So I was looking through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this complex analysis video by Math$505$ which asked us to solve the damped harmonic oscillator equation $\color{black}{\ddot{x}+2\gamma\dot{x}+w_0^2x=0}$ using Laplace Transform, which I thought that I might be able to do. Here is my attempt at solving the damped harmonic oscillator equation using Laplace Transform:
$$\mathbf{\text{Restrictions that I am placing upon myself}}$$
- I cannot use online resources
- I cannot use any of my other attempts that I have done on this to make this easier on myself
- I do not necessarily have to upload my proof for this. (Since usually it is frowned upon to upload photos on really any SE site, and while none of my questions that I have done this gotten downvoted to oblivion/closed/deleted, I think it would just be with my best judgement to just not to)
- I cannot use a calculator
- I must use Laplace Transform to solve this
$$\mathbf{\text{My attempt at solving}}$$ $$\mathbf{\text{the damped harmonic oscillator problem using Laplace Transform}}$$
$$\ddot{x}+2\gamma\dot{x}+\omega_0^2x=0$$$$\mathcal{L}\{\ddot{x}+2\gamma\dot{x}+\omega_0^2x=0\}$$$$\mathcal{L}\{\ddot{x}\}+2\gamma\mathcal{L}\{\dot{x}\}+\omega_0^2\mathcal{L}\{x\}=0$$$$S^2F-Sx_0-\dot x_0+2\gamma SF-2\gamma x_0+\omega^2F=0$$Knowing that$$\mathcal{L}\{\dot x(t)\}=SF=x(0)$$ $$\mathcal{L}\{x(t)\}=F(S)=F$$ $$\mathcal{L}\{\ddot x(t)\}=S^2F$$$$\mathcal{L}\{e^{at}\cos(bt)\}=\dfrac{b}{(s-a)^2+b^2}$$And$$\mathcal{L}\{e^{at}\sin(bt)\}=\dfrac{s-a}{(s-a)^2+b^2}$$$$F\text{ can be rewritten as }\dfrac{x_0(S+2\gamma)}{S^2+2\gamma S+\omega_0^2}+\dfrac{x_0}{S^2+2\gamma S+\omega_0^2}$$Which means we can rewrite our original equation as$$F=\dfrac{x_0(S+\gamma)}{(S+\gamma)^2+\omega^2}+(x_0\gamma+x_0)\cdot\dfrac{1}{(S+\gamma)^2+\omega^2}$$Now, going back to our Laplace Transform equations that got us here (the $\mathcal{L}\{e^{at}\cos(bt)\}$ and $\mathcal{L}\{e^{at}\sin(bt)\}$), we get$$a=-\gamma$$$$b=\omega$$$$\implies F=x_0\mathcal{L}\{e^{-\gamma t}\sin(\omega t)\}+\dfrac{x_0\gamma+\dot x_0}{\omega}\mathcal{L}\{e^{-\gamma t}\cos(\omega t)\}$$$$\therefore x(t)=x_0e^{-\gamma t}\sin(\omega t)+\dfrac{x_0\gamma+\dot x_0}{\omega}\cos(\omega t)$$Which can be simplified as$$x(t)=e^{-\gamma t}\left(x_0\sin(\omega t)+\dfrac{x_0\gamma+\dot x_0}{\omega}\cos(\omega t)\right)$$
My question
Is the solution that I have arrived at correct, or what could I do to attain the correct solution more easily?
Mistakes I might have made
- Laplace Transforms are pretty easy for me, yet I think that the fact that I can simplify them fast could probably also be the thing that makes my question incorrect.
- The second half of my answer (I was sort of sketchy with how I solved it)