SHM equation

start with the simple mechanical model of a spring attached to a wall based on FBD we have

where $-b\dot{\theta }$ refers to the drag force while $-k\theta$ refers to the elastic force. both of which resists the motion. We can then rarrange to get

as we learnt from basic differential equations we try a trial solution of $z(t)=e^{i\alpha t}$where $\alpha$ is some complex number

so by quadratic formula we obtain

it can be shown that there 3 key scenarios

1. Underdamped motion where ${\omega }_0^2>\frac{{\mathrm{\Gamma }}^2}{4}$that is we have a small drag force so defining ${\omega }^2=.w_0^2-\frac{{\mathrm{\Gamma }}^2}{4}$ we find that $\alpha =\frac{i\mathrm{\Gamma }}{2}\pm \omega$ so we have

which we can take average of these to obtain

and taking over the average over $2i$ will get you

and so we get the general solution of

2. Critically damped motion where we have ${\omega }_0^2=\frac{{\mathrm{\Gamma }}^2}{4}$ then we have the solution(repeated roots) $\alpha =\frac{i\mathrm{\Gamma }}{2}$ so $\omega =0$ so we have the general solution

3. Overdamped if ${\omega }_0^2<\frac{{\mathrm{\Gamma }}^2}{4}$ then we get $\alpha =i(\frac{\mathrm{\Gamma }}{2}\pm \sqrt{\frac{{\mathrm{\Gamma }}^2}{4}-{\omega }_0^2})$where we get

Now consider the scenario where we add a driven force so now have

this time instead of $z=e^{-\alpha it}$ like we did before to solve the homogenous version of the standard form we use $z=A{e}^{i({\omega }_{d}t-\delta )}$

Normal modes

as usual we first begin with FBD for each and notice that