. If we let \( u = 2-x \), then \( du = -dx \) and we have
\[ \begin{aligned} \int_{-\infty}^\infty x^2 \delta(2-x) dx = \int_{+\infty}^{-\infty} (2-u)^2 \delta(u) (-du) \end{aligned} \]
being careful to plug in \( u \) instead of \( x \) for the limits of integration, which flips their signs. The \( \delta(u) \) will set \( u=0 \), leaving us with
\[ \begin{aligned} (...) = 2^2 = 4, \end{aligned} \]
the positive version since the minus sign in \( -du \) cancels the extra minus sign from flipping the limits of integration back around.
We could have seen this answer coming without going through all the math if we think visually. We know that \( \delta(x) \) is always positive, so \( \delta(2-x) \) has to actually be the same thing as \( \delta(x-2) \), at which point we use the defining identity to just plug in \( 2^2 = 4 \). Since \( \delta(x-2) \) and \( x^2 \) are both positive, we can also easily see that the integral has to be positive, ruling out the negative-sign answers.
One more comment on the delta function, bringing us back towards physics. We use the delta function often to represent "point particles", which we imagine as being concentrated at a single point in space - not totally realistic, but often a useful approximation. In fact, one example where this is realistic is if we want to describe the charge density of a fundamental charge like an electron. Suppose we have an electron sititng on the \( x \)-axis at \( x=2 \). The one-dimensional charge density describing this would be
\[ \begin{aligned} \lambda(x) = -e \delta(x-2). \end{aligned} \]
So the electron has infinite charge density at point \( x=2 \), and everywhere else the charge density is zero. This might seem unphysical, but it's the only consistent way to define density for an object that has finite charge but no finite size. If we ask sensible questions like "what is the total charge", we get good answers back:
\[ \begin{aligned} Q = \int_{-\infty}^\infty \lambda(x) dx = -e \int_{-\infty}^\infty \delta(x-2) dx = -e \end{aligned} \]
which is exactly the answer we expect! One final thing to point out here: if we think about units, 1-d charge density should have units of charge/distance, which means the units of the delta function are inverse to whatever is inside of it. This makes sense, because of the identity
\[ \begin{aligned} \int dt \delta(t) = 1. \end{aligned} \]
Since \( dt \) has units of time, \( \delta(t) \) here has to have units of 1/time.
Fourier transforms and solving the damped, driven oscillator
Let's go back to our non-periodic driving force example, the impulse force, and apply the Fourier transform to it. Recall that our function for the force is
\[ \begin{aligned} F(t) = \begin{cases} F_0, & t_0 \leq t < t_0 + \tau, \\ 0, & {\rm elsewhere} \end{cases} \end{aligned} \]
The Fourier transform of this function is very straightforward to find. Since this isn't even around \( t=0 \), let's use the most general complex exponential form:
\[ \begin{aligned} G(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-i\omega t} F(t) dt \\ = \frac{F_0}{2\pi} \int_{t_0}^{t_0 + \tau} e^{-i\omega t} dt \\ = \left. \frac{-F_0}{2\pi i \omega} e^{-i \omega t} \right|_{t_0}^{t_0 + \tau} \\ = -\frac{F_0}{2\pi i \omega} \left[ e^{-i\omega (t_0+\tau)} - e^{-i\omega t_0} \right] \\ = -\frac{F_0}{2\pi i \omega} e^{-i\omega(t_0 + \tau/2)} \left[ e^{-i\omega \tau/2} - e^{i \omega \tau/2} \right] \\ = \frac{F_0}{\pi \omega} e^{-i\omega(t_0 + \tau/2)} \sin (\omega \tau/2). \end{aligned} \]
What good does this do us? Let's recall the original equation that we're trying to solve,
\[ \begin{aligned} \ddot{x} + 2\beta \dot{x} + \omega_0^2 x = \frac{F(t)}{m}. \end{aligned} \]
Now, although \( F(t) \) and \( G(\omega) \) are equivalent - they carry equal information about a single function - we can't just plug in \( G(\omega) \) on the right-hand side, because to get that from \( F(t) \) we have to do a Fourier transform, i.e. take an integral over \( t \). That means we have to Fourier transform both sides of the equation. Let's define the Fourier transform of \( x \) as well,
\[ \begin{aligned} \tilde{x}(\alpha) = \frac{1}{2\pi} \int_{-\infty}^\infty x(t) e^{-i\alpha t} dt \end{aligned} \]
What happens if I have derivatives of \( x(t) \) under the integral sign? We can use integration by parts to move the integral over, and it's not difficult to prove the general result
\[ \begin{aligned} \tilde{x^{(n)}}(\alpha) = (i \alpha)^n \tilde{x}(\alpha), \end{aligned} \]
i.e. the Fourier transform of a time derivative of \( x(t) \) is just \( i\alpha \) times the Fourier transform of the original. Now we know enough to take the Fourier transform of both sides above:
\[ \begin{aligned} (-\alpha^2 + 2i \beta \alpha + \omega_0^2) \tilde{x}(\alpha) = \frac{G(\alpha)}{m} \end{aligned} \]
(where the same frequency \( \alpha \) appears on both sides, since we're applying the same Fourier transform to both sides!) This is a very powerful result, because there is no differential equation left at all: we can simply write down the answer,
\[ \begin{aligned} \tilde{x}(\alpha) = \frac{G(\alpha)/m}{-\alpha^2 + 2i \beta \alpha + \omega_0^2} \end{aligned} \]
and take one more Fourier transform to get the time-domain solution,
\[ \begin{aligned} x(t) = \int_{-\infty}^\infty \tilde{x}(\alpha) e^{i\alpha t} d\alpha = \int_{-\infty}^\infty \frac{G(\alpha)/m}{-\alpha^2 + 2i \beta \alpha + \omega_0^2} e^{i\alpha t} d\alpha. \end{aligned} \]
This is a powerful and general way to deal with certain linear, non-homogeneous ODEs - you can see immediately that this will extend really nicely even to higher than second order, since we still get a simple algebraic equation! This method is sometimes referred to as "solving in frequency space", because we transform from considering time to frequency using the Fourier transform and the equation simplifies drastically.
The bad news is that even for a relatively simple driving force like our impulse, this integral is a nightmare to actually work out! Evaluating difficult integrals like this, particularly ones where complex numbers are already involved, is a job for the branch of mathematics known as complex analysis. Such methods can be combined with another method called the method of Green's functions, which uses the delta function to allow us to do the difficult integral just once, and end up with simpler integrals to describe solutions with more complicated driving forces. Both of these powerful methods are beyond the scope of this class, but they are widely used and important in many physics systems, so you will likely encounter them sooner or later in other physics or engineering classes!