[Mathematics][MIT 18.03] Detailed Explanation of the Frequency Problems in Second-Order Differential Equation of Oscillation System

　　Well, to begin with, I'd like to say thank you to MIT open courses twice. It's their generosity that gives me the priviledge to enjoy the most outstanding education resources.

　　Okay, come to the point -- The Frequency in Oscillation System.

　　In general, there are two cases, undamped and damped. Let's begin with the first and easier one.

　　Case 1. Undamped Oscillation System.

　　　　In this system, the general form of the equation is $y'' + \omega_{0}^{2}y=f(x)$, in which $f(x)$ refers to the input or, more practically, driving or forcing term.

　　　　In this case, there is only one frequency: The Natural Frequency of this system $\omega_{0}$, AND if $f(x)$ is a trigonometric function, to maximize the amplitude or give rise to authentic resonance, we just need to let the frequency of $f(x)$ equals to $\omega_{0}$.

　　　　Okay, let's do the inferences. First of all, if we want to get the natural frequency, we have to come back to the reduced or associated homogeneous equation(just let $f(x)$ equal to $0$).

　　　　So, we get $y'' + \omega_{0}^{2}y=0$. It's easy to solve by set $y = e^{rt}$, in which $t$ is the usually independent variable, specifically time in practical usage, and $r$ is the unknown constant.

　　　　This process is quite simple. So I just omit it and present the result. We finally get $y = c_{1}\cos\omega_{0} t + c_{2}\sin\omega_{0} t$. Let's choose a simple particular solution.

　　　　$y = \cos\omega_{0} t$

　　　　So, coming to this step, it's quite clear to see that this system possess a natural frequency $\omega_{0}$.

　　　　But, further, with the driving term $\cos\omega t$, the solution will be the real part of $\displaystyle \tilde{y} = \frac{e^{i\omega t}}{\omega_{0}^{2}-\omega^{2}}$. And, by some analysis, we will get the conclusion that the resonance will occur if $\omega \to \omega_{0}$.

　　Case  2. Damped Oscillation System.

　　　　In the same way, the general form of the equation is $y'' + 2py' + \omega_{0}^{2}y = f(x)$, in which $f(x)$ refers to the input or, more practically, driving or forcing term and $p \ne 0$.

　　　　In this case, there exists three frequency: The Natural Frequency of the associated undamped oscillation system $\omega_{0}$, The Damping Frequency, which is a pseudo frequency because actually the solution is not a pure trigonometric function, $\omega_{1}$, which equals to $\sqrt{\omega_{0}^{2}-p^{2}}$, The Frequency which driving force needs to maximize its pseudo amplitude or give rise to pseudo resonance, $\omega_{r}$, which equals to $\sqrt{\omega_{0}^{2}-2p^{2}}$ or $\sqrt{\omega_{1}^{2}-p^{2}}$.

　　　　Okay, so let's begin with the simplest one, the natural frequency. Just delect the first-order term and remove the $f(x)$ and according to the case 1, we will get the result. That's simple, so it does not deserve detailed explanation.

　　　　Secondly, the damping frequency. Above all, I have to clarify that in order to generate oscillation, this system has to be underdamped, which means that $p^{2}-\omega_{0}^{2}<0$, and we need to go back to the associated homogeneous equation too. So, by using the characteristic function, we get the roots are $-p \pm \sqrt{\omega_{0}^{2}-p^{2}}i$. That's why we set $\omega_{1}=\sqrt{\omega_{0}^{2}-p^{2}}$. As a result, we will finally get one simple particular solution $y = e^{-pt}\cos\omega_{1}t$. And, that's pseudo frequency because despite the behaviors it possesses, like frequency, period, oscillation, its amplitude actually fades with time. So the oscillation actually decays with time.

　　　　Thirdly, we come to the most difficult part, which frequency maximizes its pseudo amplitude or gives rise to pseudo resonance. Above all, by same practice, we set $f(x) =\cos\omega t$ and assume that $p(\alpha) \ne 0$. (NOTE: 1. $p(\alpha)$ means we substitude every $D$(the note of differentiation) in the polynomial with $\alpha$, in this specific case, the polynomial is $D^{2}+2pD+\omega_{0}^{2}$ 2. $\alpha$ is the complex index number of the exponential form of driving term after we complexify the $f(x)$.) So we get the complex solution $\displaystyle \tilde{y} = \frac {e^{i\omega t}}{-\omega^{2}+\omega_{0}^{2}+2p\omega i}$.

　　　　Above all, we need to simplify the problem.

　　　　Step 1. We need to do some fixes on the denominator. $\omega_{0}^{2}+2p\omega i - \omega^{2} = \omega_{1}^{2}+p^{2}+2p\omega i-\omega^{2}$. Let's say $\bar{\omega}^{2} = \omega_{1}^{2}+p^{2}-\omega^{2}$(NOTE: bar here doesn't mean conjugation). So we have $\displaystyle \tilde{y} = \frac {e^{i\omega t}}{\bar{\omega}^{2}+2p\omega i}$.

　　　　Step 2. Extract the real part. $\displaystyle \tilde{y} = \frac {e^{i\omega t}}{\bar{\omega}^{2}+2p\omega i}=\frac {e^{i\omega t}(\bar{\omega}^{2}-2p\omega i)}{\bar{\omega}^{4}+4p^{2}\omega^{2}}$, so $\displaystyle y = \frac {\bar{\omega}^{2}}{\bar{\omega}^{4}+4p^{2}\omega^{2}}\cos\omega t + \frac{2p\omega}{\bar{\omega}^{4}+4p^{2}\omega^{2}}\sin\omega t$.

　　　　Step 3. So the amplitude is $\displaystyle \sqrt{(\frac {\bar{\omega}^{2}}{\bar{\omega}^{4}+4p^{2}\omega^{2}})^{2} + (\frac{2p\omega}{\bar{\omega}^{4}+4p^{2}\omega^{2}})^{2}}$, which is $\displaystyle \frac{1}{\sqrt{\bar{\omega}^{4} + 4p^{2}\omega^{2}}} = \frac{1}{\sqrt{(\omega_{1}^{2}+p^{2}-\omega^{2})^{2} + 4p^{2}\omega^{2}}}$.

　　　　Step 4. In order to get the maximize amplitude, we need to minimize the denominator. And using some senior high school's knowledge, this would be $\omega^{2} = \omega_{1}^{2}-p^{2}$. And that is $\omega_{r}$.

　　　　And there remains an assumption to deal with. $p(\alpha) = 0$, is this possible? Let us see the denominator $\omega_{1}^{2}+p^{2}+2p\omega i-\omega^{2}$. If it's zero, then $p = 0$ or $\omega = 0$. Based on the condition, we know that $p \ne 0$ and $\omega = 0$ is meaningless. So we can safely exclude this case.