Derivation of SHM equation

every time i read a physics book (especially the textbooks) and i get to the part about simple harmonic motion (SHM), i get nervous. this is because every single time i have read the section, the author NEVER explains how to - or says it is too difficult to - derive the equation for SHM! everyone knows that it is

x(t) = A cos(ωt) ,

or some of its other variants (like with the +φ, or the sin or what-have-you). but this is just GIVEN to us! and the author never tells us how we got there! sure, we know that it is simple enough to show that it satisfies Hooke's condition:
F = -kx, and can be equated with Newton's Second Law:

F = ma = m (d²x/dt²), yielding:
m (d²x/dt²) = -kx, or
x'' + (k/m) x = 0, and letting k/m = c²
x'' + c²x = 0

but NO ONE has ever explained the process of deriving this! and i know that, at this level of physics, that sort of mathematics isn't necessary, but it shouldn't be difficult to whip together a part in the appendix which has the solution for those who are curious!

it is also worth noting that, way before i could solve these differential equations myself, i found another solution to this one. it was actually because i had been staring so long and so hard at this solution and the equation, trying to find out how it was done, but i came with it (which, with some small effort, can actually be seen as equivalent to the more common solution):

x = A e^(±ict). taking derivatives, we see that

x' = ±icA e^(±ict), and

x'' = -c²A e^(±ict), or

x'' = -c²x, which satisfies our condition!

so this was one cool (and complex!) solution the differential equation which, although not very useful (because of the imaginary component) was still exciting at the time!

so naturally i, not being able to find the solution anywhere, had to derive it myself.

i did this a couple ways (and at different times). the first time was in the infancy of my Diff. equ. class when we had just learned how to handle differential equation with second order derivatives. PERFECT! i made the necessary substitutions and solved the equation that way. it actually was not that difficult and could have easily been done in one section (or appendix) and have been understandable by anyone with a calculus background.

but this isn't the solution i'm going to present today; the one i have was derived with some knowledge of linear operators and some more basic calculus (and, of course, our dear friend Euler). i might also post my first solution up here if i get around to re-working that proof.

so here goes!

we start with our fundamental differential equation derived from Hooke's Law:

F = -kx = ma = m d²x/dt²

rearranging, we get:

d²x/dt² + (k/m) x = 0, and i will make the substitution c² = k/m

d²x/dt² + c²x = 0

note that this is a homogeneous differential equation with constant coefficients. it can also be written this way:

D²x + c²x = 0, or

(D² + c²)(x) = 0

now recall what we learned about in the last post about these types of differential equations. we have to write the auxiliary equation and find the roots:

r² + c² = 0, solving this for r:

r² = -c²

r = ±√(-c²)

r = ± ic, where i = √(-1). so we have

r₁ = ic, r₂ = -ic.

note that this is of the more general form r = α ± iβ (where α = 0, β = c)

now that we have this solution, we find that our solution for this case is of the form:

x = e^(αt) [ C₁e^(iβt) + C₂e^(-iβt) ],
where C₁ and C₂ are constants of integration.

and don't think, for a second, that i would just present an esoteric solution to a complicated problem and expect you to blindly accept it! remember that i did post how to arrive at this solution in my previous post. so if you have any questions about how we got there, go one blog post back.

and so, substituting our results into the solution, we get:

x = C₁e^(ict) + C₂e^(-ict).

but this doesn't look at all like the solution we know and love! but if we are just patient, we can do something very interesting with 'i' in the exponent of 'e' - where, again, Euler will save us. and i'm sure you've seen it before. it is the identity:

e^(iθ) = cosθ + i sinθ.

substituting ct (and -ct) for θ, we get:

e^(ict) = cos(ct) + i sin(ct), and

e^(-ict) = cos(-ct) + i sin(-ct). but we can actually use our knowledge of the odd and even qualities of sin and cos, to simplify this second equation. we will use the fact that cos(-θ) = cos(θ), and sin(-θ) = -sin(θ):

e^(-ict) = cos(ct) - i sin(ct)

so then, substituting back in, our solution looks like this:

x = C₁[ cos(ict) + i sin(ict) ]+ C₂[ cos(ict) - i sin(ict) ]. and rearranging, we can write it this way:

x(t) = (C₁ + C₂) cos(ct) + i (C₁ - C₂) sin(ct)

huzzah! this is our general solution! wonderful, isn't it? we can simplify it a little bit by giving the initial value conditions that x(0) = A, and x'(0) = 0. where A is the amplitude, or maximum value of the oscillation. by doing this we've said that at time t=0, the pendulum (or spring, or what-have-you) is at it's maximum displacement, and that it has 0 velocity at that time and place (which logically, would have to be the case).
this gives us the following:

x(0) = (C₁ + C₂) cos(0) + i (C₁ - C₂) sin(0) = (C₁ + C₂) = A

x'(0) = -c(C₁ + C₂)sin(0) + ci(C₁ - C₂)cos(0) = ci(C₁ - C₂) = 0

solving this system of C₁ and C₂, we get:
C₁ = C₂ = A/2
and plugging these into out equation:

x(t) = A cos(ct), c² = k/m

it is obvious now that this solution is EQUIVALENT to the commonly known solution! so not only have we derived it completely, we have derived it correctly! and with all steps shown and no gimmicks! it's hard to imagine that a physics textbook couldn't fit this somewhere into the text! even a mention in the appendix is not that hard to do!

i hope that, at the very least, people now have some idea that this equation CAN be derived - and rather simply too! (i've actually realized that first solution i worked out for this problem might be even simpler and easier to see than this one, so i will go ahead and post that one too).