because a SHM is periodic, we will be dealing with sines and cosines. unfortunately that's just the way of things. they're periodic functions (which I don't think comes as a surprise), and their derivatives come in especially handy.

we can use Hooke's law, it applies here or so I've been told. combining it with Newton's second law and rearranging terms a bit, we're left with -k/m y(t) = y''(t). to make it a little easier on the eyes, we define c=sqrt(k/m) such that we have y''(t) + c^2 y(t) = 0, lovely, she's beautiful.

she's also a second-order linear ODE (ordinary differential equation), which is an absolute mouthful but boils down to basically nothing. second-order refers to the highest derivative being y'', the second derivative of y, it's linear because the equation is in the form of a polynomial, and ordinary (which, rude) just means we're working with a single independent variable t (thank god)

how do we solve this, though? we use the exponential function! hurray! why? because it has a sick property, the derivative of e^x is e^x, riveting I know. let y=e^rt, then y''=r^2 e^rt, substituting and simplifying gives us something called the characteristic equation, r^2 + c^2=0, which can be factorized as (r+ci)(r-ci)=0, I hope you were hungry for imaginary numbers! so the roots here are r=+-ci, which tells us that the general solution to the ODE is given by

y(t) = a cos(ct) + b sin(ct) where a and b are arbitrary constants. (I hope you'll take my word for this. there is of course a reason why this is the general solution, but I feel like my answer is already long enough. it uses Euler's formula, trig identities and the fact that we're only interested in the real solutions, not the imaginary ones. physics is Real Life after all)

we aren't done! a general solution is.. too general. however, if we want to find what's called the particular solution, we need two initial conditions; the initial position, y(0) (so at t=0) and the velocity of the mass(?) which is y'(0), sick. let's see what this does.

y(0) = a cos(0) + b sin(0) = a (since cos(0)=1, sin(0)=0)

y'(0) = -ac sin(0) + bc cos(0) = bc, so b = y'(0)/c

thus, y(t) = y(0) cos(ct) + y'(0)/c sin(ct), awesome!

I've also seen the general solution written in the form A cos(ct-phi), I'm not entirely sure why, maybe to plot it? whatever the reason, this has to do with polar coordinates. (basically, if we're working on the real plane we use cartesian coordinates, x and y usually, polar coordinates are more commonly used in the complex plane. it's interesting for sure! but I don't know if it's particularly relevant)