Simple way to derive the equation of simple harmonic motion

Why this exists

In the physics lecture I attended in my university, the professor said he would begin from the necessary condition $F = -kx$ for simple harmonic motion and solve the differential equation, giving us $x = A\,\sin(\omega\,t)$. His approach was fine, but there were some things missing in his solution that left me unsatified::

1. He solved it assuming that, for $t_0=0, x_0=0$, meaning he solved a special case and didn't get a general solution.
2. He did say that he chose the constant $A$ so that it doesn't violate the previous equation, but he didn't go that much in depth.
3. He more-or-less said that the ±s which we will see later cancelled out, but he didn't go that much in depth.
4. There was a point where the same cosine is at both sides of an equation and he cancelled them out, but didn't justify why that's fine, considering that these cosines might as well be equal to zero

Truth be told, I am studying engineering, and a couple of my professors have told my class that we don't have to prove things as rigously as mathematicians; this is their job, not our.

Nevertheless, here is my attempt at covering these gaps. I did use mathematical concepts I don't quite understand (or have properly learned) in some points, but overall, even if this isn't an accepted solution, it should be good enough to show someone why the solution to the equation is what it is.

And yes, this is more-or-less an excuse to mess with $\LaTeX$.

Solution

Let there be a body moving along the x-axis, such that for $t_0 = 0$, $x = x_0$, and $U = u_0$, and upon which the total force exerted is:

$$\begin{aligned} F &= -kx \\ ma &= -kx \\ a &= -\frac{k}{m}x \\ \frac{du}{dt} &= -\frac{k}{m}x \\ \frac{du}{dx}\frac{dx}{dt} &= -\frac{k}{m}x \end{aligned}$$

And since $\frac{dx}{dt} = u$:

$$\begin{aligned} \frac{du}{dx}u &= -\frac{k}{m}x \\ u\,du &= -\frac{k}{m}x\,dx \\ \int_{u_0}^{u} u\,du &= \int_{x_0}^{x} -\frac{k}{m}x\,dx \\ \frac{1}{2}(u^2 - u_0^2) &= -\frac{1}{2}\frac{k}{m}(x^2 - x_0^2) \quad (1) \\ \frac{dx}{dt} &= \pm\sqrt{u_0^2 - \frac{k}{m}(x^2 - x_0^2)} \quad (2) \end{aligned}$$

We substitute $x = A\sin\theta$, and by differentiating we get $dx = A\cos\theta\,d\theta$.

Then, (2) becomes:

$$\frac{A\cos\theta\,d\theta}{dt} = \pm\sqrt{u_0^2 - \frac{k}{m}(A^2\sin^2\theta - x_0^2)} \quad (3)$$

To continue further, we need to set $A$ so that we can write that square root in
the form $B\sqrt{1 - \sin^2\theta}$:

$$\begin{aligned} B\sqrt{1 - \sin^2\theta} &= \pm\sqrt{u_0^2 - \frac{k}{m}(A^2\sin^2\theta - x_0^2)} \\ B^2(1 - \sin^2\theta) &= u_0^2 - \frac{k}{m}(A^2\sin^2\theta - x_0^2) \\ B^2 - B^2\sin^2\theta &= u_0^2 + \frac{k}{m}x_0^2 - \frac{k}{m}A^2\sin^2\theta \end{aligned}$$

Comparing both sides gives:

$$\begin{cases} B^2 = u_0^2 + \frac{k}{m}x_0^2 \\ B^2 = \frac{k}{m}A^2 \end{cases}$$

$$\gdef\Asqr{\frac{m}{k}u_0^2 + x_0^2} A^2 = \Asqr$$

We will pick $A$ such that it is positive, so we get:

$$A = \sqrt{\Asqr}$$

Equation (3) becomes:

$$\begin{aligned} \gdef\Csqr{u_0^2+\frac{k}{m}x_0^2} \frac{A\cos\theta\,d\theta}{dt} &= \pm\sqrt{(\Csqr)(1 - \sin^2\theta)} \\ \frac{A\cos\theta\,d\theta}{dt} &= \pm\sqrt{\Csqr}(\pm \cos\theta) \end{aligned}$$

Since we picked $A$ to be positive, and with $C=\sqrt{\Csqr}$, this simplifies to:

$$A\cos\theta\,d\theta = C\cos\theta\,dt$$

For $\cos\theta \neq 0$, we get:

$$A\,d\theta = C\,dt$$

Since $\cos\theta \neq 0$ almost everywhere, we can integrate per Lebesgue (or at least so I think. If you think I am wrong, contact me):

$$\begin{aligned} \int_{\theta_0}^{\theta} A\,d\theta &= \int_{0}^{t} C\,dt \\ (\theta - \theta_0)\sqrt{\Asqr} &= C\,t \\ \theta &= \frac{C}{\sqrt{\Asqr}}t + \theta_0 \end{aligned}$$

However, the term in from of $t$ simplifies to:

$$\sqrt{\frac{\Csqr}{\Asqr}} = \\ \def\comterm{\frac{m}{k}u_0^2+x_0^2} \sqrt{\frac{\frac{k}{m}(\comterm)}{\comterm}} = \sqrt{\frac{k}{m}} \gdef\angle{\sqrt{\frac{k}{m}}t+\theta_0}$$

And we get:

$$x = \left(\sqrt{\Asqr}\right)\,\sin(\angle)$$

Comparing this with the standard form of the simple harmonic motion equation:

$$x = A\sin(\omega t + \phi_0) \quad (4)$$

We get that:

$$\begin{aligned} \text{amplitude:}\,A &= \sqrt{\Asqr} \\ \text{angular velocity:}\,\omega &= \sqrt{\frac{k}{m}} \\ \text{initial phase:}\,\phi_0 &= \arcsin\frac{x_0}{A}, \,(4)\,\text{for}\,t=t_0=0, x=x_0 \end{aligned}$$

Note

We have solved for the case where for $x_0 = 0$, the body is at equilibrium ($F = -k x_0 \harr F = 0$). If that wasn't the case and the point of equilibrium was at another point for $x = x_1$, we could either change our point of reference so that $x_0 = 0$, or solve the whole equation again for that case.

Either way, the solution would be:

$$x = x_1 + A\sin(\omega t + \phi_0)$$